在表达式中减去NASM宏的意外结果

Question

我编写了以下代码：

 section .text
    %define len msg-4
    global _start 
    msg: db "Thank you"
    var: dd 0x31323334

_start:
    mov ecx, msg
    debug:
    mov edx, var-len ; **** the problem is here
    mov ebx, 1
    mov eax, 4
    int 80h
    mov eax, 1 
    mov ebx, 1
    int 80h ; exit         

我期望edx保存值13，因为var-len= var-msg+4= 13 ( var地址与msg的距离为9，因为msg为9字节)。因此，我想这段代码会打印“谢谢”。

但是，edx得到了5，“谢谢”被打印出来了。

为什么edx得到5而不是13

Answer 1

%define len msg-4是一个文本替代，见下文。

通常的方法是用$ - start计算对象后面的长度。简单地给结束贴上标签也会奏效。

section .rodata           ; groups read-only together inside the .text section
    msg: db "Thank you"
    var: db 0x34, 0x33, 0x32, 0x31   ; dd 0x31323334  ; Since you're passing these bytes to write(2), writing them separately is probably less confusing.  (x86 is little-endian, so they will come out "backwards")

    ;; apparently you want to include var as part of the message, for some reason
    msglen equ $ - msg    ; $ is the current position

    ;; msgend:            ; alternative: label the end.  There doesn't have to be a db or anything; it's fine to have multiple labels for the same address


section .text
    global _start 
_start:
    mov    edx, msglen        ; message length
    ;; mov edx, msgend - msg  ; alternative

为什么你有5

%define 是一个文本替代，类似于C预处理器。如果你使用(msg-4)的话，它会像你预期的那样工作。在CPP宏中大量使用()的所有原因也适用于这里。

 %define len msg-4        ; first of all, this is a terrible name: it's not the length!
 mov    edx, var - len    ; expands to var-msg-4,  not var - (msg-4)

 %define msg_minus_varlen  (msg-4)
 mov    edx, var - msg_minus_varlen;  expands to var - (msg-4)

这仍然是一种令人费解的方法。从实际打印的缓冲区中减去两个位置，再减去相同的值，就可以得到长度。

Answer 2

看看最上面的台词-你说

 section .text
    %define len msg-4
    global _start 
    msg: db "Thank you"
    var: dd 0x31323334

MSG-4，而不是msg + 4，这就是你所看到的不同之处。