Pointer-to-a-pointer in C throws segmentation fault

Question

This might sound pretty old school, but I'm still unable to figure out why the following program throws a segmentation fault. Any help would be great

#include <stdio.h>
pointer(char **x)
{
    printf ("Before %c",*x[0]);
    *x[0] = 'a';   // segmentation fault here!!
    printf ("After %c", *x[0]);
}
int main()
{
    char *x = "Hello";
    pointer(&x);
}

Look here, it's explained in the top answer: http://stackoverflow.com/questions/1704407/what-is-the-difference-between-char-s-and-char-s-in-c — AntonH, Aug 12 '14 at 20:08

score 1 · Answer 1 · edited May 23 '17 at 12:20

1

It's explained in the answer to this question.

TL;DR: the memory pointed to by char *x = "Hello"; is read only. Trying to write to it is illegal, and will result in a segmentation fault.

edited May 23 '17 at 12:20

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answered Aug 12 '14 at 20:10

AntonH

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Why on earth was this downvoted, 20 days after the answer was posted? – AntonH Sep 01 '14 at 19:47

score 1 · Accepted Answer · edited May 23 '17 at 10:26

1

char *x = "Hello";

This declaration makes it read-only. Writing to it the way you tried is illegal.

See this for more information

edited May 23 '17 at 10:26

Community

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answered Aug 12 '14 at 20:10

Asics

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Pointer-to-a-pointer in C throws segmentation fault

2 Answers2