Solution for Exercise 19-23

ex1923

#include <stdio.h>

void swap(int *a, int *b);

int main()
{
    int x=100,y=33;

    printf("Before swap: x=%d, y=%d\n",x,y);
    swap(&x,&y);
    printf("After swap: x=%d, y=%d\n",x,y);

    return(0);
}

void swap(int *a, int *b)
{
    int temp;

    temp = *a;
    *a = *b;
    *b = temp;
}

Output

Before swap: x=100, y=33
After swap: x=33, y=100

Notes

* The swap() function is prototyped at Line 3 with the * operator for each of its arguments. Both are pointers, memory locations.

* The swap() function requires pointers, addresses, so at Line 10 it's called with variables a and b prefixed by the address-of operator, &. If you forget this step, the compiler reminds you rather rudely.

* It's also possible to declare pointer variables in the main() function and pass these arguments directly in the swap() function.

* Pointer notation must be used for the swap action at Lines 20, 21, and 22 in the swap() function. The temp variable, however, need not be a pointer; it just temporarily holds the values during the swap process.

Exercise ex2001 →