Why some work and some not ( Pointer and Non )-CodePudding

#include <stdio.h>

void swap1(int a, int b)
{
    int tmp;

    tmp = a;
    a = b;
    b = tmp;
    return;
}

int main()
{
    int x = 1, y = 2;

    swap1(x,y);
    printf("%d %d",x,y);
    return 0;
}

I don't understand why without Pointer these will not work but with Pointer it will work normally. I have search about it on Google and saw a few question about it but I still don't understand it after reading all the answers.

but this code work normally without using Pointer

#include <stdio.h>

int padovan(int n);

int main()
{
    int n;
    printf("Enter number : ");
    scanf("%d",&n);
    printf("Result : %d",padovan(n));
}

int padovan(int x)   
{
    if(x == 0 || x == 1 || x == 2)
        return 1;
    else if(x > 2)
        return padovan(x-2)   padovan(x-3);
}

CodePudding user response：

void swap1(int a, int b)
{
    int tmp
    tmp = a; 
    a = b;
    b = tmp;
    return; 
}

You're passing a and b by value, which means, when you pass a and b, the compiler makes a copy of a and b, not changing the actual value of a and b

If you want to make this to work, you have to use some pointers:

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

   int tmp;
   tmp = *a;    // save the value at address x 
   *a = *b;     // put y into x 
   *b = tmp ;   // put tmp into y 
  
   return;
}

int padovan(int x)   
{
    if(x == 0 || x == 1 || x == 2)
        return 1;
    else if(x > 2)
        return padovan(x-2)   padovan(x-3);
}

This code is fine. padovan() returns an int and you're printing that. You're not modifying any arguments in padovan().