如何查找变量是在堆栈中还是在堆中分配？

Question

Stumbled upon this interview question somewhere,

In C, Given a variable x , how do you find out if the space for that variable is allocated on the stack or heap?

(Is there any way to find it out programatically and not having to go through the symbol table, etc? And does finding if the space is allocated in stack or heap has any practical implications?)

Answer 1

No, not in general.

Do you know of gcc -fsplit-stack ?

It is up to the implementation to decide whether to allocate a contiguous stack or a stack where blocks are interleaved with heap blocks in memory. Good luck figuring out whether a block was allocated for the heap or the stack when the latter is split.

Answer 2

If you are working on an architecture that stores the stack on a larger address than the heap, you could compare the variable address with the bottom of the stack. Using the pthread threading API, this comparison would look like this:

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <inttypes.h>

int is_stack(void *ptr)
{
  pthread_t self = pthread_self();
  pthread_attr_t attr;
  void *stack;
  size_t stacksize;
  pthread_getattr_np(self, &attr);
  pthread_attr_getstack(&attr, &stack, &stacksize);
  return ((uintptr_t) ptr >= (uintptr_t) stack
          && (uintptr_t) ptr < (uintptr_t) stack + stacksize);
}

The test:

int main()
{
  int x;
  int *p1 = malloc(sizeof(int));
  int *p2 = &x;

  printf("%d %d\n", is_stack(p1), is_stack(p2));
  return 0;
}

...prints 0 1 , as expected.

The above code will not detect storage from stacks in other threads. To do that, the code would need to track all created threads.

Answer 3

This is NOT guaranteed by any standard BUT

on most platforms the stack grows down from highest address available, and heap grows up from the bottom if the most significant byte of the address is in the top half of the available memory space for your platform, and you haven't allocated gigabytes of memory, it's a pretty good bet it's on the stack.

#include <iostream>
#include <stdlib.h>
int main()
{
int x = 0;
int* y = new int;

unsigned int a1 = (int) &x;
unsigned int a2 = (int) y;

std::cout<<std::hex<<a1<<"  "<<a2<<std::endl;
}

gives the output ffbff474 21600 on the machine I'm typing this.

Answer 4

It might be a trick question. Variables have either automatic or static storage duration[*]. You can fairly safely say that automatics are allocated "on the stack", at least assuming they aren't optimized into registers. It's not a requirement of the standard that there be "a stack", but a conforming C implementation must maintain a call stack and associate automatic variables with the levels of the call stack. So whatever the details of what it actually does, you can pretty much call that "the stack".

Variables with static storage duration generally inhabit one or more data sections. From the POV of the OS, data sections might be allocated from the heap before the program starts, but from the POV of the program they have no relation to the "free store".

You can tell the storage duration of a variable by examining its definition in the source -- if it's in function scope then it's automatic unless marked static . If it's not in function scope then it has static duration regardless of whether or not it is marked static (since the static keyword means something different there).

There is no portable way to tell the storage duration of a variable from its address, but particular implementations might provide ways to do it, or tools that you can use with greater or lesser reliability to take a guess.

Objects can also have dynamic storage duration (which generally is what "allocated on the heap" is intended to mean), but such objects are not variables , so that would be the trick if there is one.

[*] Or thread-local in C11 and C++11.

Answer 5

I don't think it has solutions. The code may adjust var's address by stack(heap) address scope, but it's would not be an exact way. At most, the code can only run in some certain platforms.

Answer 6

不，不可能通过内存位置来确定，编译器必须使用 isstack() 支持它才能移植。