What happens to the memory which is freed after being allocated by malloc()?

Question

What really happens to the memory that is allocated using malloc() after being freed? Suppose I do the following...

int main(){
    int * arr;
    arr=(int*) malloc(sizeof(int)*20);
    int i;
    for(i=0;i<20;i++) arr[i]=2*i+1;
    int * tmp=arr;
    for(i=0;i<20;i++) printf("%d ",*(tmp+i));
    printf("\n");
    free(arr);
    for(i=0;i<20;i++) printf("%d ",*(tmp+i));
    return 0;
}

I get the output...

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 
0 0 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39

Why do the first two entries change(and the others don't)?

once memory's freed, it can be re-used. just because your own code doesn't explicitly change that memory, doesn't mean that something else deeper down inside C can't do its own malloc and get some/all of your freed block. — Marc B, Apr 22 '16 at 15:05
If you have the time, please read this http://stackoverflow.com/a/6445794/434551. Understanding pointers will be a lot easier after that. — R Sahu, Apr 22 '16 at 15:11
The freed block probably gets loaded with heap-management metadata by the free() call.. — Martin James, Apr 22 '16 at 15:14
"Why do the first two entries change" --> Aside from curiosity, why are you seeking to know what happens? Why is it important if the data stays the same or changes? — chux - Reinstate Monica, Apr 22 '16 at 15:16
Note that "Undefined Behaviour" doesn't mean "random". It means "that depends on some code we **shouldn't care of**". Yes, there could be a reason behind that issue, but it's not reasonable to think of it. — Matt, Apr 22 '16 at 15:21

Sourav Ghosh · Accepted Answer · 2016-04-22T15:07:49.750

2

Why do the first two entries change(and the others don't)?

TL;DR undefined behavior.

Once you've called free() on a pointer previously returned by malloc(), that pointer is not valid anymore in your program context. Attempt to make use of it invokes undefined behavior.

Coming to the point of what happens to the actual memory, well, that is also environment dependent. Calling free() is just a way to inform the lower layer (OS / memory manager) that it is OK to reclaim and reuse the memory if need be. There is nothing mandated that the memory location has to be cleaned (zeroed) or alike.

edited Apr 22 '16 at 15:07

answered Apr 22 '16 at 15:04

Sourav Ghosh

133,132
16
183
261

So there is a chance that the memory will remain unchanged? – samurdhilbk Apr 22 '16 at 15:07
@samurdhilbk I've updated my answer, please have a look. If you have any doubt, let me know. – Sourav Ghosh Apr 22 '16 at 15:10
Yes, the memory is quite likely to be unchanged--why would the system waste time doing anything with it at all? The only thing important is that it *doesn't belong to you* anymore, so you can't use it. – Lee Daniel Crocker Apr 22 '16 at 17:02

score 0 · Answer 2 · answered Apr 22 '16 at 15:09

The malloc, free, and realloc functions manage an area of memory traditionally called the "heap". malloc and realloc pick an area of the heap, mark it as being "in use", and return a pointer to the memory to you. free returns memory to the heap, for use by future malloc and free calls.

When free returns memory to the heap, and marks it as being no longer in use by your program, it may be that one of the ways it marks it involves setting bits in the memory itself. In your case, that's why some of your array values changed.

(But you can't depend on this, of course. Other implementations might have left the now-freed memory completely unchanged. Others might have erased it entirely. And it's also possible that your program could have crashed when trying to print out the memory after freeing it, since of course you're not supposed to do that.)

What happens to the memory which is freed after being allocated by malloc()?

2 Answers2