Understanding 16-byte padding and function prolog in x64 assembly

Question

In x64 assembly, the stack frame, according to Microsoft, should be 16-byte aligned

The stack will always be maintained 16-byte aligned, except within the prolog (for example, after the return address is pushed), and except where indicated in Function Types for a certain class of frame functions.

Assume we have the following function:

void foo() {
    long long int foo;
    long long int bar;
}

The stack would look something like this:

|-----------|
|   rbp     |    
|-----------|
|   foo     |    
|-----------|
|   bar     |    
|-----------|
| undefined |    
|-----------|

So, the stack would need to allocate 20h bytes. And the assembly instruction would look like:

push rbp
mov rbp, rsp
sub 20h        ; 32 bytes needed to fulfill alignment requirements

Is my understanding of this correct, or am I way off? I'm also assuming no optimizations.

Answer 1

I have been trying to figure this out myself and I believe I understand what is needed here.

Assuming the CALLER of foo() ensured 16-byte alignment before the call then:

• The CALL statement (by the caller of your function foo() ) puts 8-bytes on the stack i.e. the return address. [Stack not 16 Byte Aligned]
• Then your function pushed RBP on the stack (8-bytes). [Stack is 16 Byte Aligned]
• You make space for the FOO & BAR local variables adding 2x 8-bytes. [Stack is still 16-Byte Aligned]

In this scenario I see no need for extra padding, however, you cannot guarantee that the CALLER of your function ensured 16-byte alignment before the call (unless you are working in an environment where it is guaranteed) so you need to make sure by adding the following line of code:

and   rsp, -16

so your final code should be:

push rbp
mov  rbp, rsp
sub  rsp, 10h     ; make space for locals foo and bar
and  rsp, -16     ; force 16-byte alignment by zeroing last four bits

Now you are sure that the state of the stack is 16-byte aligned before you make any call to a function that may require it.

Note that if you know that the functions you call from your foo() procedure do not need 16-byte alignment then there should be no need to go thru' these hoops.