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declare as function pointer and initialize in the same line

6

In C++ how do we do the following

// fundamental language construct 
type name = value; 

// for example 
int x = y;

with function pointers?

 typedef (char)(*FP)(unsigned);

 // AFAIK not possible in C++
 FP x = y ;

I can use lambdas

 FP x = []( unsigned k) -> char return char(k); 

But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:

void whatever () 
typedef void (*FP) (void); 
FP x = whatever ;

The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.

c++

share|improve this question

edited 2 hours ago

Guillaume Racicot

15.8k53568

asked 4 hours ago

emma brainemma brain

863

New contributor

emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  You could always stick with auto x = &the_function;'.
  
  – François Andrieux
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The name of a function pointer variable appears between the return type and the arguments It won't look like type name = value;.
  
  – François Andrieux
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You're missing the & before whatever. FP x = &whatever ;
  
  – dave
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the (char), which should be char in your typedef)
  
  – andreee
  4 hours ago
  
  

  
  
  
  

add a comment | 

6

In C++ how do we do the following

// fundamental language construct 
type name = value; 

// for example 
int x = y;

with function pointers?

 typedef (char)(*FP)(unsigned);

 // AFAIK not possible in C++
 FP x = y ;

I can use lambdas

 FP x = []( unsigned k) -> char return char(k); 

But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:

void whatever () 
typedef void (*FP) (void); 
FP x = whatever ;

The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.

c++

share|improve this question

edited 2 hours ago

Guillaume Racicot

15.8k53568

asked 4 hours ago

emma brainemma brain

863

New contributor

emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  You could always stick with auto x = &the_function;'.
  
  – François Andrieux
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  The name of a function pointer variable appears between the return type and the arguments It won't look like type name = value;.
  
  – François Andrieux
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You're missing the & before whatever. FP x = &whatever ;
  
  – dave
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @dave: That's the same for functions, you don't need the ampersand. @emma: Why should that not be possible in C++? This should work fine (except for the (char), which should be char in your typedef)
  
  – andreee
  4 hours ago
  
  

  
  
  
  

add a comment | 

In C++ how do we do the following

// fundamental language construct 
type name = value; 

// for example 
int x = y;

with function pointers?

 typedef (char)(*FP)(unsigned);

 // AFAIK not possible in C++
 FP x = y ;

I can use lambdas

 FP x = []( unsigned k) -> char return char(k); 

But I do not know how to do this without lambda. Any ideas? This is not an answer. we know this works:

void whatever () 
typedef void (*FP) (void); 
FP x = whatever ;

The question is if one can do this in one line in C++. Like one can do it in one line in C++ with every other type kind.

c++

share|improve this question

edited 2 hours ago

Guillaume Racicot

15.8k53568

asked 4 hours ago

emma brainemma brain

863

New contributor

emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

// fundamental language construct 
type name = value; 

// for example 
int x = y;

 typedef (char)(*FP)(unsigned);

 // AFAIK not possible in C++
 FP x = y ;

 FP x = []( unsigned k) -> char return char(k); 

void whatever () 
typedef void (*FP) (void); 
FP x = whatever ;

share|improve this question

edited 2 hours ago

Guillaume Racicot

15.8k53568

asked 4 hours ago

emma brainemma brain

863

New contributor

emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 2 hours ago

Guillaume Racicot

15.8k53568

asked 4 hours ago

emma brainemma brain

863

New contributor

emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 hours ago

Guillaume Racicot

15.8k53568

edited 2 hours ago

Guillaume Racicot

15.8k53568

asked 4 hours ago

emma brainemma brain

863

New contributor

emma brain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 4 hours ago

emma brainemma brain

863

2 Answers
2

active

oldest

votes

7

Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.

Example:

 // typedef
 typedef char(*FP)(unsigned);
 FP x = y ;

 // no typedef
 char(*x)(unsigned) = y;

Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.

share|improve this answer

answered 4 hours ago

n.m.n.m.

73.7k884171

add a comment | 

8

You can use auto:

auto fptr = &f;

It skips the need of a typedef and conserve a nice syntax.

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

15.8k53568

add a comment | 

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7

Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.

Example:

 // typedef
 typedef char(*FP)(unsigned);
 FP x = y ;

 // no typedef
 char(*x)(unsigned) = y;

Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.

share|improve this answer

answered 4 hours ago

n.m.n.m.

73.7k884171

add a comment | 

7

Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.

Example:

 // typedef
 typedef char(*FP)(unsigned);
 FP x = y ;

 // no typedef
 char(*x)(unsigned) = y;

Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.

share|improve this answer

answered 4 hours ago

n.m.n.m.

73.7k884171

add a comment | 

Whenever you can write a typedef, you can also write a variable declaration with no typedef, with almost identical syntax.

Example:

 // typedef
 typedef char(*FP)(unsigned);
 FP x = y ;

 // no typedef
 char(*x)(unsigned) = y;

Remove the typedef keyword, and you have a variable declaration. Slap an initialisation on it if you want.

share|improve this answer

answered 4 hours ago

n.m.n.m.

73.7k884171

 // typedef
 typedef char(*FP)(unsigned);
 FP x = y ;

 // no typedef
 char(*x)(unsigned) = y;

share|improve this answer

answered 4 hours ago

n.m.n.m.

73.7k884171

answered 4 hours ago

n.m.n.m.

73.7k884171

answered 4 hours ago

n.m.n.m.

73.7k884171

8

You can use auto:

auto fptr = &f;

It skips the need of a typedef and conserve a nice syntax.

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

15.8k53568

add a comment | 

8

You can use auto:

auto fptr = &f;

It skips the need of a typedef and conserve a nice syntax.

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

15.8k53568

add a comment | 

You can use auto:

auto fptr = &f;

It skips the need of a typedef and conserve a nice syntax.

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

15.8k53568

auto fptr = &f;

share|improve this answer

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

15.8k53568

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

15.8k53568

answered 4 hours ago

Guillaume RacicotGuillaume Racicot

15.8k53568