Pfthb

int fun (short a,long b, int c)short q; char s[3]; int y; 

My stack frame looks like this: (I'm not sure)

Two questions:

1. Is it good?


2. In this convention stack is with increasing addresses and heap with decreasing?

There is not a single x86 calling convention and the layout of the local variables is not defined by the calling convention anyway.

The question is thus not framed correctly, even before taking into consideration how the C types are implemented.

If we assume:

• 
  stdcall or cdecl calling convention.


• The compiler is using a frame pointer.


• The compiler will align data at least on their natural boundaries.


• 
  short, int, long are 16, 32, 32 bits respectively and aligned on 2, 4 and 4 bytes respectively.

then I'd expect this stack (note that I use the hex notation):

[EBP +10h] c
[EBP +0Ch] b
[EBP +08h] a
[EBP +04h] Return address
[EBP +00h] Old EBP
[EBP -02h] q
[EBP -05h] s
[EBP -0ch] y

That's what Clang does without optimisations except that it preserves 3 registers thereby shifting the offsets down by 12.

a, b and c are pushed on the stack in reverse order, all types must occupy a number of bytes multiple of 4.

Locals allocation is not absolute, it depends on the surrounding code (usage of non volatile registers, calls, temporaries, stack protections and so on) and in this case, where there is no code or context at all, the question is a bit vacuous.

Using only the last two assumptions¹ we can put q at [EBP -02h] since q is 2 bytes and need to be at an address multiple of 2; s is at [EBP -05h] since it is 3 bytes and is byte-aligned; finally y could go at [EBP -09h] from a size-only perspective but since it must be aligned on 4-bytes, it goes at [EBP -0ch].

Note that all the reasoning above exploit the fact that EBP is aligned on 4-bytes.

¹ We also assume that locals are laid out in reverse order: the last one in order of declaration is laid out at a lower address.