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
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。 –