Changing pointer address - function

Question

I have problem with pointers. This is working fine -

int main(void){
    char *w;
    w = calloc(20, sizeof(char));
    w = "ab";

    printf("%c",*w);
    w = w + sizeof(char);
    printf("%c",*w);

    return 0;
}

but if i use function like:

void por(char *t){
    t = t + sizeof(char);
}

and

int main(void){
    char *w;
    w = calloc(20, sizeof(char));
    w = "ab";
    printf("%c",*w);
    por(w);
    printf("%c",*w);

    return 0;
}

then it prints "aa" instead of "ab". I know its probably pretty stupid question, but i don't know what is going and how to solve that issue.

sizeof (char) is per definition always 1. So no need to write it. So incrementing a pointer can be written simply `p++` — Patrick Schlüter, Nov 15 '11 at 14:11
exactly, you actually got a memory leak here because you make w point away to the other place than where it is allocated — ComfortablyNumb, Nov 15 '11 at 14:17
There is also a memory leak because free() isn't called anywhere... — Lundin, Nov 15 '11 at 14:45

score 4 · Accepted Answer · answered Nov 15 '11 at 14:01

4

In your por function, t will not be changed. You need change it

void por(char **t){
 *t = *t + sizeof(char);
}

and call it with por(&w)

answered Nov 15 '11 at 14:01

ComfortablyNumb

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Thank you. Problem solved, but can you explain it ? :) &w - passes adress of pointer, yes ? I cant understand what **t meaning. (sory for my english) – kejmil01 Nov 15 '11 at 14:07
1

You need to dig a bit deeper about how function call takes place. Basically, you parameter will be copied in a function, so the original one will not be changed after the call. To change it, you need manipulate its address. Here what you want to change is a pointer, so you need pass the parameter as the address of pointer: the pointer of pointer. Now can you accept my answer so that I earn some reputation. :) – ComfortablyNumb Nov 15 '11 at 14:14

score 1 · Answer 2 · answered Nov 15 '11 at 14:06

1

Try this:

static char *por(char *t)
{
    return t + sizeof(char);
}

int main(void)
{
    char *w = "ab";
    printf("%c",*w);
    w = por(w);
    printf("%c",*w);

    return 0;
}

answered Nov 15 '11 at 14:06

trojanfoe

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score 0 · Answer 3 · answered Nov 15 '11 at 15:09

0

you are incrementing the copy which is local to the function.

answered Nov 15 '11 at 15:09

Bazooka

1,428
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Changing pointer address - function

3 Answers3