Hfrxdjt

In math, if z = x+y/2, then z will always change whenever we replace the value of x and y. Can we do that in programming without having to specifically updating z whenever we change the value of x and y?

I mean something like that won't work, right?

int x;
int y;
int zx + y;
cin >> x;
cin >> y;
cout << z;

You can get close to this with by using a lambda in C++. Generally, when you set a variable like

int x;
int y;
int zx + y;

z will only be the result of x + y at that time. You'd have to do z = x + y; every time you chnage x or y to keep it update.

If you use a lambda though, you can have it capture what objects it should refer to, and what calculation should be done, and then every time you access the lambda it will give you the result at that point in time. That looks like

int x;
int y;
auto z = [&]() return x + y; ;
cin >> x;
cin >> y;
cout << z();

and now z() will have the correct value instead of the uninitialized garbage that the original code had.

If the computation is very expensive you can even add some caching to the lambda to make sure you aren't running the computation when you don't need to. That would look like

auto z = [&]() static auto cache_x = x; 
 static auto cache_y = y; 
 static auto cache_result = x + y;
 if (x == cache_x && y == cache_y)
 return cache_result;
 else
 
 cache_x = x; 
 cache_y = y; 
 cache_result = x + y;
 reutrn cache_result;
 
;

The closest you probably can get is to create a functor:

#include <iostream>

int main() 
 int x;
 int y;

 auto z = [&x, &y] return x + y; ; // a lambda capturing x and y

 while(true) 
 std::cin >> x;
 std::cin >> y;
 std::cout << z() << "n";
 

You mean something like this:

class Z

 int& x;
 int& y;
public:
 Z(int& x, int& y) : x(x), y(y) 
 operator int() return x + y; 
;

The class delays calculation of the result until casted as int. As cast operator is not explicit, Z can be used whenever an int is required. As there's an overload of operator<< for int, you can use it with e. g. std::cout directly:

int x, y;
Z z(x, y);
std::cin >> x >> y;
if(std::cin) // otherwise, IO error! (e. g. bad user input)
 std::cout << z << std::endl;

You can define the following lambda z which always returns the current value of x+y because x and y are captured by reference:

DEMO

int x;
int y;
const auto z = [&x, &y]() return x+y; ;

1. You create a function for that.


2. You call the function with the appropriate arguments when you need the value.

int z(int x, int y)

 return (x + y);



int x;
int y;

// This does ot work
// int zx + y;

cin >> x;
cin >> y;
cout << z(x, y);