Question

    #include<stdio.h>
    void swap(char *, char *);

    int main()
    {
        char *pstr[2] = {"Hello", "IndiaBIX"};
        swap(&pstr[0],&pstr[1]);
        printf("%s\t%s", pstr[0], pstr[1]);
        return 0;
    }
    void swap(char **t1, char **t2)
    {
         char *t;
         t=*t1;
         *t1=*t2;
         *t2=*t;
    }

I don’t understand why can’t swap the pointer of strings by calling the function like this:

swap(pstr[0],pstr[1]);

I was in a dilemma why I should not use that.Some one please help me.

thanks..

Answer 1

Actually, you have mainly two errors:

• in the instruction *t2 = *t, *t2 is a string whereas *t is a single character;
• the declaration of swap is different from its definition.

Also, pstr[0] and pstr[1] can be pointers to read-only strings, so declare them as pointers to const char is considered as a good practice.

In that case, the following code works fine (it swaps just the value of the both pointers, not the strings themselves).

#include <stdio.h>

static void 
swap(const char ** const p, const char ** const q)
{
    const char * const pTmp = *p;
    *p = *q;
    *q = pTmp;
}

int main(void)
{
    const char *p[] = { "Hello", "IndiaBIX" };
    printf("%s - %s\n", p[0], p[1]);
    swap(p, p + 1);
    printf("%s - %s\n", p[0], p[1]);
    return 0;
}