While inspecting disassembled code on my Macbook Pro, I couldn’t figure out why the stack allocated so much space when I only requested a little. After some investigation into GCC I figured it out.
According to Apple’s developer documentation on GCC, “If
-mpreferred-stack-boundary
is not specified, the default is 4 (16 bytes or 128 bits).” That means that the stack will always allocate chunks in factors of 16 (2^4) unless you specify otherwise.
That means that the following code
int main(int argc, char **argv) {
char buf[1];
return 0;
}
Translates to allocating 16 bytes on the stack even though it’s only 1 byte.
... 0x0000000100000ef2 : sub rsp,0x10 # 16 bytes ...
Update
This is not specific to Mac, all 64 bit platforms use this stack alignment. Here is the Windows documentation and here is the Linux documentation. Thanks Pascal!
I was doing the same just to see and I got this
0x7fff71022d18 : “”
0x7fff71022d19 : “”
0x7fff71022d1a : “”
0x7fff71022d1b : “”
0x7fff71022d1c : “”
0x7fff71022d1d : “”
0x7fff71022d1e : “”
0x7fff71022d1f : “”
which are only 8 bytes. I just compiled with the option
gcc -g3 test.c
did you made something special ?
Hi Phil, are you running in 64 bit mode?
yes, i am, I checked after to be sure that GDB was well configured as well.
(gdb) show architecture
The target architecture is set automatically (currently i386:x86-64)
Hmm, interesting. What do your stack allocation instructions look like? e.g. sub rsp …
Otherwise I read mostly the entire blog,
but you should sometimes be a bit more precise, there are some stuff that you come up with .. without explaining anything … which is hard for the reader to understand .. but nice work.
Great, thanks for the feedback. Any specifics that I can explain more?