change a string in a function passing a pointer

Question

Hello I have some trouble with pointer !

#include <stdio.h>
#include <stdlib.h>

void test(char **str)
{
    *str = (char *)malloc(sizeof(char) * 2);
    *str[0] = 'b';
    *str[1] = '\0';
}

int     main(void)
{
    char *str;

    test(&str);
    printf("%s\n", str);
    return (0);
}

So I pass a pointer of an non allocated string in my function test, then I allocate my string in the function and try to manipulate it but this code segfault so I guess I miss something on my pointer lesson :)

Can you help me to figure out what is happening here ? thank you a lot !

edit : when I remove my return (0) in the main, the code compile and display my char * ! super weird

To avoid array indexing try using this (**str = 'b'; *(*str+1) = '\0';) in your code — Sudipta Kumar Sahoo, Feb 29 '16 at 18:44
Maybe you were taught otherwise, but [don't cast the result of `malloc()`](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). Also, checking the return value is a good idea. — e0k, Feb 29 '16 at 18:47

score 3 · Answer 1 · answered Feb 29 '16 at 18:44

The array subscript operator [] has higher precedence than the indirection operator *. So the assignment to str is effectively this which is incorrect:

*(str[0]) = 'b';
*(str[1]) = '\0';

You need to add parenthesis:

(*str)[0] = 'b';
(*str)[1] = '\0';

score 2 · Accepted Answer · answered Feb 29 '16 at 18:43

2

You must write:

(*str)[0]= 'b';

You must first dereference the str and now the indexing will work on the target, the memory allocated by malloc.

answered Feb 29 '16 at 18:43

Paul Ogilvie

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thank you ! It's working fine. can you explain to me how *str[0] is interpreted ? because *str[] and (*str)[] seems pretty similar to me – decarte Feb 29 '16 at 18:48
@hadewsMM *str[0] means `*(str[0])` – user253751 Feb 29 '16 at 20:05

score 0 · Answer 3 · answered Feb 29 '16 at 19:45

Actually, just 'str' itself means he address of character array. But 'str' means the value of str. So the str[0] contains the value of 'b' but as you added '' before str[0] (*str[0]), computer trying to find the value where the address is 'b'. But this address is not allowed for this process. So this program returns segment fault. Thank you.

change a string in a function passing a pointer

3 Answers3