Why segmentation fault occurs in the following

Question

I am trying to remove two consecutive duplicate elements from the string.I am getting segmentation fault in the line 16 .even the commented 17th line also get me the same error.

dont worry about my logic of my program.it may be wrong...but i am struck with this error..help me out...explain why i am getting this error

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

int main()
{
    char *st="fvvbbyyr";
    int i=0,j,len;
    for(len=0;st[len]!='\0';len++);
    for(i=0;i<len;i++)
    {
        if(st[i]==st[i+1])
        {
            for(j=i+2;j<len;j++)
               {
                   *(st+j-2)=*(st+j);
                   //st[j-2]=st[j];
               }
            len = len-2;
            i=-1;
        }
    }
    return 0;
}

You're not using your compiler warnings. Make sure to exhaust your locally available tools before turning to SO. — Kerrek SB, Feb 26 '13 at 17:31
@Pascal Cuoq it is a duplicate of the above mentioned one ... — karthik, Feb 26 '13 at 17:38

score 4 · Accepted Answer · answered Feb 26 '13 at 17:31

char *st = "fvvbbyyr";

st points to the anonymous string "fvvbbyyr", which can be located in read-only memory.

*(st+j-2) = *(st+j);

Attempting to modify such value leads to an undefined behavior.

A good habit is to declare the pointer as const char * because a string litteral behaves in this way.

const char *st = "fvvbbyyr";

Then your compiler should print some warnings/errors. Use rather an array:

char st[] = "fvvbbyyr";

BTW, the array subscripting operator [] may make your code more readable.

st[j - 2] = st[j];

Moreover you are accessing to st[i+1] == st[len] once in your loop.

thanks for the explanation...now the error is not happening..thanks — karthik, Feb 26 '13 at 17:40

Why segmentation fault occurs in the following

1 Answers1