Why below code gives segmentation fault problem?

Question

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Why does simple C code receive segmentation fault?

I have code below which will remove trailing spaces from a string but i don't what going in this code so as it gives segmentation fault problem??

 void main(void);

char* rtrim(char*);

void main(void)
 { 
 char* trail_str = "This string has trailing spaces in it.               ";


 printf("Before calling rtrim(), trail_str is '%s'\n", trail_str);

 printf("and has a length of %d.\n", strlen(trail_str));



 rtrim(trail_str);



 printf("After calling rtrim(), trail_str is '%s'\n", trail_str);

 printf("and has a length of %d.\n", strlen(trail_str));

 }


  char* rtrim(char* str)
  {
   int n = strlen(str) - 1;    

    while (n>0)            
   {
       if (*(str+n) != ' ')    
      {

           *(str+n+1) = '\0'; 



           break;             
      }

      else

      n--;
 }

 return str;      

 }

char trail_str[] = "This string has trailing spaces in it. "; Duplicates http://stackoverflow.com/questions/164194/why-does-simple-c-code-receive-segmentation-fault — Gilbert, Jun 19 '11 at 11:58

score 3 · Accepted Answer · answered Jun 19 '11 at 11:58

3

trail_str points to a constant area in the memory and thus cannot be changed in *(str+n+1) = '\0'

when initializing

char* trail_str = "This string has trailing spaces in it.               ";

you actualy generate a constant string: "This string has trailing spaces in it. " and tell trail_str to point to it.

answered Jun 19 '11 at 11:58

amit

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r u saying declaring like char trail_str[]=""This string has trailing spaces in it. "; will solve this problem?? – Amit Singh Tomar Jun 19 '11 at 12:02
(ironic: @amit is answering @AMIT): yes, because str[]="..." is actually str[] = { 'T','h',...} (creating array of chars on stack and initializing it) while *str = "..." creates a const char[] on memory, and points to it. – amit Jun 19 '11 at 12:06
Thanks @AMIT i GOT your point.but still am not getting desired result ,may be logic have some problem. – Amit Singh Tomar Jun 19 '11 at 12:09
note that for a string containing only whitespaces `(" ")` the function will give seg fault, because `n` will be negative! (you will not enter the if condition and break, and try to approach str-1, which will result in undefined behavior.) – amit Jun 19 '11 at 12:26

score 2 · Answer 2 · answered Jun 19 '11 at 11:58

2

char* trail_str = "This string has trailing spaces in it.               ";

The string pointed to by tail_str can be stored in read-only memory. You can't modify it.

If you want to modify it, you'll need to allocate storage for it and copy that string constant .

(Also, main should return an int, not void.)

answered Jun 19 '11 at 11:58

Mat

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The type of `"blablabla"` is `const char[]` which decays to `const char*`, not `char*`. – rubenvb Jun 19 '11 at 11:59
2

@rubenvb: not in C. The type of a string literal is `char[]` (and string literals decay to type `char*`). It is, however, undefined behaviour to modify the contents of a literal string. Some people like to **add** `const` when declaring pointers to string literals and get help from the compiler in case they try to modify it – pmg Jun 19 '11 at 12:41
Then what does C99 6.4.5/5: `The multibyte character sequence is then used to initialize an array of static storage duration and length just sufficient to contain the sequence` and 6.7.8/4: `All the expressions in an initializer for an object that has static storage duration shall be constant expressions or string literals.` mean? One references the other, and I can't make out if the char* array of static storage duration used for the string literal will need to be a constant expression. This is confusing wording... – rubenvb Jun 19 '11 at 13:34

Why below code gives segmentation fault problem?

2 Answers2