Pointer Confusion: swap method in c

Question

#include
void swap(int *a,int *b){
    int p=*b;
    *b=*a;
    *a=p;

    /*int *p=b;
    b=a;
    a=p;
    */
}

int main(){
    int a,b;
    scanf(\"%d         


        

Answer 1

Assuming you meant

void swap(int *a,int *b){
int *p=b;
b=a;
a=p;
}

That code just swaps the value of the pointers in the swap() function. That won't swap the addresses around in main() because, as you said, "parameter int *a,int *b is local variable".

When you call the function swap() like this

swap(&a,&b);

the addresses of a and b are passed and become local variables in the swap() function. You can't change the address of a or b - they have a location in memory.

In the code that works

void swap(int *a,int *b){
int p=*b;
*b=*a;
*a=p;
}

you don't change the value of the pointers, you change the values in the memory the pointers point to, which is why that works.

While C is pass-by-value, if you pass the address of something as the value a function can modify something outside its scope because you told the function where that variable is.