Segmentation fault (11) when using malloc inside a function to form a string

Question

I'm trying to use a function to assign space and fill that space (or at least some of it) with characters to form a string. Within the function I make a call to malloc, and within the same function I assign characters to the given space. The following code gives the general gist of what I'm doing:

#define INITIAL 10

int func(char **s);

int
main(int argc, char **argv) {
    char *s;
    int n;

    n = func(&s);
    printf("Done\n");

    return 0;
}

int
func(char **s) {
    int i;

    *s = (char*)malloc(INITIAL*sizeof(char));
    assert(*s);

    for (i=0; i<5; i++) {
        printf("i=%d\n", i);
        *s[i] = 'a'; /*'a' is an arbitrary char for this example */  
    }
    return i;    
}

The output of this code is:

i=0
i=1
i=2
Segmentation fault: 11

The reason I have my function return an int is because I ultimately want the function to return the length of the string I have formed.
I'm completely unsure why I am getting a segmentation fault; it seems I have assigned enough space to fit the next char in. It also seems weird to me that it stops at i=2. If anyone could identify the mistakes I have made I would greatly appreciate it!

Can't you use strdup or asprintf (on Linux, you could)? Don't forget to compile your code with all warnings & debug ino (gcc -Wall -Wextra -g if using GCC...). Use a debugger (gdb) & valgrind; avoid undefined behavior — Basile Starynkevitch
– Basile Starynkevitch, Commented Oct 3, 2015 at 14:20
You should add free(s); before return 0; or your program will create a memory leak — Mike -- No longer here
– Mike -- No longer here, Commented Oct 3, 2015 at 16:36
Curious: what suggest using *sizeof(char) in malloc(INITIAL*sizeof(char)) as sizeof(char) is always 1? Nice: alternative: *s = malloc(INITIAL* sizeof *(*s)); — chux
– chux, Commented Oct 3, 2015 at 21:39

Vaughn Cato · Accepted Answer · 2015-10-03 14:17:42Z

5

Instead of

*s[i] = 'a';

you want

(*s)[i] = 'a';

*s[i] is equivalent to *(s[i]). That is, it treats s as an array of strings and gives you the first character of the string at index i.

answered Oct 3, 2015 at 14:17

Vaughn Cato

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1 Comment

Motmot Over a year ago

Thanks a lot! All fixed :)

MikeCAT · Accepted Answer · 2015-10-03 14:17:49Z

2

*s[i] first calculate s[i], which won't be valid place for i!=0, then dereference it and try to put 'a' there. It may cause Segmentation Fault.

Try changing *s[i] to (*s)[i].

answered Oct 3, 2015 at 14:17

MikeCAT

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1 Comment

Motmot Over a year ago

Thanks for your help!

John Bode · Accepted Answer · 2015-10-03 15:35:53Z

1

Postfix [] has higher precedence than unary *, so *s[i] is being parsed as *(s[i]), which isn't what you want; you want to dereference s and index into the result, so you need to explicitly group the * operator with s: (*s)[i].

answered Oct 3, 2015 at 15:35

John Bode

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1 Comment

Motmot Over a year ago

Thanks a lot John, appreciate the help :)

score 0 · Accepted Answer · 2015-10-03 18:14:06Z

0

You may want to use size_t instead of an int. Or ssize_t if you need the function to return a negative value:

#include <stdio.h>
#include <stdlib.h>
#define INITIAL 10

ssize_t func(char **);

int main(void)
{
    char *s;

    if((func(&s)) == -1)
    {
        printf("An error occurred\n");
        return 1;
    }

    printf("Done\n");

    free(s);
    return 0;
}


ssize_t func(char **s)
{
    size_t i = 0;

    if ( INITIAL < 1 )
            return -1;

    if (!(*s = malloc(INITIAL*sizeof(char))))
            return -1;

    for (i=0; i< 5; i++) {
        printf("i=%zu\n", i);
        (*s)[i] = 'a';; /*'a' is an arbitrary char for this example */
    }
    return i;
}

edited Oct 3, 2015 at 18:14

answered Oct 3, 2015 at 17:04

user4832408

2 Comments

chux Over a year ago

Note: ssize_t is not defined in standard C. Common extension though.

Motmot Over a year ago

If you don't mind me asking what would be the benefit of doing this? I'm fairly new to C and my textbook only briefly mentions size_t, and doesn't specify how it is different to int.

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