CODEWITHSUNDEEP
c
edwin
edwin

Reputation: 107

Swap function returning array

Trying to understand how a pointer works in a function that returns an array.

When the temp array is returned to the function, why is it that p[0] is 1 and p[1] is 3? Since x and y variable are swapped within the function and temp[0] and temp[1] are not swapped.

int *swap(int *x, int *y){
  static int temp[2];

  temp[0] = *x; 
  temp[1] = *y;
  *x = temp[1];
  *y = temp[0];
  return temp;
}


int main() {

  int x = 3;
  int y = 1;
  int *p = swap(&x, &y);
    GPIO_PORTF_AHB_DATA_BITS_R[LED_RED] = LED_RED;//turn on red led
    delay(p[0]);

    GPIO_PORTF_AHB_DATA_BITS_R[LED_RED] = 0;//turn off red led
    delay(p[1]);

  }

Upvotes: 0

Views: 174

Answers (1)

user149341
user149341

Reputation:

why is it that p[0] is 1 and p[1] is 3

It isn't.

Replacing your microcontroller-specific code with:

printf("p[0] = %d, p[1] = %d\n", p[0], p[1]);

and running your code on a computer gives me the output:

p[0] = 3, p[1] = 1

as expected.

Upvotes: 3

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