2013-09-26 28 views
0

我使用這個lex程序來分析每個寄存器的用法,無論它們是用作源,目的地還是用作間接尋址,這些都是我的代碼,我必須分析所有12個登記冊......是否還有其他有效的方法來做這些反對宣佈36個變量和36個規則? (我是學生)。使用lex程序來分析一個asm文件

%{ 
#include <stdio.h> 

int q,w,e,r,t,y,u,i,o,p,a,s = 0; 

%} 

%% 
[push|dec]?+[ ]+(eax|EAX)+[,] q++ ; 
[,| ]+(eax|EAX)+[ ]? w++ ; 
['[']+(eax|EAX)+[ ]? e++ ; 
[push|dec]?+[ ]+(ebx|EBX)+[,] r++ ; 
[,| ]+(ebx|EBX)+[ ]? t++ ; 
['[']+(ebx|EBX)+[ ]? y++ ; 
[push|dec]?+[ ]+(ecx|ECX)+[,] u++ ; 
[,| ]+(ecx|ECX)+[ ]? i++ ; 
['[']+(ecx|ECX)+[ ]? o++ ; 
[push|dec]?+[ ]+(edx|EDX)+[,] p++ ; 
[,| ]+(edx|EDX)+[ ]? a++ ; 
['[']+(edx|EDX)+[ ]? s++ ; 




. ; 



%% 

main() 

{ 

yylex() ; 

printf("EAX as source: %d\n", q) ; 
printf("EAX as destination: %d\n", w) ; 
printf("EAX as indirect address: %d\n", e) ; 
printf("EBX as source: %d\n", r) ; 
printf("EBX as destination: %d\n", t) ; 
printf("EBX as indirect address: %d\n", y) ; 
printf("ECX as source: %d\n", u) ; 
printf("ECX as destination: %d\n", i) ; 
printf("ECX as indirect address: %d\n", o) ; 
printf("EDX as source: %d\n", p) ; 
printf("EDX as destination: %d\n", a) ; 
printf("EDX as indirect address: %d\n", s) ; 
} 

這是我的輸出:

EAX as source: 56 
EAX as destination: 71 
EAX as indirect address: 0 
EBX as source: 25 
EBX as destination: 38 
EBX as indirect address: 0 
ECX as source: 26 
ECX as destination: 30 
ECX as indirect address: 0 
EDX as source: 30 
EDX as destination: 38 
EDX as indirect address: 0 
[email protected]:~$ ^C 

我試圖分析該文件是在這裏this is where the asm file is

+0

好像您可以將'Register'定義爲'EAX,EBX,ECX'等之一,然後用'Register'替換表達式中的特定寄存器名稱。所以你只有一組表達式。但是,我使用lex的時間太長了,所以我不記得那是如何完成的。 –

回答

1

正如我說我的意見,這是一個很長的時間,因爲我有法合作。不過,這個想法是定義你的寄存器,然後定義一個使用該定義的表達式。

因此,舉例來說,你的lex文件將是這樣的:

% 

R EAX|eax|EBX|ebx|ECX|ecx|EDX|edx 

%% 

[push|dec]?+[ ]+({R})+[,] q++ ; 
[,| ]+({R})+[ ]? w++ ; 
['[']+({R})+[ ]? e++ ; 

%% 

main() 
... 

我的語法可能是不完美的,但這是一般的想法。查找一些簡單的lex定義示例。