Unable to change values in a function

Question

I\'m starting with developing, sorry about this newbie question.

I need to create a function that swap values between 2 vars.

I have wrote this code, but no chan

Answer 1

You need to understand that by default, C++ use a call-by-value calling convention.

When you call swap_values, its stack frame receives a copy of the values passed to it as parameters. Thus, the parameters int x, int y are completely independent of the caller, and the variables int a, b.

Fortunately for you, C++ also support call-by-reference (see wikipedia, or a good programming language design textbook on that), which essentially means that the variables in your function are bound to (or, an alias of) the variables in the caller (this is a gross simplification).

The syntax for call-by-reference is:

void swap_values( int &x, int &y ){
    // do your swap here
}

Answer 2

You need to pass the variables by reference:

void swap_values( int& x, int& y )
{
   int z;
   z = y;
   y = x;
   x = z;
}

pass-by-value and pass-by-reference are key concepts in major programming languages. In C++, unless you specify by-reference, a pass-by-value occurs.

It basically means that it's not the original variables that are passed to the function, but copies.

That's why, outside the function, the variables remained the same - because inside the functions, only the copies were modified.

If you pass by reference (int& x, int& y), the function operates on the original variables.

Answer 3

you are passing by value. you can still pass by value but need to work with pointers.

here is the correct code needed:

void swap(int *i, int *j) {
   int t = *i;
   *i = *j;
   *j = t;
}

void main() {
   int a = 23, b = 47;
   printf("Before. a: %d, b: %d\n", a, b);
   swap(&a, &b);
   printf("After . a: %d, b: %d\n", a, b);
}

also a small document that explains "by reference" vs "by value" : http://www.tech-recipes.com/rx/1232/c-pointers-pass-by-value-pass-by-reference/