Why do you think segmentation error occurs at that code?

Question

#include <stdio.h>

void reverse(char *str){
    char * end = str;
    char tmp;
    if (str){
        while(*end){
            ++end;
        }
        --end;
        while (str < end){
            tmp = *str;
            *str++ = *end; // segmentation error
            *end-- = tmp;
        }
    }
}
  int main() 
  {
    char *name = "erogol"
    reverse(name);
    //printf("%s\n", name );

    return 0;  
  }

Why do you think segmentation error occurs at the place I singed with comment?

David Ranieri · Accepted Answer · 2014-01-17T07:48:00.203

2

Because name (as string literal) is for read only.

Use

char name[] = "erogol";

Seems that you are using a debugger, consider also using valgrind (sometimes it gives more details):

==4040== Process terminating with default action of signal 11 (SIGSEGV)
==4040==  Bad permissions for mapped region at address 0x40065C
==4040==    at 0x400551: reverse (demo.c:16)
==4040==    by 0x40058E: main (demo.c:24)

edited Jan 17 '14 at 07:48

answered Jan 17 '14 at 07:20

David Ranieri

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what is the technical difference between those two – erogol Jan 17 '14 at 07:25
@erogol, http://stackoverflow.com/a/12795948/1606345 – David Ranieri Jan 17 '14 at 07:35

score 1 · Answer 2 · answered Jan 17 '14 at 07:19

1

Because you modify a string literal.

Even though their type is char[] and not char const[] you don't have the right to modify them.

answered Jan 17 '14 at 07:19

Jens Gustedt

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Why do you think segmentation error occurs at that code?

2 Answers2