Thread: free on modified pointer

Hi,

here there's an example code that creates a memory leak.
My problem is to call free on modified pointer.

Code:

  int num=10;
  char *test;
  int i=0;

  if ((test = calloc(num, sizeof(char *))) == NULL) {
    puts("allocation error!");
    return 1;
  }
  puts("Mem allocated");
  
  printf("Insert name: ");
  scanf("%s",test);
  printf("your name is %s\n", test);
  //NEED TO PERFORM SOME OPERATIONS
  test=&test[1];   <-------- pointer changed
  printf("%s\n",test);

 //free(test);

  return 0;
}

I mean if I need to perform some operations on 'test' so I change it to
point to the next char there's no way to free up the memory correctly? So Do I always have to use a backup pointer (a temporary pointer) without change the original one?

Thanks in advance

Hi,
thx for reply.

Originally Posted by Dedalus

My problem is to call free on modified pointer.
...
I mean if I need to perform some operations on 'test' so I change it to
point to the next char there's no way to free up the memory correctly? So Do I always have to use a backup pointer (a temporary pointer) without change the original one?

If you want free() to work, yes you do. And you don't need to change pointers to strings... simply use array notation to access the next character as in...

Code:

char *str = malloc(100 * sizeof(char));
char chr;
strcpy (str, "This is a silly old test to learn from");

chr = str[8];

free (str);

There is seldom a real need to disturb the root pointer of an allocated memory block.

Every malloc/calloc needs a free on unchanged initial value (from malloc/calloc).

Code:

  int num=10;
  char *test;
  int i=0;

  if ((test = malloc(num)) == NULL) {
    puts("allocation error!");
    return 1;
  }
  puts("Mem allocated");
  
  printf("Insert name: ");
  scanf("%9s",test);
  printf("your name is %s\n", test);

  printf("%s\n",test+0);
  printf("%s\n",test+1);
  printf("%s\n",test+2);
  printf("%s\n",test+3);
  printf("%s\n",test+8);
  printf("%s\n",test+9);

  free(test);

  return 0;
}

Hi
thx all for reply.
I tried to work out a solution.
If I would remove the first letter of a string I could then copy back the rest of the string without modify the pointer and miss the reference.
Here is the code:

Code:

  int num=10;
  char *test;
  int i=0;

  if ((test = calloc(num, sizeof(char *))) == NULL) {
    puts("allocation error!");
    return 1;
  }
  puts("Mem allocated");
  printf("Insert name: ");
  scanf("%s",test);
  printf("your name is %s\n", test);
  memmove(test,&test[1],strlen(&test[1]));
  memset(&test[strlen(test)-1],'\0', 1);
  printf("%s\n",test);

 free(test);

  return 0;
}

Do you think is clean act like that?
In this way I don't have to use a backup pointer.
Thx in advance

Hi

thx for the advice
I see your point. So the cleanest way seems to be:

Code:

  int num=10;
  char *test, *tmp;
  int i=0;

  if ((test = calloc(num, sizeof(char *))) == NULL) {
    puts("allocation error!");
    return 1;
  }
  tmp=test;
  puts("Mem allocated");
  printf("Insert name: ");
  scanf("%s",test);
  printf("your name is %s\n", test);
  test=&test[1];
  printf("%s\n",test);

 free(tmp);

  return 0;
}