Segmentation Fault in Code which replaces string in position

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This is a code i am trying in which string is repalced by its even position character . Let say if string is "abcde" the code should give output "acebd". I haven't implemented the odd position strinng replacement.

What my code do is it stores the even position character continously and when it is done ,it stores "a" to fill the array . I.e output expected by this code is "aceaa" . But it gives segmentation fault at line "str[j]=str[i]" .
Please do help me .

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

void moveplace(char *str)
{
  int len=strlen(str);
  int i=0,j=0;
  while(str[i]!='\0')
    {
      str[j]=str[i];
      i=i+2;
      j++;
    }

  while(j<len)
    {
      str[j]="a";
      j++;    
}

}

int main()
{
  int i;
  char *str="abcde";
  moveplace(str);
  for(i=0;strlen(str);i++)
    {
 printf("%c ",str[i]);
    }
}

Posted 9-Sep-13 19:21pm

Member 10200096

Updated 9-Sep-13 19:27pm

v3

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CPallini · Answer 1 · 2013-09-09T22:07:00

Solution 3

The following code should satisfy your requirements:

C

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

void moveplace(char * str)
{
    int len = strlen(str);
    int lim = len / 2 + len % 2;
    int i;
    for (i=0; i<lim; ++i)
    {
        char k = str[i];
        str[i] = str[i*2];
        str[i*2] = k;
    }
}

int main()
{
  int i;
  char str[]="abcde";
  moveplace(str);
  for(i=0; str[i] != '\0' ;++i)
        printf("%c ",str[i]);
    printf("\n");
}

Posted 9-Sep-13 22:07pm

CPallini

Comments

Member 10200096 10-Sep-13 4:16am

yes it does the same thing as needed :) but you don't have to use extra memory like you did storing characters in char K. And even giving the condition in while loop i

CPallini 10-Sep-13 4:19am

What's the point? You already use 'extra memory' for 'j' variable, don't you?

H.Brydon 10-Sep-13 12:30pm

+5. I didn't try it but it looks good.

CPallini 10-Sep-13 12:46pm

Thank you.

H.Brydon · Answer 2 · 2013-09-09T20:20:00

Solution 2

There may be other problems, but your first loop is defective. You check if the current index location is '\0' (fine) but increment i by 2. So if the NULL is in any odd location (1, 3, 5, 7...), your index will skip over it past the end of the string. The string you pass has 6 characters with the NULL at index 5.

Posted 9-Sep-13 20:20pm

H.Brydon

Comments

nv3 10-Sep-13 3:08am

+5 - still up that late at night?

H.Brydon 10-Sep-13 12:27pm

Sadly, yes. I just can't tear myself away from the keyboard. :-(

CPallini 10-Sep-13 3:51am

5.

nv3 · Answer 3 · 2013-09-09T20:18:00

Solution 1

There are several bugs in your code.

C++

while(str[i]!='\0')
  {
    str[j]=str[i];
    i=i+2;
    j++;
  }

By incrementing i in steps of 2 you will run over the null terminator at the end. And that probably produces you segmentation fault.

C++

while(j<len)>
  {
    str[j]="a";
    j++;
  }

You certainly meant

C++

str[j] = 'a';

[EDIT]
And there is a third one:

C++

for(i=0;strlen(str);i++)

This loop runs forever.

Posted 9-Sep-13 20:18pm

nv3

Updated 9-Sep-13 21:10pm

v2

Comments

Member 10200096 10-Sep-13 3:04am

that can be corrected using while(str[i]!='\0' && i<len)> but it does not work even then

nv3 10-Sep-13 3:11am

See the amendment to my solution.

CPallini 10-Sep-13 3:52am

5.

H.Brydon 10-Sep-13 12:30pm

+5 back... I didn't realize you posted a solution just before mine. :-)