简单的装配问题 [英] Simple assembly questions

问题描述

; int __stdcall wWinMain(HINSTANCE hInstance, HINSTANCE hPrevInstance, LPWSTR lpCmdLine, int nShowCmd)
_wWinMain@16 proc near

var_4= dword ptr -4
hInstance= dword ptr  4
hPrevInstance= dword ptr  8
lpCmdLine= dword ptr  0Ch
nShowCmd= dword ptr  10h

$ b b从我可以看到，最后4个变量是传递给WinMain函数的参数。另外，var_4必须是稍后在函数体中声明的int变量。现在，我有几个问题：

From what I can see, the last 4 variables are the parameters passed to the WinMain function. Also, the var_4 must be a int variable I declared later in the function body. Now, I have a couple of questions:

a）32位Windows程序上的字的大小是多少？ 4字节？ a dword 8？

a) What is the size of a word on a 32bit windows program? 4 bytes? being a dword 8?

b）为什么var_4设置为-4？为什么不从0开始？

b) Why is var_4 set to -4? Why not start at, let's say, 0?

c）在c中定义int的标准过程是

c) The standard procedure for defining ints in c is though

var_x = dword ptr y？

var_x = dword ptr y?

谢谢。

推荐答案

。 DWORD未签署，32位：请参阅此处 （旧名称，MS在Windows为16位时开始使用它。）

a. DWORD is unsigned, 32 bit: see here (old name, MS started using it back when Windows was 16-bit).

b。堆栈顶部（dword ptr 0）由返回地址

b. the top of stack (dword ptr 0) is taken up by the return-address

c占用。一个变量 y 这里不工作。无论如何， dword ptr 的系统使用是某些汇编程序和反汇编程序的特性，而不是标准过程。

c. a variable y here would not work. Anyway, this systematic use of dword ptr is characteristic of certain assemblers and disassemblers, not a "standard procedure".

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