CODEWITHSUNDEEP
csegmentation-faultmemset
AppleDash
AppleDash

Reputation: 1714

Why is this usage of memset() segfaulting?

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

int main(int argc, char *argv[]) {
    char *str = "This is a string!";
    int therealthing = sizeof(str[0]) * 4;
    memset(str, 'b', therealthing);
    printf("%s\n", str);
    return 0;
}

This code causes a segfault, any ideas why? I have already tried passing it as a memory address and as a pointer.

Upvotes: 1

Views: 299

Answers (2)

ciphermagi
ciphermagi

Reputation: 748

This is a string literal. It's immutable. Can't be changed.

char *str = "This is a string!";

You're trying to change it with memset. You could use an array of characters

char str[] = "This is a string!";

or

char * str = malloc(sizeof(char) * (strlen("This is a string!") + 1));
strcpy(str, "This is a string!");

Upvotes: 2

haccks
haccks

Reputation: 106012

You can't modify a string literal. It will invoke undefined behavior.
Try this instead 

char str[] = "This is a string!";

Upvotes: 1

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