Segmentation fault on execution of a C program related to strings

Question

I have the following code. I initialized 2 pointers, one at the beginning and another at the ending of a string. After every step, I increment the 1st pointer and decrement the second pointer. I copy the value of first pointer into the second if the value obtained by dereferencing 1st pointer is less than the value obtained by dereferencing 2nd pointer.

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

int main() {
   char *word="aacdbc";
   char *p=word;
   char *q=word+(strlen(word)-1);
   printf("\n%s\n",word);
   int i;
   for(i=1;i<=strlen(word)-1;++i) {
      if(*p<*q) {
         *q=*p;
      }
      ++p;
      --q;
   }
   printf("\n%s\n",word);
   return 0;
}

But the code shows a "Segmentation fault error". In which line did I make a mistake?

Vlad from Moscow · Accepted Answer · 2014-08-14T17:58:53.493

4

String literals in C (and C++) are immutable. So your attempt to change the string literal pointed to by the variable word

char *word = "aacdbc";

has undefined behaviour.

Change the definition from a pointer to an array

char word[] = "aacdbc";

The program could look the following way

#include <stdio.h>

int main( void )
{
   char word[] = "aacdbc";
   char *p = word;
   char *q = word+ + sizeof( worrd ) - 1;

   printf( "\n%s\n", word );

   while ( p < q && p < --q )
   {
      if ( *p < *q ) *q = *p;
      ++p;
   }

   printf( "\n%s\n", word );

   return 0;
}

edited Aug 14 '14 at 17:58

answered Aug 14 '14 at 17:51

Vlad from Moscow

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1

It might be worth also mentioning how the string literal is stored in the code segment (RO), while the array is in the stack (RW). http://stackoverflow.com/questions/16021454/difference-between-declared-string-and-allocated-string – Cloud Aug 14 '14 at 17:56
@Dogbert: That's an implementation-detail that might be so, or might not. – Deduplicator Aug 14 '14 at 17:57
@Deduplicator That type of detail is common across all ANSI-compliant C compilers, and is not limited to a specific platform or OS. – Cloud Aug 14 '14 at 20:34
@Dogbert: constant literals need neither be pooled, nor be in the code-segment, nor need they be in any type of read-only section. On systems with memory-protection or read-only memory they will be put there when possible by most compilers per default, sure, but that's not contractual. – Deduplicator Aug 14 '14 at 20:46
@Deduplicator Could you point me to some docs or posts on this? It would be interesting to read up on further (personal development and education, etc). Thanks for the insight! – Cloud Aug 14 '14 at 20:54
1

@Dogbert: Well, the best documentation for that is actually the standard. If you want a good post here which explicitly handles that, look at this answer: http://stackoverflow.com/a/4105123 – Deduplicator Aug 14 '14 at 21:13

score 1 · Answer 2 · answered Aug 14 '14 at 17:51

Because word is a pointer to a constant (read-only) characters sequence. In other words, you can't change the contents of a string literal.

If you want to change its contents, you should declare word as an array of chars:

char word[]="aacdbc";

With that small change, the segfault should disappear.

Segmentation fault on execution of a C program related to strings

2 Answers2