Swapping two string pointers

Question

i have to char[] in C and i wanted to swap between them, by only swaping the pointer to the array and not one char at a time so i wrote this code:

#include <stdio.h>
void fastSwap (char **i, char **d)
{
    char *t = *d;
    *d = *i;
    *i = t;
}
int main ()
{
    char num1[] = "012345678910";
    char num2[] = "abcdefghujk";
    fastSwap ((char**)&num1,(char**)&num2);
    printf ("%s\n",num1);
    printf ("%s\n",num2);
    return 0;
}

I get this output (note the last 4 characters)

abcdefgh8910
01234567ujk

When I expect:

abcdefghujk
012345678910

NOTE: I am working on a 64 bit Linux System.

Answer 1

Change fastSwap to:

void fastSwap (char *i, char *d)
{
   while ((*i) && (*d))
   {
      char t = *d;
      *d = *i;
      *i = t;

      i ++;
      d ++;
   }
}

then call fastSwap (num1, num2);

Answer 2

Arrays are not pointers. While they decay to pointers when you call your fastSwap(), these pointers are not the actual arrays. The fact that you need a cast should give you a hint that something is wrong.

This would work:

void fastSwap (const char **i, const char **d)
{
    const char *t = *d;
    *d = *i;
    *i = t;
}

const char* num1 = "012345678910";
const char* num2 = "abcdefghujk";
fastSwap (&num1,&num2);
printf ("%s\n",num1);
printf ("%s\n",num2);

Answer 3

You can't modify the addresses of num1 and num2, your code should work if your test was instead:

int main ()
{
    char num1[] = "012345678910";
    char num2[] = "abcdefghujk";
    char *test1 = num1;
    char *test2 = num2;
    fastSwap (&test1,&test2);
    printf ("%s\n",test1);
    printf ("%s\n",test2);
    return 0;
}

Answer 4

This will work:

int main ()
{
    char *num1 = "012345678910";
    char *num2 = "abcdefghujk";
    fastSwap (&num1,&num2);
    printf ("%s\n",num1);
    printf ("%s\n",num2);
    return 0;
}

Answer 5

Your fastSwap only seems to work. You're invoking undefined behavior by casting '&num1' and '&num2' (which are pointers to the characters of num1 and num2) to pointers to pointers of characters (char**).

char *t = *d

t will point to whatever d's contents are pointing to, however d is pointing to the actually characters of num2 ("abcdefghujk" or 0x61 0x62 0x63 0x64 0x65 0x66 0x67 0x68 0x75 0x6B 0x00). This means that '*d' is actually copying the contents of 'num2' and not the pointer to num2 as you probably expected.

't' then is a bad pointer however since it is never dereferenced you avoid a crash/segment fault.

Because you're on a 64 bit machine/OS pointers are 8 bytes the value of 't' is now the first 8 bytes of 'num2' and this is what gets put into num1 after

*i = t

If you intend to swap pointers you must first create pointer variables as Mark did

char *test1 = num1;
char *test2 = num2;
fastSwap (&test1,&test2);

Or change num1 and num2 into pointers (char *) rather than arrays (char[]) as sb1/Karl did

char *num1 = "012345678910";
char *num2 = "abcdefghujk";
fastSwap (&num1,&num2);

Answer 6

num1 is an array, and &num1 is the address of the array itself - it is not an address of a pointer.

The address of an array itself is the same location in memory as the address of the first element of the array, but it has a different type. When you cast that address to char **, you are claiming that it points at a char * value - but it does not. It points at a block of 13 chars. Your swap function then accesses that array of 13 chars as if it were a char * - since the latter is the same size as 8 chars on your platform, you end up swapping the first 8 chars of each array.