Hello,
Consider the following program:
void toto(int *tab);
void foo1(void)
{
int tab[N];
toto(tab);
}
I checked how much space gcc allocates on the stack for various values of N:
for ((N=1; N<=12; ++N)); do
echo -n N = $N
gcc-7 -O3 -march=haswell -Wall -S -fno-stack-protector -D N=$N alloc.c
grep 'subq' alloc.s
done
N = 1 subq $24, %rsp
N = 2 subq $24, %rsp
N = 3 subq $24, %rsp
N = 4 subq $24, %rsp
N = 5 subq $40, %rsp
N = 6 subq $40, %rsp
N = 7 subq $40, %rsp
N = 8 subq $40, %rsp
N = 9 subq $56, %rsp
N = 10 subq $56, %rsp
N = 11 subq $56, %rsp
N = 12 subq $56, %rsp
The call instruction will push the return address (8 bytes)
on the stack, therefore gcc is allocating
{1,2,3,4} = 24 + 8 = 32
{5,6,7,8} = 40 + 8 = 48
{9,10,11,12} = 56 + 8 = 64
16-byte alignment, because SIMD, right?
However, it seems gcc could allocate less in some cases:
{1,2} = 8 + 8 = 16
{3,4,5,6} = 24 + 8 = 32
{7,8,9,10} = 40 + 8 = 48
What am I missing?
For example, why is gcc allocating 40 bytes for N=5 and N=6?
Hmmm... reading up on https://wiki.osdev.org/System_V_ABI I see that
"The stack is 16-byte aligned just before the call instruction is called."
IIUC, it's the callee's responsibility to align the stack.
I suppose that doesn't change the reasoning above, though.
Regards.
