14

On my Debian GNU/Linux 9 system, when a binary is executed,

  • the stack is uninitialized but
  • the heap is zero-initialized.

Why?

I assume that zero-initialization promotes security but, if for the heap, then why not also for the stack? Does the stack, too, not need security?

My question is not specific to Debian as far as I know.

Sample C code:

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

const size_t n = 8;

// --------------------------------------------------------------------
// UNINTERESTING CODE
// --------------------------------------------------------------------
static void print_array(
  const int *const p, const size_t size, const char *const name
)
{
    printf("%s at %p: ", name, p);
    for (size_t i = 0; i < size; ++i) printf("%d ", p[i]);
    printf("\n");
}

// --------------------------------------------------------------------
// INTERESTING CODE
// --------------------------------------------------------------------
int main()
{
    int a[n];
    int *const b = malloc(n*sizeof(int));
    print_array(a, n, "a");
    print_array(b, n, "b");
    free(b);
    return 0;
}

Output:

a at 0x7ffe118997e0: 194 0 294230047 32766 294230046 32766 -550453275 32713 
b at 0x561d4bbfe010: 0 0 0 0 0 0 0 0 

The C standard does not ask malloc() to clear memory before allocating it, of course, but my C program is merely for illustration. The question is not a question about C or about C's standard library. Rather, the question is a question about why the kernel and/or run-time loader are zeroing the heap but not the stack.

ANOTHER EXPERIMENT

My question regards observable GNU/Linux behavior rather than the requirements of standards documents. If unsure what I mean, then try this code, which invokes further undefined behavior (undefined, that is, as far as the C standard is concerned) to illustrate the point:

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

const size_t n = 4;

int main()
{
    for (size_t i = n; i; --i) {
        int *const p = malloc(sizeof(int));
        printf("%p %d ", p, *p);
        ++*p;
        printf("%d\n", *p);
        free(p);
    }
    return 0;
}

Output from my machine:

0x555e86696010 0 1
0x555e86696010 0 1
0x555e86696010 0 1
0x555e86696010 0 1

As far as the C standard is concerned, behavior is undefined, so my question does not regard the C standard. A call to malloc() need not return the same address each time but, since this call to malloc() does indeed happen to return the same address each time, it is interesting to notice that the memory, which is on the heap, is zeroed each time.

The stack, by contrast, had not seemed to be zeroed.

I do not know what the latter code will do on your machine, since I do not know which layer of the GNU/Linux system is causing the observed behavior. You can but try it.

UPDATE

@Kusalananda has observed in comments:

For what it's worth, your most recent code returns different addresses and (occasional) uninitialised (non-zero) data when run on OpenBSD. This obviously does not say anything about the behaviour that you are witnessing on Linux.

That my result differs from the result on OpenBSD is indeed interesting. Apparently, my experiments were discovering not a kernel (or linker) security protocol, as I had thought, but a mere implementational artifact.

In this light, I believe that, together, the answers below of @mosvy, @StephenKitt and @AndreasGrapentin settle my question.

See also on Stack Overflow: Why does malloc initialize the values to 0 in gcc? (credit: @bta).

  • 2
    For what it's worth, your most recent code returns different addresses and (occasional) uninitialised (non-zero) data when run on OpenBSD. This obviously does not say anything about the behaviour that you are witnessing on Linux. – Kusalananda Mar 28 at 21:07
  • Please do not change the scope of your question, and do not try to edit it in order to make answers and comments redundant. In C, the "heap" is nothing else but the memory returned by malloc() and calloc(), and only the latter is zeroing out the memory; the new operator in C++ (also "heap") is on Linux just a wrapper for malloc(); the kernel doesn't know nor care what the "heap" is. – mosvy Mar 28 at 21:16
  • 3
    Your second example is simply exposing an artifact of the malloc implementation in glibc; if you do that repeated malloc/free with a buffer larger than 8 bytes, you will clearly see that only the first 8 bytes are zeroed. – mosvy Mar 28 at 21:16
  • @Kusalananda I see. That my result differs from the result on OpenBSD is indeed interesting. Apparently, you and Mosvy have shown that my experiments were discovering not a kernel (or linker) security protocol, as I had thought, but a mere implementational artifact. – thb Mar 28 at 22:05
  • @thb I believe that this may be a correct observation, yes. – Kusalananda Mar 28 at 22:07
27

The storage returned by malloc() is not zero-initialized. Do not ever assume it is.

In your test program, it's just a fluke: I guess the malloc()just got a fresh block off mmap(), but don't rely on that, either.

For an example, if I run your program on my machine this way:

$ echo 'void __attribute__((constructor)) p(void){
    void *b = malloc(4444); memset(b, 4, 4444); free(b);
}' | cc -include stdlib.h -include string.h -xc - -shared -o pollute.so

$ LD_PRELOAD=./pollute.so ./your_program
a at 0x7ffd40d3aa60: 1256994848 21891 1256994464 21891 1087613792 32765 0 0
b at 0x55834c75d010: 67372036 67372036 67372036 67372036 67372036 67372036 67372036 67372036

Your second example is simply exposing an artifact of the malloc implementation in glibc; if you do that repeated malloc/free with a buffer larger than 8 bytes, you will clearly see that only the first 8 bytes are zeroed, as in the following sample code.

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

const size_t n = 4;
const size_t m = 0x10;

int main()
{
    for (size_t i = n; i; --i) {
        int *const p = malloc(m*sizeof(int));
        printf("%p ", p);
        for (size_t j = 0; j < m; ++j) {
            printf("%d:", p[j]);
            ++p[j];
            printf("%d ", p[j]);
        }
        free(p);
        printf("\n");
    }
    return 0;
}

Output:

0x55be12864010 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 0:1 
0x55be12864010 0:1 0:1 1:2 1:2 1:2 1:2 1:2 1:2 1:2 1:2 1:2 1:2 1:2 1:2 1:2 1:2 
0x55be12864010 0:1 0:1 2:3 2:3 2:3 2:3 2:3 2:3 2:3 2:3 2:3 2:3 2:3 2:3 2:3 2:3 
0x55be12864010 0:1 0:1 3:4 3:4 3:4 3:4 3:4 3:4 3:4 3:4 3:4 3:4 3:4 3:4 3:4 3:4
  • 2
    Well, yes, but this is why I have asked the question here rather than on Stack Overflow. My question was not about the C standard but about the way modern GNU/Linux systems typically link and load binaries. Your LD_PRELOAD is humorous but answers another question than the question I had meant to ask. – thb Mar 28 at 19:56
  • 19
    I'm happy I made you laugh, but your assumptions and prejudices aren't funny at all. On a "modern GNU/Linux system", binaries are typically loaded by a dynamic linker, which is running constructors from dynamic libraries before getting to the main() function from your program. On your very Debian GNU/Linux 9 system, both malloc() and free() will be called more than once before the main() function from your program, even when not using any preloaded libraries. – mosvy Mar 28 at 20:11
23

Regardless of how the stack is initialised, you’re not seeing a pristine stack, because the C library does a number of things before calling main, and they touch the stack.

With the GNU C library, on x86-64, execution starts at the _start entry point, which calls __libc_start_main to set things up, and the latter ends up calling main. But before calling main, it calls a number of other functions, which causes various pieces of data to be written to the stack. The stack’s contents aren’t cleared in between function calls, so when you get into main, your stack contains leftovers from the previous function calls.

This only explains the results you get from the stack, see the other answers regarding your general approach and assumptions.

  • Note that by the time main() is called, initialization routines may very well have modified memory returned by malloc() - especially if C++ libraries are linked in. Assuming the "heap" is initialized to anything is a really, really bad assumption. – Andrew Henle Mar 28 at 20:26
  • Your answer together with the Mosvy's settle my question. The system unfortunately allows me to accept only one of the two; otherwise, I would accept both. – thb Mar 28 at 22:26
17

In both cases, you get uninitialized memory, and you can't make any assumptions about its contents.

When the OS has to apportion a new page to your process (whether that's for its stack or for the arena used by malloc()), it guarantees that it won't expose data from other processes; the usual way to ensure that is to fill it with zeros (but it's equally valid to overwrite with anything else, including even a page worth of /dev/urandom - in fact some debugging malloc() implementations write non-zero patterns, to catch mistaken assumptions such as yours).

If malloc() can satisfy the request from memory already used and released by this process, its contents won't be cleared (in fact, the clearing is nothing to do with malloc() and it can't be - it has to happen before the memory is mapped into your address space). You may get memory that has previously been written by your process/program (e.g. before main()).

In your example program, you're seeing a malloc() region that hasn't yet been written by this process (i.e. it's direct from a new page) and a stack that has been written to (by pre-main() code in your program). If you examine more of the stack, you'll find it's zero-filled further down (in its direction of growth).

If you really want to understand what's happening at the OS level, I recommend that you bypass the C Library layer and interact using system calls such as brk() and mmap() instead.

  • 1
    A week or two ago, I tried a different experiment, calling malloc() and free() repeatedly. Though nothing requires malloc() to reuse the same storage recently freed, in the experiment, malloc() did happen to do that. It happened to return the same address each time, but also nulled the memory each time, which I had not expected. This was interesting to me. Further experiments have led to today's question. – thb Mar 28 at 20:16
  • 1
    @thb, Perhaps I'm not being clear enough - most implementations of malloc() do absolutely nothing with the memory they hand you - it's either previously-used, or freshly-assigned (and therefore zeroed by the OS). In your test, you evidently got the latter. Similarly, the stack memory is given to your process in the cleared state, but you don't examine it far enough to see parts your process hasn't yet touched. Your stack memory is cleared before it's given to your process. – Toby Speight Mar 28 at 21:23
  • 2
    @TobySpeight: brk and sbrk are obsoleted by mmap. pubs.opengroup.org/onlinepubs/7908799/xsh/brk.html says LEGACY right at the top. – Joshua Mar 28 at 21:56
  • 2
    If you need initialized memory using calloc might be an option (instead of memset) – eckes Mar 29 at 0:58
  • 2
    @thb and Toby: fun fact: new pages from the kernel are often lazily allocated, and merely copy-on-write mapped to a shared zeroed page. This happens for mmap(MAP_ANONYMOUS) unless you use MAP_POPULATE as well. New stack pages are hopefully backed by fresh physical pages and wired up (mapped in the hardware page tables, as well as the kernel's pointer/length list of mappings) when growing, because normally new stack memory is being written when touched for the first time. But yes, the kernel must avoid leaking data somehow, and zeroing is the cheapest and most useful. – Peter Cordes Mar 29 at 22:29
8

Your premise is wrong.

What you describe as 'security' is really confidentiality, meaning that no process may read another processes memory, unless this memory is explicitly shared between these processes. In an operating system, this is one aspect of the isolation of concurrent activities, or processes.

What the operating system is doing to ensure this isolation, is whenever memory is requested by the process for heap or stack allocations, this memory is either coming from a region in physical memory that is filled whith zeroes, or that is filled with junk that is coming from the same process.

This ensures that you're only ever seeing zeroes, or your own junk, so confidentiality is ensured, and both heap and stack are 'secure', albeit not necessarily (zero-)initialized.

You're reading too much into your measurements.

  • 1
    The question's Update section now explicitly references your illuminating answer. – thb Mar 29 at 15:46

protected by Kusalananda Mar 29 at 15:35

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