I compiled with gcc -O0 -S option and the compiler produces the code that I cannot quite follow the stack frame layout strategy in gcc.
First of all, I think you'll find in more complicated functions that the stack layout for -O3 is dramatically different. This is partly because many variables do not need to end up on the stack, and the effect of other optimizations.
Hopefully this is just for educational purposes, and you're not planning on writing code that relies on the layout of the stack frame! I can guarantee you that this will change. If you just want to find the return address, there are easier ways...
Anyway, on to your questions:
question1: Why the stack frame size is 56? observation: (1) compiler add 16 bytes padding before allocating storage for array buffer1 (2) buffer1 need 5 bytes. However, due to alignment issue, they seem to add 3 extra bytes. Thus, -24(%ebp) should point to buffer1[0].
Then, why they adding 16 bytes padding?
I can't say for sure, but remember that the stack frame has more than just variables on it. There are also saved registers, such as the caller's frame pointer, and a number of other things. Also, many machines work best if they 32 bit alignment for memory loads.
question2: Why gcc makes the stack frame bigger before
the function call printf?
subl $8, %esp
Does it related to printf? If it does, then could you explain why?
This is to make room for the parameters to printf(); there are two, and each is a pointer, and hence four bytes, so the total space required is eight bytes.
Niko
