Question

how to free pointer after doing pointer-arithmetic?

Asked by: gcssolve

If you move the char-pointer as done in the code below how would you free the memory originally assigned to the pointer? The call to the free-function makes the program crash.

regards,
Thorbjørn


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


int main(void)
{
   FILE *fhin;
   char *LinePtr;
   char line[1024];

   while (1) {
      if (!(fhin = fopen("/tmp/testfile", "r"))) {
         printf("error");
         exit(0);
      }

      while (fgets(line, sizeof(line), fhin) != NULL) {
         printf("got line %s\n", line);
         LinePtr = (char *) calloc(sizeof(line), sizeof(char));
         strcpy(LinePtr, line);
         *LinePtr++;
         free(LinePtr);

      }
      fclose(fhin);
      sleep(5);
   }
}

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Asked On
2007-01-02 at 08:35:55ID22108335
Tags

free

,

programming

Topic

C Programming Language

Participating Experts
6
Points
125
Comments
11

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Answers

 

by: grg99Posted on 2007-01-02 at 09:14:33ID: 18229059

why do you have a LinePtr++ ??     You don't need it,and it makes it impossible to free the pointer.

btw you probably wanted to malloc the length of the string, not the size of the array.  use strlen()+1 instead of sizeof()

 

by: AxterPosted on 2007-01-02 at 10:10:51ID: 18229509

Hi gcssolve,
> how would you free the memory originally assigned to the pointer?

Keep a copy of the original pointer, so later, you can free it.

David Maisonave (Axter)
Cheers!

 

by: _iskywalker_Posted on 2007-01-03 at 06:24:45ID: 18235182

*LinePtr++;
is hard to use, since (i do not know the right one) it could be:
*(LinePtr++);
which would change the address making LinePtr maybe invalid,
or
(*LinePtr)++;
which would just make the first char of LinePtr increase the value by 1.
The biggest problem is see is that you use sizeof(line).
I would use something like:
#define LINELENGTH 1024
...
char line[LINELENGTH];
...
fgets(line, LINELENGTH, fhin)
...
LinePtr = (char *) calloc(LINELENGTH, sizeof(char));
...

 

by: nripPosted on 2007-01-04 at 01:52:41ID: 18241845

Simply move   *LinePtr++;  to after  the free
 
i.e.
replace

         *LinePtr++;
         free(LinePtr);

with

         free(LinePtr);
         *LinePtr++;

 

by: _iskywalker_Posted on 2007-01-04 at 02:44:05ID: 18242033

nrip... test it before you to say it... LinePtr will be 0x0 after free, so *LinePtr++; will segfault!

 

by: DineshJolaniaPosted on 2007-01-12 at 03:15:13ID: 18300337

I don't see any utility in  "*LinePtr++;" statement.

It's a clear cut example of intentional  memory leak;

 

by: shajithchandranPosted on 2007-01-17 at 11:31:16ID: 18335104

>>nrip... test it before you to say it... LinePtr will be 0x0 after free, so *LinePtr++; will segfault!

LinePtr will not be 0x0 (NULL) because free was called. How will free change the value of LinePtr when it is passed by value.

Going back to the original question: You can't do it. You will have to save a original copy of the pointer.

Or you will have to implement a smarter free() which can find out to which allocated block does the pointer belong too and then free that block!!!

 

by: _iskywalker_Posted on 2007-01-18 at 01:48:53ID: 18339333

shajithchandran  after you use free it must not be 0x0, but it is VERY dangerous to still use it, normally save code would set free(LinePtr); LinePtr= NULL;
since using LinePtr without allocating memory for it could crash the program.

 

by: shajithchandranPosted on 2007-01-18 at 02:41:53ID: 18339527

Yes, you are correct. Its a normal practice to set the pointer to NULL after free. I was just trying to convey that free doesnt set the pointer to NULL (infact it cant).

Thanks
Shaj

20120131-EE-VQP-002

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