程序的虚拟地址空间有多大？

Question

I was reading Operating Systems: Three Easy Pieces . To learn how the virtual address space for a program look like, I run the following code.

#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
    printf("location of code : %p\n", (void *) main);
    printf("location of heap : %p\n", (void *) malloc(1));
    int x = 3;
    printf("location of stack : %p\n", (void *) &x);
    return x;
}

Its output is:

location of code : 0x564eac1266fa
location of heap : 0x564ead8e5670
location of stack : 0x7fffd0e77e54

Why the code segment's location is 0x564eac1266fa ? What does so large a (virtual) space before it use for? Why doesn't it start from or near 0x0 )

And, why the program's virtual address is so large?(from the stack location, it's 48 bits wide) What's the point of it?

Answer 1

The possible virtual address space organizations are defined by the hardware you are using, specifically the MMU it supports. The OS may then use any organization that the hardware can be coerced into using, but generally it just uses it directly (possibly with some subsetting), as that is most efficient.

The x86_64 architecture defines a 48-bit virtual address space ¹ , and most OSes reserve half of that for system use, so user programs see a 47 bit address space. Within that address space, most OSes will randomize the addresses used for any given program, so as to make exploiting bugs in the programs harder.

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¹ _{Strictly speaking, the architecture defines a 64-bit virtual address space, but then reserves all addresses that do not have the top 17 bits all 0 or all 1.}

Answer 2

You are barking up the wrong tree with what you are trying to do here. A process has multiple stacks, may have multiple heaps, and main might not be the start of the code. Viewing an address space as a code segment, stack segment, heap segment, ... as horrible operating systems books do is only going to get you confused.

Because of logical addressing, the memory mapped into the address space does not have to be contiguous.

Why the code segment's location is 0x564eac1266fa? What does so large a (virtual) space before it use for? Why doesn't it start from or near 0x0)

The start of code in your process would well be at 0x564eac1266f8. The fact that you have a high address does not mean the lower addresses have been mapped into the process address space.

And, why the program's virtual address is so large?(from the stack location, it's 48 bits wide) What's the point of it?

Stacks generally start high and grow low.