CODEWITHSUNDEEP
cpointerscharvoid-pointerschar-pointer
sam0101
sam0101

Reputation: 375

Generic swap function using pointer to char in C

I don't understand so well how this code works:

#include <stdio.h>
void gswap(void* ptra, void* ptrb, int size)
{
 char temp;
 char *pa = (char*)ptra;
 char *pb = (char*)ptrb;
 for (int i = 0 ; i < size ; i++) {
   temp = pa[i];
   pa[i] = pb[i];
   pb[i] = temp;
 }
}

int main()
{
    int a=1, b=5;
    gswap(&a, &b, sizeof(int));
    printf("%d , %d", a, b)
}

What I understand is that char has 1 byte(size) in memory and we are using pointers to swap each byte of the int value(4 bytes).
But in the end, how it is possible to dereference a char pointer to int value?

Upvotes: 2

Views: 1193

Answers (1)

Bernardo Meurer
Bernardo Meurer

Reputation: 2335

Let's try and figure this out, step by step, with code comments

#include <stdio.h>

//gswap() takes two pointers, prta and ptrb, and the size of the data they point to
void gswap(void* ptra, void* ptrb, int size)
{
    // temp will be our temporary variable for exchanging the values
    char temp;
    // We reinterpret the pointers as char* (byte) pointers
    char *pa = (char*)ptra;
    char *pb = (char*)ptrb;
    // We loop over each byte of the type/structure ptra/b point too, i.e. we loop over size
    for (int i = 0 ; i < size ; i++) {
        temp = pa[i]; //store a in temp
        pa[i] = pb[i]; // replace a with b
        pb[i] = temp; // replace b with temp = old(a)
    }
}

int main()
{
    // Two integers
    int a=1, b=5;
    // Swap them
    gswap(&a, &b, sizeof(int));
    // See they've been swapped!
    printf("%d , %d", a, b);
}

So, basically, it works by going over any given datatype, reinterpreting as bytes, and swapping the bytes.

Upvotes: 4

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