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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.
There are lots of things wrong with your code.
1. main returns int, not void
2. You don't include several header files.
3. You cast the result of malloc - see the FAQ
4. You access memory after you free it.
> i still able to access y.field1 value even after free it. Is it anything wrong with my code ?
That just makes you lucky, that's all.
Tomorrow, with another program, and another compiler, you're asking "what does segfault mean".
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.
Allocating memory is like visiting somebody's home. You have the privilege to be inside. Deallocating it is like being kicked out. Just because you no longer have the privilege to be inside, doesn't make the contents of the home suddenly disappear.
Accessing memory after you've freed it is like trespassing. You might get busted.
Thanks for all feedback.
base on statement above am i right to say code below will not work correctly ?Quote:
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.
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.
It will work correctly as long as you terminate the second malloc with a semicolon (in red).Quote:
Originally Posted by kakarato
After freeing memory allocated through the first malloc you can malloc a fresh chunk of storage from the heap.
What, the first malloc() call doesn't need a semicolon or something? :PQuote:
It will work correctly as long as you terminate the second malloc with a semicolon (in red).