Could anyone help me understand what is the gcc
strategy to prepare the stack frame?
For the following function,
void function(int a, int b, int c)
{
char buffer1[5];
char buffer2[10];
int *ret;
ret = &buffer1[0]+28;
printf("0x%x=return address, *ret);
}
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.
function:
pushl %ebp
movl %esp, %ebp
subl $56, %esp // question 1
leal -24(%ebp), %eax
addl $28, %eax
movl %eax, -44(%ebp)
subl $8, %esp // question 2
movl -44(%ebp), %eax
pushl (%eax)
pushl $.LC0
call printf
Here are my 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?
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?
Thanks in advance.
Best regards,
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