Why does calling sbrk(0) twice give a different value?

问题

I'm trying to understand the sbrk() function.

From what I know:
sbrk(0) returns the current address of the break and doesn't increment it.
sbrk(size) increments the address of the break by size bytes and returns the previous address of the break.

So I created something to test it:

#include <unistd.h>
#include <stdio.h>

int main(void)
{
    printf("sbrk(0) = %p\n", sbrk(0)); // should return value x
    printf("sbrk(0) = %p\n", sbrk(0)); // should return value x
    printf("sbrk(5) = %p\n", sbrk(5)); // should return value x
    printf("sbrk(0) = %p\n", sbrk(0)); // should return value x + 5
}

So I'm expecting to see a result looking like this:

sbrk(0) = 0x1677000 // x value
sbrk(0) = 0x1677000 // x value
sbrk(5) = 0x1677000 // x value
sbrk(0) = 0x1677005 // x value + 5

but instead I'm getting this:

sbrk(0) = 0x1677000 // x value
sbrk(0) = 0x1698000 // y value
sbrk(5) = 0x1698000 // y value
sbrk(0) = 0x1698005 // y value + 5

Why don't the first two calls of sbrk(0) return the same value? What happens between those two calls that changes the break address?

EDIT : Storing addresses in variables solves the problem:

int main(void)
{
    void *toto1 = sbrk(0);
    void *toto2 = sbrk(0);
    void *toto3 = sbrk(5);
    void *toto4 = sbrk(0);

    printf("sbrk(0) = %p\n", toto1);
    printf("sbrk(0) = %p\n", toto2);
    printf("sbrk(5) = %p\n", toto3);
    printf("sbrk(0) = %p\n", toto4);
}

回答1:

Your program performs the following sequence of calls:

sbrk()
printf()
sbrk()
printf()
...

The first call to printf calls malloc internally to allocate a buffer for stdout (stdout is line buffered by default, but the buffer is created on demand the first time you print to it).

That's why the second call to sbrk returns a different value.

(This answer is not directly related, but the error messages from valgrind expose the existence of the underlying malloc call hidden inside printf.)

Your second example performs all sbrk calls up front, so there are no surprises from other functions calling malloc behind your back.

来源：https://stackoverflow.com/questions/54294724/why-does-calling-sbrk0-twice-give-a-different-value

标签

memory

sbrk

brk