Why does this code snippet give segmentation fault

Question

This piece of code throws seg fault.Please help me identify the reason for the same

#include<stdio.h>

int main() {
        char* str;
        str = "abcd";
        str[0] = 'r';
        printf("%c\n" , str[0]);
        return 0;
}

thanks

cnicutar · Accepted Answer · 2011-10-05 19:19:28Z

6

Well explained in C FAQ 1.32. It's illegal to modify string literals.

6.4.5/6

It is unspeciﬁed whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undeﬁned.

answered Oct 5, 2011 at 19:19

cnicutar

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1 Comment

Rajan Over a year ago

thanks for clearing this with detailed explanation and C FAQ is definitely a treasure

Jerry Coffin · Accepted Answer · 2011-10-05 19:19:59Z

3

    str = "abcd";
    str[0] = 'r';

This attempts to modify a string literal. Officially, that's undefined behavior. On most modern systems, however, the memory holding string literals will be marked read-only, so attempting to modify them will give a fault.

answered Oct 5, 2011 at 19:19

Jerry Coffin

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Community · Accepted Answer · 2017-05-23 12:03:35Z

0

This is similar Question related to Segmentation Fault

Refer this for more info.

edited May 23, 2017 at 12:03

CommunityBot

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answered Oct 5, 2011 at 19:30

Siva Charan

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