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declaring a variable twice in IIFE


Scope of Default function parameters in javascriptWhy would a JavaScript variable start with a dollar sign?How do you use a variable in a regular expression?What is the scope of variables in JavaScript?How do you check if a variable is an array in JavaScript?How do I declare a namespace in JavaScript?How to determine if variable is 'undefined' or 'null'?Check if a variable is a string in JavaScriptRead environment variables in Node.jsJavaScript check if variable exists (is defined/initialized)Is there a standard function to check for null, undefined, or blank variables in JavaScript?






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15















I came through this fun quiz on the internet.



console.log((function(x, f = (() => x))
var x;
var y = x;
x = 2;
return [x, y, f()]
)(1))


and the choices were:



  1. [2,1,1]


  2. [2, undefined, 1]


  3. [2, 1, 2]


  4. [2, undefined, 2]


I picked solution 2 TBH, basing that on that x has been redefined, y was declared and defined with no value, and that f has a different scope hence getting the global x memory spot than function x memory spot.



However, I tried it in jsbin.com



and I found it was solution 1, while I was not sure why that happened I messed with the function body and I removed var x from the function body, I found that the response changed to #3 which makes sense as x value changed and hence it showed x and f as 2 and y as 1 which was declared globally.



but still I can't get why it shows 1 instead of undefined.










share|improve this question

















  • 1





    I find the best way to figure these things out is to step through them line by line with a debugger, and/or print out the values after each line.

    – Heretic Monkey
    13 hours ago











  • var x; doesn't define a new variable within the function scope, where var x = somevalue; would

    – Alex
    13 hours ago






  • 2





    @alex it does. ...

    – Jonas Wilms
    13 hours ago











  • @JonasWilms I see, would have expected x to be undefined then, implicitly copy the function formal value looks really odd to me.

    – Alex
    12 hours ago






  • 3





    I'm glad this question was asked :)

    – Pointy
    12 hours ago

















15















I came through this fun quiz on the internet.



console.log((function(x, f = (() => x))
var x;
var y = x;
x = 2;
return [x, y, f()]
)(1))


and the choices were:



  1. [2,1,1]


  2. [2, undefined, 1]


  3. [2, 1, 2]


  4. [2, undefined, 2]


I picked solution 2 TBH, basing that on that x has been redefined, y was declared and defined with no value, and that f has a different scope hence getting the global x memory spot than function x memory spot.



However, I tried it in jsbin.com



and I found it was solution 1, while I was not sure why that happened I messed with the function body and I removed var x from the function body, I found that the response changed to #3 which makes sense as x value changed and hence it showed x and f as 2 and y as 1 which was declared globally.



but still I can't get why it shows 1 instead of undefined.










share|improve this question

















  • 1





    I find the best way to figure these things out is to step through them line by line with a debugger, and/or print out the values after each line.

    – Heretic Monkey
    13 hours ago











  • var x; doesn't define a new variable within the function scope, where var x = somevalue; would

    – Alex
    13 hours ago






  • 2





    @alex it does. ...

    – Jonas Wilms
    13 hours ago











  • @JonasWilms I see, would have expected x to be undefined then, implicitly copy the function formal value looks really odd to me.

    – Alex
    12 hours ago






  • 3





    I'm glad this question was asked :)

    – Pointy
    12 hours ago













15












15








15


5






I came through this fun quiz on the internet.



console.log((function(x, f = (() => x))
var x;
var y = x;
x = 2;
return [x, y, f()]
)(1))


and the choices were:



  1. [2,1,1]


  2. [2, undefined, 1]


  3. [2, 1, 2]


  4. [2, undefined, 2]


I picked solution 2 TBH, basing that on that x has been redefined, y was declared and defined with no value, and that f has a different scope hence getting the global x memory spot than function x memory spot.



However, I tried it in jsbin.com



and I found it was solution 1, while I was not sure why that happened I messed with the function body and I removed var x from the function body, I found that the response changed to #3 which makes sense as x value changed and hence it showed x and f as 2 and y as 1 which was declared globally.



but still I can't get why it shows 1 instead of undefined.










share|improve this question














I came through this fun quiz on the internet.



console.log((function(x, f = (() => x))
var x;
var y = x;
x = 2;
return [x, y, f()]
)(1))


and the choices were:



  1. [2,1,1]


  2. [2, undefined, 1]


  3. [2, 1, 2]


  4. [2, undefined, 2]


I picked solution 2 TBH, basing that on that x has been redefined, y was declared and defined with no value, and that f has a different scope hence getting the global x memory spot than function x memory spot.



However, I tried it in jsbin.com



and I found it was solution 1, while I was not sure why that happened I messed with the function body and I removed var x from the function body, I found that the response changed to #3 which makes sense as x value changed and hence it showed x and f as 2 and y as 1 which was declared globally.



but still I can't get why it shows 1 instead of undefined.







javascript iife






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 13 hours ago









Hamza MohamedHamza Mohamed

689423




689423







  • 1





    I find the best way to figure these things out is to step through them line by line with a debugger, and/or print out the values after each line.

    – Heretic Monkey
    13 hours ago











  • var x; doesn't define a new variable within the function scope, where var x = somevalue; would

    – Alex
    13 hours ago






  • 2





    @alex it does. ...

    – Jonas Wilms
    13 hours ago











  • @JonasWilms I see, would have expected x to be undefined then, implicitly copy the function formal value looks really odd to me.

    – Alex
    12 hours ago






  • 3





    I'm glad this question was asked :)

    – Pointy
    12 hours ago












  • 1





    I find the best way to figure these things out is to step through them line by line with a debugger, and/or print out the values after each line.

    – Heretic Monkey
    13 hours ago











  • var x; doesn't define a new variable within the function scope, where var x = somevalue; would

    – Alex
    13 hours ago






  • 2





    @alex it does. ...

    – Jonas Wilms
    13 hours ago











  • @JonasWilms I see, would have expected x to be undefined then, implicitly copy the function formal value looks really odd to me.

    – Alex
    12 hours ago






  • 3





    I'm glad this question was asked :)

    – Pointy
    12 hours ago







1




1





I find the best way to figure these things out is to step through them line by line with a debugger, and/or print out the values after each line.

– Heretic Monkey
13 hours ago





I find the best way to figure these things out is to step through them line by line with a debugger, and/or print out the values after each line.

– Heretic Monkey
13 hours ago













var x; doesn't define a new variable within the function scope, where var x = somevalue; would

– Alex
13 hours ago





var x; doesn't define a new variable within the function scope, where var x = somevalue; would

– Alex
13 hours ago




2




2





@alex it does. ...

– Jonas Wilms
13 hours ago





@alex it does. ...

– Jonas Wilms
13 hours ago













@JonasWilms I see, would have expected x to be undefined then, implicitly copy the function formal value looks really odd to me.

– Alex
12 hours ago





@JonasWilms I see, would have expected x to be undefined then, implicitly copy the function formal value looks really odd to me.

– Alex
12 hours ago




3




3





I'm glad this question was asked :)

– Pointy
12 hours ago





I'm glad this question was asked :)

– Pointy
12 hours ago












2 Answers
2






active

oldest

votes


















17















but still I can't get why it shows 1 instead of undefined.




It's not just you. This is a deep, dark part of the specification. :-)



The key here is that there are two xs. Yes, really. There's the parameter x, and there's the variable x.



A parameter list containing expressions (like f's default value) has its own scope separate from the function body's scope. But prior to parameter lists possibly having expressions, having var x within a function with an x parameter had no effect (x was still the parameter, with the parameter's value). So to preserve that, when there's a parameter list with expressions in it, a separate variable is created and the value of the parameter is copied to the variable at the beginning of the function body. Which is the reason for this seemingly-odd (no, not just seemingly) odd behavior. (If you're the kind who likes to dive into the spec, this copying is Step 28 of FunctionDeclarationInstantiation.)



Since f's default value, () => x, is created within the parameter list scope, it refers to the parameter x, not the var.



So the first solution, [2, 1, 1] is correct, because:




  • 2 was assigned to the var x in the function body. So at the end of the function, the var x is 2.


  • 1 was assigned to y from the var x before x got the value 2, so at the end of the function, y is 1.

  • The parameter x's value has never changed, so f() results in 1 at the end of the function

It's as though the code were written like this instead (I've removed unnecessary parens and added missing semicolons):






console.log(function(param_x, f = () => param_x) 
var var_x = param_x;
var y = var_x;
var_x = 2;
return [var_x, y, f()];
(1));






...I removed var x from the function body, I found that the response changed to #3...




#3 is [2, 1, 2]. That's correct, because when you remove the var x from the function, there's only one x, the parameter (inherited by the function body from the parmeter list). So assigning 2 to x changes the parameter's value, which f returns.



Taking the earier example with param_x and var_x, here's what it looks like if you remove the var x; from it:






console.log(function(param_x, f = () => param_x) 
var y = param_x;
param_x = 2;
return [param_x, y, f()];
(1));






Here's an annotated description of the original code (with the extraneous parentheses removed and missing semicolons added):



// /---- the parameter "x"
// v vvvvvvvvvvv--- the parameter "f" with a default value
console.log(function(x, f = () => x)
var x; // <=== the *variable* x, which gets its initial value from the
// parameter x
var y = x; // <=== sets y to 1 (x's current value)
x = 2; // <=== changes the *variable* x's value to 2
// +---------- 2, because this is the *variable* x
// (1));



Final note regarding your title:




declaring a variable twice in IIFE




The variable is only declared once. The other thing is a parameter, not a variable. The distinction is rarely important...this being one of those rare times. :-)






share|improve this answer




















  • 1





    This is now in my list of "new weird things about JavaScript" alongside the semantics of (not yet really standard) instance property initialization expressions in class declarations.

    – Pointy
    12 hours ago











  • @Pointy - Oh, this is much weirder than field declarations. :-D

    – T.J. Crowder
    12 hours ago











  • Well field declarations are jarring to me because this is the instance, not the this of the surrounding scope as with object initializers. I guess it's not really "weird" if you think of class declarations as being something akin to macro expansion.

    – Pointy
    12 hours ago











  • @Pointy - Yeah. I think of them as code that's relocated to the beginning of the constructor (just after the super call if it's a subclass). That's what Java does with instance field initializers and instance initialization blocks (and they're literally copied to the beginning of each constructor in Java; thankfully in JavaScript we just have the one). JavaScript took effectively the same approach.

    – T.J. Crowder
    12 hours ago



















1














The tricky part of that code is that the => function is created as part of a default parameter value expression. In parameter default value expressions, the scope includes the parameters declared to the left, which in this case includes the parameter x. Thus for that reason the x in the => function is in fact the first parameter.



The function is called with just one parameter, 1, so when the => function is called that's what it returns, giving [2, 1, 1].



The var x declaration, as Mr Crowder points out, has the (somewhat weird, at least to me) effect of making a new x in the function scope, into which is copied the value of the parameter x. Without it, there's only the one (the parameter).






share|improve this answer

























  • That still does not explain why removing var x; results in 2, 1, 2 ...

    – Jonas Wilms
    13 hours ago











  • @JonasWilms you're right; the var declaration must ... do something but it doesn't seem obvious to me what that is. It's as if the var declaration creates a new x in the function scope, but it clearly gets the value of the parameter x anyway (which I'd expect), leaving the parameter apparently in its own scope.

    – Pointy
    13 hours ago











  • I guess the values are somewhat copied from the default initializers scope to the bodies scope. Im already digging into the spec

    – Jonas Wilms
    13 hours ago











  • It makes sense. As noted in the spec: NOTE: A separate Environment Record is needed to ensure that closures created by expressions in the formal parameter list do not have visibility of declarations in the function body. thats reasonable.

    – Jonas Wilms
    12 hours ago











  • @Pointy - It's a very deep, dark part of the spec. :-)

    – T.J. Crowder
    12 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









17















but still I can't get why it shows 1 instead of undefined.




It's not just you. This is a deep, dark part of the specification. :-)



The key here is that there are two xs. Yes, really. There's the parameter x, and there's the variable x.



A parameter list containing expressions (like f's default value) has its own scope separate from the function body's scope. But prior to parameter lists possibly having expressions, having var x within a function with an x parameter had no effect (x was still the parameter, with the parameter's value). So to preserve that, when there's a parameter list with expressions in it, a separate variable is created and the value of the parameter is copied to the variable at the beginning of the function body. Which is the reason for this seemingly-odd (no, not just seemingly) odd behavior. (If you're the kind who likes to dive into the spec, this copying is Step 28 of FunctionDeclarationInstantiation.)



Since f's default value, () => x, is created within the parameter list scope, it refers to the parameter x, not the var.



So the first solution, [2, 1, 1] is correct, because:




  • 2 was assigned to the var x in the function body. So at the end of the function, the var x is 2.


  • 1 was assigned to y from the var x before x got the value 2, so at the end of the function, y is 1.

  • The parameter x's value has never changed, so f() results in 1 at the end of the function

It's as though the code were written like this instead (I've removed unnecessary parens and added missing semicolons):






console.log(function(param_x, f = () => param_x) 
var var_x = param_x;
var y = var_x;
var_x = 2;
return [var_x, y, f()];
(1));






...I removed var x from the function body, I found that the response changed to #3...




#3 is [2, 1, 2]. That's correct, because when you remove the var x from the function, there's only one x, the parameter (inherited by the function body from the parmeter list). So assigning 2 to x changes the parameter's value, which f returns.



Taking the earier example with param_x and var_x, here's what it looks like if you remove the var x; from it:






console.log(function(param_x, f = () => param_x) 
var y = param_x;
param_x = 2;
return [param_x, y, f()];
(1));






Here's an annotated description of the original code (with the extraneous parentheses removed and missing semicolons added):



// /---- the parameter "x"
// v vvvvvvvvvvv--- the parameter "f" with a default value
console.log(function(x, f = () => x)
var x; // <=== the *variable* x, which gets its initial value from the
// parameter x
var y = x; // <=== sets y to 1 (x's current value)
x = 2; // <=== changes the *variable* x's value to 2
// +---------- 2, because this is the *variable* x
// (1));



Final note regarding your title:




declaring a variable twice in IIFE




The variable is only declared once. The other thing is a parameter, not a variable. The distinction is rarely important...this being one of those rare times. :-)






share|improve this answer




















  • 1





    This is now in my list of "new weird things about JavaScript" alongside the semantics of (not yet really standard) instance property initialization expressions in class declarations.

    – Pointy
    12 hours ago











  • @Pointy - Oh, this is much weirder than field declarations. :-D

    – T.J. Crowder
    12 hours ago











  • Well field declarations are jarring to me because this is the instance, not the this of the surrounding scope as with object initializers. I guess it's not really "weird" if you think of class declarations as being something akin to macro expansion.

    – Pointy
    12 hours ago











  • @Pointy - Yeah. I think of them as code that's relocated to the beginning of the constructor (just after the super call if it's a subclass). That's what Java does with instance field initializers and instance initialization blocks (and they're literally copied to the beginning of each constructor in Java; thankfully in JavaScript we just have the one). JavaScript took effectively the same approach.

    – T.J. Crowder
    12 hours ago
















17















but still I can't get why it shows 1 instead of undefined.




It's not just you. This is a deep, dark part of the specification. :-)



The key here is that there are two xs. Yes, really. There's the parameter x, and there's the variable x.



A parameter list containing expressions (like f's default value) has its own scope separate from the function body's scope. But prior to parameter lists possibly having expressions, having var x within a function with an x parameter had no effect (x was still the parameter, with the parameter's value). So to preserve that, when there's a parameter list with expressions in it, a separate variable is created and the value of the parameter is copied to the variable at the beginning of the function body. Which is the reason for this seemingly-odd (no, not just seemingly) odd behavior. (If you're the kind who likes to dive into the spec, this copying is Step 28 of FunctionDeclarationInstantiation.)



Since f's default value, () => x, is created within the parameter list scope, it refers to the parameter x, not the var.



So the first solution, [2, 1, 1] is correct, because:




  • 2 was assigned to the var x in the function body. So at the end of the function, the var x is 2.


  • 1 was assigned to y from the var x before x got the value 2, so at the end of the function, y is 1.

  • The parameter x's value has never changed, so f() results in 1 at the end of the function

It's as though the code were written like this instead (I've removed unnecessary parens and added missing semicolons):






console.log(function(param_x, f = () => param_x) 
var var_x = param_x;
var y = var_x;
var_x = 2;
return [var_x, y, f()];
(1));






...I removed var x from the function body, I found that the response changed to #3...




#3 is [2, 1, 2]. That's correct, because when you remove the var x from the function, there's only one x, the parameter (inherited by the function body from the parmeter list). So assigning 2 to x changes the parameter's value, which f returns.



Taking the earier example with param_x and var_x, here's what it looks like if you remove the var x; from it:






console.log(function(param_x, f = () => param_x) 
var y = param_x;
param_x = 2;
return [param_x, y, f()];
(1));






Here's an annotated description of the original code (with the extraneous parentheses removed and missing semicolons added):



// /---- the parameter "x"
// v vvvvvvvvvvv--- the parameter "f" with a default value
console.log(function(x, f = () => x)
var x; // <=== the *variable* x, which gets its initial value from the
// parameter x
var y = x; // <=== sets y to 1 (x's current value)
x = 2; // <=== changes the *variable* x's value to 2
// +---------- 2, because this is the *variable* x
// (1));



Final note regarding your title:




declaring a variable twice in IIFE




The variable is only declared once. The other thing is a parameter, not a variable. The distinction is rarely important...this being one of those rare times. :-)






share|improve this answer




















  • 1





    This is now in my list of "new weird things about JavaScript" alongside the semantics of (not yet really standard) instance property initialization expressions in class declarations.

    – Pointy
    12 hours ago











  • @Pointy - Oh, this is much weirder than field declarations. :-D

    – T.J. Crowder
    12 hours ago











  • Well field declarations are jarring to me because this is the instance, not the this of the surrounding scope as with object initializers. I guess it's not really "weird" if you think of class declarations as being something akin to macro expansion.

    – Pointy
    12 hours ago











  • @Pointy - Yeah. I think of them as code that's relocated to the beginning of the constructor (just after the super call if it's a subclass). That's what Java does with instance field initializers and instance initialization blocks (and they're literally copied to the beginning of each constructor in Java; thankfully in JavaScript we just have the one). JavaScript took effectively the same approach.

    – T.J. Crowder
    12 hours ago














17












17








17








but still I can't get why it shows 1 instead of undefined.




It's not just you. This is a deep, dark part of the specification. :-)



The key here is that there are two xs. Yes, really. There's the parameter x, and there's the variable x.



A parameter list containing expressions (like f's default value) has its own scope separate from the function body's scope. But prior to parameter lists possibly having expressions, having var x within a function with an x parameter had no effect (x was still the parameter, with the parameter's value). So to preserve that, when there's a parameter list with expressions in it, a separate variable is created and the value of the parameter is copied to the variable at the beginning of the function body. Which is the reason for this seemingly-odd (no, not just seemingly) odd behavior. (If you're the kind who likes to dive into the spec, this copying is Step 28 of FunctionDeclarationInstantiation.)



Since f's default value, () => x, is created within the parameter list scope, it refers to the parameter x, not the var.



So the first solution, [2, 1, 1] is correct, because:




  • 2 was assigned to the var x in the function body. So at the end of the function, the var x is 2.


  • 1 was assigned to y from the var x before x got the value 2, so at the end of the function, y is 1.

  • The parameter x's value has never changed, so f() results in 1 at the end of the function

It's as though the code were written like this instead (I've removed unnecessary parens and added missing semicolons):






console.log(function(param_x, f = () => param_x) 
var var_x = param_x;
var y = var_x;
var_x = 2;
return [var_x, y, f()];
(1));






...I removed var x from the function body, I found that the response changed to #3...




#3 is [2, 1, 2]. That's correct, because when you remove the var x from the function, there's only one x, the parameter (inherited by the function body from the parmeter list). So assigning 2 to x changes the parameter's value, which f returns.



Taking the earier example with param_x and var_x, here's what it looks like if you remove the var x; from it:






console.log(function(param_x, f = () => param_x) 
var y = param_x;
param_x = 2;
return [param_x, y, f()];
(1));






Here's an annotated description of the original code (with the extraneous parentheses removed and missing semicolons added):



// /---- the parameter "x"
// v vvvvvvvvvvv--- the parameter "f" with a default value
console.log(function(x, f = () => x)
var x; // <=== the *variable* x, which gets its initial value from the
// parameter x
var y = x; // <=== sets y to 1 (x's current value)
x = 2; // <=== changes the *variable* x's value to 2
// +---------- 2, because this is the *variable* x
// (1));



Final note regarding your title:




declaring a variable twice in IIFE




The variable is only declared once. The other thing is a parameter, not a variable. The distinction is rarely important...this being one of those rare times. :-)






share|improve this answer
















but still I can't get why it shows 1 instead of undefined.




It's not just you. This is a deep, dark part of the specification. :-)



The key here is that there are two xs. Yes, really. There's the parameter x, and there's the variable x.



A parameter list containing expressions (like f's default value) has its own scope separate from the function body's scope. But prior to parameter lists possibly having expressions, having var x within a function with an x parameter had no effect (x was still the parameter, with the parameter's value). So to preserve that, when there's a parameter list with expressions in it, a separate variable is created and the value of the parameter is copied to the variable at the beginning of the function body. Which is the reason for this seemingly-odd (no, not just seemingly) odd behavior. (If you're the kind who likes to dive into the spec, this copying is Step 28 of FunctionDeclarationInstantiation.)



Since f's default value, () => x, is created within the parameter list scope, it refers to the parameter x, not the var.



So the first solution, [2, 1, 1] is correct, because:




  • 2 was assigned to the var x in the function body. So at the end of the function, the var x is 2.


  • 1 was assigned to y from the var x before x got the value 2, so at the end of the function, y is 1.

  • The parameter x's value has never changed, so f() results in 1 at the end of the function

It's as though the code were written like this instead (I've removed unnecessary parens and added missing semicolons):






console.log(function(param_x, f = () => param_x) 
var var_x = param_x;
var y = var_x;
var_x = 2;
return [var_x, y, f()];
(1));






...I removed var x from the function body, I found that the response changed to #3...




#3 is [2, 1, 2]. That's correct, because when you remove the var x from the function, there's only one x, the parameter (inherited by the function body from the parmeter list). So assigning 2 to x changes the parameter's value, which f returns.



Taking the earier example with param_x and var_x, here's what it looks like if you remove the var x; from it:






console.log(function(param_x, f = () => param_x) 
var y = param_x;
param_x = 2;
return [param_x, y, f()];
(1));






Here's an annotated description of the original code (with the extraneous parentheses removed and missing semicolons added):



// /---- the parameter "x"
// v vvvvvvvvvvv--- the parameter "f" with a default value
console.log(function(x, f = () => x)
var x; // <=== the *variable* x, which gets its initial value from the
// parameter x
var y = x; // <=== sets y to 1 (x's current value)
x = 2; // <=== changes the *variable* x's value to 2
// +---------- 2, because this is the *variable* x
// (1));



Final note regarding your title:




declaring a variable twice in IIFE




The variable is only declared once. The other thing is a parameter, not a variable. The distinction is rarely important...this being one of those rare times. :-)






console.log(function(param_x, f = () => param_x) 
var var_x = param_x;
var y = var_x;
var_x = 2;
return [var_x, y, f()];
(1));





console.log(function(param_x, f = () => param_x) 
var var_x = param_x;
var y = var_x;
var_x = 2;
return [var_x, y, f()];
(1));





console.log(function(param_x, f = () => param_x) 
var y = param_x;
param_x = 2;
return [param_x, y, f()];
(1));





console.log(function(param_x, f = () => param_x) 
var y = param_x;
param_x = 2;
return [param_x, y, f()];
(1));






share|improve this answer














share|improve this answer



share|improve this answer








edited 12 hours ago

























answered 13 hours ago









T.J. CrowderT.J. Crowder

699k12312431341




699k12312431341







  • 1





    This is now in my list of "new weird things about JavaScript" alongside the semantics of (not yet really standard) instance property initialization expressions in class declarations.

    – Pointy
    12 hours ago











  • @Pointy - Oh, this is much weirder than field declarations. :-D

    – T.J. Crowder
    12 hours ago











  • Well field declarations are jarring to me because this is the instance, not the this of the surrounding scope as with object initializers. I guess it's not really "weird" if you think of class declarations as being something akin to macro expansion.

    – Pointy
    12 hours ago











  • @Pointy - Yeah. I think of them as code that's relocated to the beginning of the constructor (just after the super call if it's a subclass). That's what Java does with instance field initializers and instance initialization blocks (and they're literally copied to the beginning of each constructor in Java; thankfully in JavaScript we just have the one). JavaScript took effectively the same approach.

    – T.J. Crowder
    12 hours ago













  • 1





    This is now in my list of "new weird things about JavaScript" alongside the semantics of (not yet really standard) instance property initialization expressions in class declarations.

    – Pointy
    12 hours ago











  • @Pointy - Oh, this is much weirder than field declarations. :-D

    – T.J. Crowder
    12 hours ago











  • Well field declarations are jarring to me because this is the instance, not the this of the surrounding scope as with object initializers. I guess it's not really "weird" if you think of class declarations as being something akin to macro expansion.

    – Pointy
    12 hours ago











  • @Pointy - Yeah. I think of them as code that's relocated to the beginning of the constructor (just after the super call if it's a subclass). That's what Java does with instance field initializers and instance initialization blocks (and they're literally copied to the beginning of each constructor in Java; thankfully in JavaScript we just have the one). JavaScript took effectively the same approach.

    – T.J. Crowder
    12 hours ago








1




1





This is now in my list of "new weird things about JavaScript" alongside the semantics of (not yet really standard) instance property initialization expressions in class declarations.

– Pointy
12 hours ago





This is now in my list of "new weird things about JavaScript" alongside the semantics of (not yet really standard) instance property initialization expressions in class declarations.

– Pointy
12 hours ago













@Pointy - Oh, this is much weirder than field declarations. :-D

– T.J. Crowder
12 hours ago





@Pointy - Oh, this is much weirder than field declarations. :-D

– T.J. Crowder
12 hours ago













Well field declarations are jarring to me because this is the instance, not the this of the surrounding scope as with object initializers. I guess it's not really "weird" if you think of class declarations as being something akin to macro expansion.

– Pointy
12 hours ago





Well field declarations are jarring to me because this is the instance, not the this of the surrounding scope as with object initializers. I guess it's not really "weird" if you think of class declarations as being something akin to macro expansion.

– Pointy
12 hours ago













@Pointy - Yeah. I think of them as code that's relocated to the beginning of the constructor (just after the super call if it's a subclass). That's what Java does with instance field initializers and instance initialization blocks (and they're literally copied to the beginning of each constructor in Java; thankfully in JavaScript we just have the one). JavaScript took effectively the same approach.

– T.J. Crowder
12 hours ago






@Pointy - Yeah. I think of them as code that's relocated to the beginning of the constructor (just after the super call if it's a subclass). That's what Java does with instance field initializers and instance initialization blocks (and they're literally copied to the beginning of each constructor in Java; thankfully in JavaScript we just have the one). JavaScript took effectively the same approach.

– T.J. Crowder
12 hours ago














1














The tricky part of that code is that the => function is created as part of a default parameter value expression. In parameter default value expressions, the scope includes the parameters declared to the left, which in this case includes the parameter x. Thus for that reason the x in the => function is in fact the first parameter.



The function is called with just one parameter, 1, so when the => function is called that's what it returns, giving [2, 1, 1].



The var x declaration, as Mr Crowder points out, has the (somewhat weird, at least to me) effect of making a new x in the function scope, into which is copied the value of the parameter x. Without it, there's only the one (the parameter).






share|improve this answer

























  • That still does not explain why removing var x; results in 2, 1, 2 ...

    – Jonas Wilms
    13 hours ago











  • @JonasWilms you're right; the var declaration must ... do something but it doesn't seem obvious to me what that is. It's as if the var declaration creates a new x in the function scope, but it clearly gets the value of the parameter x anyway (which I'd expect), leaving the parameter apparently in its own scope.

    – Pointy
    13 hours ago











  • I guess the values are somewhat copied from the default initializers scope to the bodies scope. Im already digging into the spec

    – Jonas Wilms
    13 hours ago











  • It makes sense. As noted in the spec: NOTE: A separate Environment Record is needed to ensure that closures created by expressions in the formal parameter list do not have visibility of declarations in the function body. thats reasonable.

    – Jonas Wilms
    12 hours ago











  • @Pointy - It's a very deep, dark part of the spec. :-)

    – T.J. Crowder
    12 hours ago















1














The tricky part of that code is that the => function is created as part of a default parameter value expression. In parameter default value expressions, the scope includes the parameters declared to the left, which in this case includes the parameter x. Thus for that reason the x in the => function is in fact the first parameter.



The function is called with just one parameter, 1, so when the => function is called that's what it returns, giving [2, 1, 1].



The var x declaration, as Mr Crowder points out, has the (somewhat weird, at least to me) effect of making a new x in the function scope, into which is copied the value of the parameter x. Without it, there's only the one (the parameter).






share|improve this answer

























  • That still does not explain why removing var x; results in 2, 1, 2 ...

    – Jonas Wilms
    13 hours ago











  • @JonasWilms you're right; the var declaration must ... do something but it doesn't seem obvious to me what that is. It's as if the var declaration creates a new x in the function scope, but it clearly gets the value of the parameter x anyway (which I'd expect), leaving the parameter apparently in its own scope.

    – Pointy
    13 hours ago











  • I guess the values are somewhat copied from the default initializers scope to the bodies scope. Im already digging into the spec

    – Jonas Wilms
    13 hours ago











  • It makes sense. As noted in the spec: NOTE: A separate Environment Record is needed to ensure that closures created by expressions in the formal parameter list do not have visibility of declarations in the function body. thats reasonable.

    – Jonas Wilms
    12 hours ago











  • @Pointy - It's a very deep, dark part of the spec. :-)

    – T.J. Crowder
    12 hours ago













1












1








1







The tricky part of that code is that the => function is created as part of a default parameter value expression. In parameter default value expressions, the scope includes the parameters declared to the left, which in this case includes the parameter x. Thus for that reason the x in the => function is in fact the first parameter.



The function is called with just one parameter, 1, so when the => function is called that's what it returns, giving [2, 1, 1].



The var x declaration, as Mr Crowder points out, has the (somewhat weird, at least to me) effect of making a new x in the function scope, into which is copied the value of the parameter x. Without it, there's only the one (the parameter).






share|improve this answer















The tricky part of that code is that the => function is created as part of a default parameter value expression. In parameter default value expressions, the scope includes the parameters declared to the left, which in this case includes the parameter x. Thus for that reason the x in the => function is in fact the first parameter.



The function is called with just one parameter, 1, so when the => function is called that's what it returns, giving [2, 1, 1].



The var x declaration, as Mr Crowder points out, has the (somewhat weird, at least to me) effect of making a new x in the function scope, into which is copied the value of the parameter x. Without it, there's only the one (the parameter).







share|improve this answer














share|improve this answer



share|improve this answer








edited 13 hours ago

























answered 13 hours ago









PointyPointy

321k45461528




321k45461528












  • That still does not explain why removing var x; results in 2, 1, 2 ...

    – Jonas Wilms
    13 hours ago











  • @JonasWilms you're right; the var declaration must ... do something but it doesn't seem obvious to me what that is. It's as if the var declaration creates a new x in the function scope, but it clearly gets the value of the parameter x anyway (which I'd expect), leaving the parameter apparently in its own scope.

    – Pointy
    13 hours ago











  • I guess the values are somewhat copied from the default initializers scope to the bodies scope. Im already digging into the spec

    – Jonas Wilms
    13 hours ago











  • It makes sense. As noted in the spec: NOTE: A separate Environment Record is needed to ensure that closures created by expressions in the formal parameter list do not have visibility of declarations in the function body. thats reasonable.

    – Jonas Wilms
    12 hours ago











  • @Pointy - It's a very deep, dark part of the spec. :-)

    – T.J. Crowder
    12 hours ago

















  • That still does not explain why removing var x; results in 2, 1, 2 ...

    – Jonas Wilms
    13 hours ago











  • @JonasWilms you're right; the var declaration must ... do something but it doesn't seem obvious to me what that is. It's as if the var declaration creates a new x in the function scope, but it clearly gets the value of the parameter x anyway (which I'd expect), leaving the parameter apparently in its own scope.

    – Pointy
    13 hours ago











  • I guess the values are somewhat copied from the default initializers scope to the bodies scope. Im already digging into the spec

    – Jonas Wilms
    13 hours ago











  • It makes sense. As noted in the spec: NOTE: A separate Environment Record is needed to ensure that closures created by expressions in the formal parameter list do not have visibility of declarations in the function body. thats reasonable.

    – Jonas Wilms
    12 hours ago











  • @Pointy - It's a very deep, dark part of the spec. :-)

    – T.J. Crowder
    12 hours ago
















That still does not explain why removing var x; results in 2, 1, 2 ...

– Jonas Wilms
13 hours ago





That still does not explain why removing var x; results in 2, 1, 2 ...

– Jonas Wilms
13 hours ago













@JonasWilms you're right; the var declaration must ... do something but it doesn't seem obvious to me what that is. It's as if the var declaration creates a new x in the function scope, but it clearly gets the value of the parameter x anyway (which I'd expect), leaving the parameter apparently in its own scope.

– Pointy
13 hours ago





@JonasWilms you're right; the var declaration must ... do something but it doesn't seem obvious to me what that is. It's as if the var declaration creates a new x in the function scope, but it clearly gets the value of the parameter x anyway (which I'd expect), leaving the parameter apparently in its own scope.

– Pointy
13 hours ago













I guess the values are somewhat copied from the default initializers scope to the bodies scope. Im already digging into the spec

– Jonas Wilms
13 hours ago





I guess the values are somewhat copied from the default initializers scope to the bodies scope. Im already digging into the spec

– Jonas Wilms
13 hours ago













It makes sense. As noted in the spec: NOTE: A separate Environment Record is needed to ensure that closures created by expressions in the formal parameter list do not have visibility of declarations in the function body. thats reasonable.

– Jonas Wilms
12 hours ago





It makes sense. As noted in the spec: NOTE: A separate Environment Record is needed to ensure that closures created by expressions in the formal parameter list do not have visibility of declarations in the function body. thats reasonable.

– Jonas Wilms
12 hours ago













@Pointy - It's a very deep, dark part of the spec. :-)

– T.J. Crowder
12 hours ago





@Pointy - It's a very deep, dark part of the spec. :-)

– T.J. Crowder
12 hours ago

















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