試圖將if語句轉換爲程序集

Question

我在做什麼錯了？

這是我寫的assmebly：

char encode(char plain){ 
__asm{ 
    mov al, plain 

    ;check for y or z status 
    cmp al, 'y' 
    je YorZ 
    cmp al, 'z' 
    je YorZ 
    cmp al, 'Y' 
    je YorZ 
    cmp al, 'Z' 
    je YorZ 

    ;check to make sure it is in the alphabet now 
    mov cl, al 
    sub cl, 'A' 

    cmp cl, 24 
    jl Other 

    sub cl, '6' ;there are six characters between 'Z' and 'a' 

    cmp cl, 24 
    jl Other 
    jmp done ;means it is not in the alphabet 



YorZ: 
    sub al, 24 
    jmp done 

Other: 
    add al, 2 
    jmp done 

done: 
    leave 
    ret 
} 
} 

，這是C代碼它應該更換，但不

char encode(char plain){ 
char code; 
if((plain>='a' && plain<='x') || (plain>='A' && plain <='X')){ 
    code = plain+2; 
}else if(plain == 'y' || plain=='z' || plain=='Y' || plain == 'y'){ 
    code = plain - 24; 
}else{ 
    code = plain; 
} 

return code; 
} 

這似乎是每一個ISN字符轉換將y，z，Y，Z變成加2等價物而不僅僅是A-Xa-x。任何想法爲什麼？

Answer 1

jl是簽名比較。

Answer 2

我認爲

sub cl, '6' 

應該

sub cl, 6 

否則，就表示有 '6' - > 0x36 - > 'Z' 和 'a' 之間的54個字符。

Answer 3

對於無符號比較，您需要'jb'而不是'jl'。