Truncating string causes segmentation fault

Question

I would like to write a function in C that truncates an input string to 32 chars, but the code below gives me a segmentation fault. Could anyone explain why it is so?

void foo (char *value){
    if (strlen(value)>32) {
        printf("%c\n", value[31]); // This works
        value[31] = '\0';          // This seg faults
    }
}

This is just one small part of a bigger function that I have written. I didn't post the rest of the function, which works when a value string is passed into it. — Hui Wang, Mar 12 '17 at 03:49
"The caller side code is just: foo ("hello");" -- you should compile your code with warnings enabled, and pay attention to them. Your compiler will very likely warn that passing `"hello"` to a function that requires a (modifiable) `char*` parameter is invalid. — Employed Russian, Mar 12 '17 at 04:01
@EmployedRussian I don't get any warnings from GCC with `-Wall -Wextra -pedantic`. — Spikatrix, Mar 12 '17 at 04:19
@CoolGuy Try with `-Wwrite-strings`. I am surprised it's not in `-Wall`. — Employed Russian, Mar 12 '17 at 06:44

score 6 · Accepted Answer · answered Mar 12 '17 at 03:52

If you call your function like this:

char str[] = "1234567890123456789012345678901234567890";
foo(str);

It will work fine. But if you call it like this:

char *str = "1234567890123456789012345678901234567890";
foo(str);

That can cause a segfault.

The difference here is that in the former case str is a char array, while in the latter case str is a pointer to a string constant. A string constant typically lives in a read only section of memory, so attempting to modify it causes the core dump.

@HuiWang Glad I could help. Feel free to [accept this answer](http://stackoverflow.com/help/accepted-answer) if you found it useful. — dbush, Mar 12 '17 at 13:25

androidFan · Answer 2 · 2017-03-12T04:38:02.967

0

Your program should be something like this :

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

void foo(char **value) {
    if (strlen(*value)>32) {
        printf("%c\n", (*value)[32]);
        (*value)[32] = '\0'; // You want the string length to be 32, so set 32th character to '\0' so 32 characters will be from 0 to 31
    }
}

int main() {
    char *str;
    str = malloc(100); // Change 100 to max possible length of string user can enter

    printf("Enter string : ");
    scanf("%s", str);

    foo(&str);

    printf("Truncated string is %s\n", str);

    free(str);

    return 0;
}

edited Mar 12 '17 at 04:38

answered Mar 12 '17 at 04:23

androidFan

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This is exactly one of the reasons why you [should not cast the result of `malloc`](http://stackoverflow.com/q/605845/3049655). See the last bullet point of [this answer](http://stackoverflow.com/a/605858/3049655). You forgot to include `stdlib.h` and the cast hides the error. Also, `free` the memory (although it would automatically be freed in this case as the program ends) – Spikatrix Mar 12 '17 at 04:30
@CoolGuy since C99 the cast does not hide the error – M.M Mar 12 '17 at 04:39
Important thing to note here is that you have to cast for C++ – androidFan Mar 12 '17 at 04:41
@M.M Thanks for letting me know! :-) – Spikatrix Mar 12 '17 at 05:57

Truncating string causes segmentation fault

2 Answers2