CODEWITHSUNDEEP
cfunctionvoid
Rocket
Rocket

Reputation: 7

How do I make this code output "After swapping: 10, 5"?

So I'm new to programming. And I found this code from Quora when searching about different uses of void in C. I first assumed it would have a result of:

"Before swapping: 5, 10 After swapping: 10, 5"

But I get an output of:

"Before swapping: 5, 10 After swapping: 5, 10"

instead.

I felt like I needed to isolate either "swap(num1, num2);" "printf("After swapping: %d, %d \n", num1, num2);"

or

"int num1= 5, num2 = 10;"

from the rest of the statements inside the main function.

I tried declaring "int num1= 5, num2 = 10 ;" outside of main and also tried putting it inside of the swap function. That didn't work obv.

// The problem

void swap( int var1, int var2 )
{
  int temp;
  temp=var1;
  var1=var2;
  var2=temp;
}
void main( )
{
  int num1= 5, num2 = 10;
  printf("Before swapping: %d, %d \n", num1, num2);
  swap(num1, num2);
  printf("After swapping: %d, %d \n", num1, num2);
}

So, what are my options to have the desired output of:

"Before swapping: 5, 10 After swapping: 10, 5"

?

Upvotes: 0

Views: 91

Answers (1)

YunYanan
YunYanan

Reputation: 538

You should pass in the addresses of num1 and num2 to the swap function

void swap(int *var1, int *var2)
{
    int temp = 0;

    temp = *var1;
    *var1 = *var2;
    *var2 = temp;
}

void main(void)
{
    int num1= 5, num2 = 10;
    printf("Before swapping: %d, %d \n", num1, num2);
    swap(&num1, &num2);
    printf("After swapping: %d, %d \n", num1, num2);
}

Upvotes: 2

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