Why is this simple piece of C code giving a segmentation fault?

Question

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

int main() {
    int size_to_alloc = sizeof(char*) * 1;
    char** p = (char**) malloc(size_to_alloc);
    p[0] = (char*) malloc (sizeof(char) * 10);
    strcpy("hello", p[0]);
    printf("%s\n", p[0]);
}

I am obviously missing something very basic but can't figure out what.

[Please see this discussion on why not to cast the return value of `malloc()` and family in `C`.](http://stackoverflow.com/q/605845/2173917). — Sourav Ghosh, Nov 30 '15 at 07:17
Thanks for the answers. Marking the first answer by Dilip as accepted. — Harsh Savla, Nov 30 '15 at 07:42
Dilip's answer is actually the last one ;-). All other answers were posted 15 minutes before, in the space of 1 minute. keithmo's comment was even faster, providing the correct answer in less than 2 minutes. — chqrlie, Nov 30 '15 at 07:52

Dilip Kumar · Answer 1 · 2015-11-30T08:38:47.773

1

strcpy() function has argument mismatch.

usage of string copy as per the man page char *strcpy(char *dest, const char *src);

So your strcpy() call has to be strcpy(p[0], "hello");

edited Nov 30 '15 at 08:38

answered Nov 30 '15 at 07:35

Dilip Kumar

1,710
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score 0 · Answer 2 · answered Nov 30 '15 at 07:20

Just make small change in strcpy function : char *strcpy(char *dest, const char *src).

dest : the destination array
src : string to be copied.

Like:

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

int main() {
    int size_to_alloc = sizeof(char*) * 1;
    char** p = (char**) malloc(size_to_alloc);
    p[0] = (char*) malloc (sizeof(char) * 10);
    strcpy(p[0],"Hello");      /* change */
    printf("%s\n", p[0]);
}

Output: Hello

score 0 · Answer 3 · answered Nov 30 '15 at 07:20

Please read the man page of strcpy(). It says

char *strcpy(char *dest, const char *src);

So, you need to have the first argument as the destination (aka, copy to), second one as source (aka copy from).

In your case, "hello" has been supplied as the copy to address and "hello" being a string literal, attempt to copy anything into it (i.e.,modification of a string literal) will result in undefined behavior, which leads to the segmentation fault.

Solution: swap the arguments of the function call.

score 0 · Accepted Answer · answered Nov 30 '15 at 07:20

0

strcpy takes its arguments in the opposite order: destination first, source second. Try:

strcpy(p[0], "hello");

answered Nov 30 '15 at 07:20

chqrlie

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Why is this simple piece of C code giving a segmentation fault?

4 Answers4