ANSI C strstr() not working with pointers

Question

I'm trying to use strstr() to find a substring in a string. Using char[] works only, char* is not working, gives segmentation fault.

So this one is working:

int main() {
    char str[] = "apple apple peach";
    char *p;

    p = strstr(str, "peach");
    if (p!= NULL) {
        strncpy(p, "apple", 5);
    }
    return 0;
}

But this one is not working:

int main() {
    char *str = "apple apple peach";
    char *p;

    p = strstr(str, "peach");
    if (p!= NULL) {
        strncpy(p, "apple", 5);
    }
    return 0;
}

This one is neither:

int main() {
    char *str = "apple apple peach";
    char *peach = "peach";
    char *p;

    p = strstr(str, peach);
    if (p!= NULL) {
        strncpy(p, "apple", 5);
    }
    return 0;
}

And neither this one:

int main() {
    char *str = "apple apple peach";
    char peach[] = "peach";
    char *p;

    p = strstr(str, peach);
    if (p!= NULL) {
        strncpy(p, "apple", 5);
    }
    return 0;
}

Is that a known bug or feature?

This is not a problem of `strstr`. You can not write into string constant. It is placed into read-only part of memory. — Marian, Nov 06 '16 at 10:07
To OP: Compile with -Wwrite-strings to get a warning about this issue. — Bjorn A., Nov 06 '16 at 10:11
or better with -Wall and -Wextra, when you use gcc or use the equivalent flags for your compiler. and check every warning. — 12431234123412341234123, Nov 06 '16 at 10:23
@12431234123412341234123. Note that write-strings is _not_ enabled by -Wall -Wextra. It used to be, but not anymore. — Bjorn A., Nov 06 '16 at 10:32

score 5 · Answer 1 · answered Nov 06 '16 at 10:07

5

Your problem is not with strstr, it's with strncpy. In the pointer case, you're using strncpy to try to write to string literal, which is constant.

answered Nov 06 '16 at 10:07

jamesdlin

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Thanks for the answer – Janekx Nov 06 '16 at 10:11

eca2ed291a2f572f66f4a5fcf57511 · Accepted Answer · 2016-11-06T10:33:50.793

string-literal is not safe to modify! The one with char[] was successful only because strncpy(p, "apple", 5); modified the copied version of string-literal, not the string-literal itself. The relevant paragraphs in the standard read:

6.4.5, paragraph 6

... The multibyte character sequence is then used to initialize an array of static storage duration and length just sufficient to contain the sequence...

6.4.5, paragraph 7

... If the program attempts to modify such an array, the behavior is undefined.

In short, string-literal is (conceptually) replaced with an array, and it's not safe to modify this. Note that string-literal is not necessarily const-qualified, although MSVC++ seems to mandate that string-literal is const char * or something that's const-qualified.

Sergey Kalinichenko · Answer 3 · 2016-11-06T10:11:20.923

It has nothing to do with strstr: it works as expected. The crash happens in strcpy, when you attempt to modify memory allocated to string literal. This is undefined behavior.

The key to solving this problem is placing the data into memory that allows writing. Your first program does so by declaring str an array:

char str[] = "apple apple peach";

That is not UB, so the program runs to completion. Another alternative would be to allocate str in dynamic memory, like this:

char *str = malloc(20);
strcpy(str, "apple apple peach");
char *p;
p = strstr(str, "peach");
if (p!= NULL) {
    strncpy(p, "apple", 5);
}
free(str);

Great, thanks. That was a point which was forgotten. :) – Janekx Nov 06 '16 at 10:10 — Janekx, Nov 06 '16 at 10:10

ANSI C strstr() not working with pointers

3 Answers3

6.4.5, paragraph 6

6.4.5, paragraph 7