This is the code why it is not returning the swapped values

Question

This code is written to return swapped values, why values are not being returned, My basic question is how can i take return value from function called, please demonstrate with any code example.

<#include 
int swap (int x,int y);
int main (void)
{
    int a = 1;
    int b = 2;
    printf("value of a and b is %i,%i respectively\n", a,b);
    printf ("Swapping!\n");
    swap (a,b);
   printf("value of a and b is %i,%i respectively\n", a,b);


}

int swap (int x,int y)
{
    int temp=x;
    x=y;
    y=temp;

 return x,y;

}>

Answer 1

When you call swap() with int a and int b you are not actually "passing" swap() the variables. Instead you are handing swap() a copy of them, more specifically a copy of the value the variables contain. So you are indeed swapping correctly, you are just not handing over the final results back over the the caller. Below is a correct swap function.

void swap(int* first, int* second)
{
    int temp = *first;
    *first = *second;
    *second = temp;
}

Here you can see I prefixed the variable with a "*", this basically means the variable is a pointer (or an address) to a value, so now it refers to int a and int b. In other words, using this code, main() calls swap() with the address of its very own int a and int b.

Do keep in mind though, now that you're passing by reference. Is to call swap() like: swap(&a, &b);, the & or "address-of" operator retrieves the address of its right operand before passing it to swap().

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CS50 covers this (pointers) in great detail -- in fact I'd say it is the primary focus of the course's middle-half -- more information is freely available all over the internet and at https://study.cs50.net/