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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;

}>

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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().


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/

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  • Thanks a lot I got it....
    – Ahmed Raza
    Sep 28 '16 at 11:31

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