I am trying to implement a swap back function that will swap the values back to the original position. This is the instruction I have to do accordingly:
Implement a second swapping function, swapBack(int v[2]), that takes the vector v = [x, y] as an input and swaps its entries. Includes the new function into the program above to swap back the values of x and y.
#include <stdio.h>
void swap(int *pX, int * pY){
int temp = *pX;
*pX = *pY;
*pY = temp;
}
void swapBack(int x, int y){
swap(&x,&y);
}
int main() {
int x = 10;
int y = 5;
swap(&x, &y);
printf("x = %d, y = %d\n", x, y);
swapBack(&x, &y);
printf("x = %d, y = %d\n", x, y);
return x + y;
}
However, both printf
produce the same results and I am not sure how to fix it? thanks
You're defining swapBack
wrong. It's supposed to take an array (which decays to a pointer) of two integers.
This should be about what you want:
void swapBack(int v[2])
{
int temp = v[0];
v[0] = v[1];
v[1] = temp;
}
A program to test this:
#include <stdio.h>
void swapBack(int v[2])
{
int temp = v[0];
v[0] = v[1];
v[1] = temp;
}
int main(void)
{
int v[2] = {10, 5};
printf("v[0] = %d, v[1] = %d\n", v[0], v[1]);
swapBack(v);
printf("v[0] = %d, v[1] = %d\n", v[0], v[1]);
printf("v[0] + v[1] = %d\n", v[0] + v[1]);
return 0;
}
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