Thread: unable to free 1 of the struc member

When i try code below i still able to access y.field1 value even after free it. Is it anything wrong with my code ?

Code:

typedef struct
{
    char* field1;
    char* field2;
}lookup;

void main(void)
{
  lookup y;
  y.field1 = (char*)malloc(4);
  strcpy(y.field1,"123");
  clrscr();
  printf("\n%s",y.field1);  //print out 123
  printf("\n%d",y.field1);
  free (y.field1);                //After try free field1 i still able to print out 123 in subsequenceline
  printf("\n%s",y.field1);  //still print out 123
}

Just because you typed "free(y.field1)" doesn't mean that your program will open the computer case and physically remove the bit of memory corresponding to that variable. y.field1 is still a pointer -- it's just pointing to memory you don't own anymore, and if the new tenant hasn't got around to knocking down your variable and putting up one of his own, your "123" will still be there.

free returns the chunk of storage obtained by malloc back to the heap but it does not clear its contents nor does it set y.field1 to NULL.
As an fyi using y.field1 after it has been freed is undefined behavior.

Thanks for all feedback.

free returns the chunk of storage obtained by malloc back to the heap but it does not clear its contents nor does it set y.field1 to NULL.
As an fyi using y.field1 after it has been freed is undefined behavior.

base on statement above am i right to say code below will not work correctly ?

Code:

y.field1 = (char*)malloc(4)
strcpy(y.field1,"123");
free (y.field1);
y.field1 = (char*)malloc(10)
strcpy(y.field1,"123456789");

That code will work perfectly fine, except for the missing semicolon after the malloc statements and the casting of malloc, which you shouldn't do in C.

Originally Posted by kakarato

Thanks for all feedback.



base on statement above am i right to say code below will not work correctly ?

Code:

y.field1 = (char*)malloc(4)
strcpy(y.field1,"123");
free (y.field1);
y.field1 = (char*)malloc(10);
strcpy(y.field1,"123456789");

It will work correctly as long as you terminate the second malloc with a semicolon (in red).
After freeing memory allocated through the first malloc you can malloc a fresh chunk of storage from the heap.