CODEWITHSUNDEEP
cpointersgenericsswap
Somjit
Somjit

Reputation: 2772

Generic swapping of "string" in C

I was looking at a video lecture about how to do a generic swap using memcpy() and strdup().

The generic swap code is as follows :

void GenericSwap(void* ptr1, void* ptr2, int size){

    char buf[size];
    memcpy(buf,ptr1,size);
    memcpy(ptr1,ptr2,size);
    memcpy(ptr2,buf,size);
}

and we call it like this :

char* h = strdup("fred");
char* w = strdup("wilma");
GenericSwap(&h,&w,sizeof(char*));
printf("h : %s\nw : %s\n\n",h,w);

This swapped "fred" and "wilma" as a whole string. That is, h points to "wilma" and w points to "fred" now.

However, when i change the code to this : 

char h1[10] = "fred";
char w1[10] = "wilma";
GenericSwap(&h1,&w1,sizeof(char*));
printf("h1 : %s\nw1 : %s\n",h1,w1);

I get "wilm" and "freda" for h1 and w1. I am passing address of stuff holding the address of the 1st element of the array, just like the proper working code up top, but this isn't working so nicely as the top one did.

Is it Something about h1 and w1 being mere pointers, and not pointer-to-pointers ?

Upvotes: 1

Views: 253

Answers (1)

unwind
unwind

Reputation: 399803

You're trying to change the address of the arrays, which you cannot do. Array variables act as const pointers to their data, they cannot be changed.

Instead, you're swapping the first sizeof (char *) (which seems to be 4 on your system) characters.

Basically for

char h[8] = "foo";

The two statements

printf("%p\n", (void *) h);

and

printf("%p\n", (void *) &h);

will print the same address, i.e. the expression &h evaluates to the address of the first element, just like h. There is no separate variable that holds the array's address, for you to change.

Upvotes: 3

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