Question

void* curbrk;

__asm__ __volatile__(
    "mov .curbrk, %%rax;"
    "mov %%rax, %0"
        : "=r" (curbrk)
        :: "%rax"
);

Can anyone explain what this simple assembly code does? Thanks.

Answer 1

It copies the value of a symbol .curbrk, probably defined somewhere on assembly or a linker script into the C variable curbrk, clobbering the RAX register in the process.

.curbrk probably points to the current end of the data segment. Glibc appears to define a similar symbol __curbrk, you are probably using some other libc (BSD?). In any case, sbrk(0) would be a more portable way of accessing that value.

After looking at the FreeBSD crossreference, I can say it indeed points to the current end of the data segment: it is used both in brk() and sbrk(), using the HIDENAME macro to prepend a ., and it appears on amd64’s System.map (no longer true on current FreeBSD).

Note, however, that in newer FreeBSD, brk() and sbrk() have been reimplemented to no longer depend of .curbrk, which was initialized from _end, which was supposed to come from the executable, but there was trouble when mixing LLVM’s ld and GNU ld. So brk() and sbrk() now use the kernel to initialize its internal curbrk, and no longer depend on _end. See FreeBSD PR228754 for details.