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下面两个代码示例中 s 和 sf 之间的实际区别是什么？

我知道堆栈相关看起来像 Mem [SP + OprndSpec] 和延迟看起来像 Mem [Mem [SP + OprndSpec]]。但是我不明白这是如何完成的。

堆栈延迟

         BR      main
a:       .BLOCK  2          ;global variable #2d
b:       .BLOCK  2          ;global variable #2d
;
;******* void swap (int& r, int& s)
r:       .EQUATE 6          ;formal parameter #2h
s:       .EQUATE 4          ;formal parameter #2h
temp:    .EQUATE 0          ;local variable #2d
swap:    SUBSP   2,i        ;allocate #temp
         LDA     r,sf       ;temp = r
         STA     temp,s
         LDA     s,sf       ;r = s
         STA     r,sf
         LDA     temp,s     ;s = temp
         STA     s,sf
         RET2               ;deallocate #temp, pop retAddr
;
;******* void order (int& x, int& y)
x:       .EQUATE 4          ;formal parameter #2h
y:       .EQUATE 2          ;formal parameter #2h
order:   LDA     x,sf       ;if (x > y)
         CPA     y,sf
         BRLE    endIf
         LDA     x,s        ;   push x
         STA     -2,s
         LDA     y,s        ;   push y
         STA     -4,s
         SUBSP   4,i        ;   push #r #s
         CALL    swap       ;   swap (x, y)
         ADDSP   4,i        ;   pop #s #r
endIf:   RET0               ;pop retAddr

;
;******* main ()
main:    STRO    msg1,d     ;cout << "Enter an integer: "
         DECI    a,d        ;cin >> a
         STRO    msg1,d     ;cout << "Enter an integer: "
         DECI    b,d        ;cin >> b
         LDA     a,i        ;push the address of a
         STA     -2,s
         LDA     b,i        ;push the address of b
         STA     -4,s
         SUBSP   4,i        ;push #x #y
         CALL    order      ;order (a, b)
ra1:     ADDSP   4,i        ;pop #y #x
         STRO    msg2,d     ;cout << "Ordered they are: "
         DECO    a,d        ;     << a
         STRO    msg3,d     ;     << ", "
         DECO    b,d        ;     << b
         CHARO   '\n',i     ;     << endl
         STOP
msg1:    .ASCII  "Enter an integer: \x00"
msg2:    .ASCII  "Ordered they are: \x00"
msg3:    .ASCII  ", \x00"
         .END

堆栈相对

         BR      main        
;
;******* int binomCoeff (int n, int k)
retVal:  .EQUATE 10          ;returned value #2d
n:       .EQUATE 8           ;formal parameter #2d
k:       .EQUATE 6           ;formal parameter #2d
y1:      .EQUATE 2           ;local variable #2d
y2:      .EQUATE 0           ;local variable #2d
binCoeff:SUBSP   4,i         ;allocate #y1 #y2
if:      LDA     k,s         ;if ((k == 0)
         BREQ    then        
         LDA     n,s         ;|| (n == k))
         CPA     k,s         
         BRNE    else        
then:    LDA     1,i         ;return 1
         STA     retVal,s    
         RET4                ;deallocate #y2 #y1, pop retAddr
else:    LDA     n,s         ;push n - 1
         SUBA    1,i         
         STA     -4,s        
         LDA     k,s         ;push k
         STA     -6,s        
         SUBSP   6,i         ;push #retVal #n #k
         CALL    binCoeff    
ra2:     ADDSP   6,i         ;pop #k #n #retVal
         LDA     -2,s        ;y1 = binomCoeff (n - 1, k)
         STA     y1,s        
         LDA     n,s         ;push n - 1
         SUBA    1,i         
         STA     -4,s        
         LDA     k,s         ;push k - 1
         SUBA    1,i         
         STA     -6,s        
         SUBSP   6,i         ;push #retVal #n #k
         CALL    binCoeff    
ra3:     ADDSP   6,i         ;pop #k #n #retVal
         LDA     -2,s        ;y2 = binomCoeff (n - 1, k - 1)
         STA     y2,s        
         LDA     y1,s        ;return y1 + y2
         ADDA    y2,s        
         STA     retVal,s    
endIf:   RET4                ;deallocate #y2 #y1, pop retAddr
;
;******* main ()
main:    STRO    msg,d       ;cout << "binCoeff (3, 1) = "
         LDA     3,i         ;push 3
         STA     -4,s        
         LDA     1,i         ;push 1
         STA     -6,s        
         SUBSP   6,i         ;push #retVal #n #k
         CALL    binCoeff    
ra1:     ADDSP   6,i         ;pop #k #n #retVal
         DECO    -2,s        ;<< binCoeff (3, 1)
         CHARO   '\n',i      ;cout << endl
         STOP                
msg:     .ASCII  "binCoeff (3, 1) = \x00"
         .END
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1 回答 1

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您的问题似乎很简单，除了它们可能包含示例之外，与代码几乎没有关系。

（我不熟悉这个指令集，但在汇编中已经编码了很长时间。）

根据您提供的定义，“堆栈相对”是指在堆栈指针确定的位置从内存中获取项目，加上可能嵌入在指令中的常量偏移量。这被大多数人称为索引寻址，特别注意它是堆栈指针索引的。

“延迟”（一个旧术语）通常意味着“间接通过内存位置”，您对此的定义与此想法一致：找到“堆栈相对”位置，读取该位置，然后将该值用作要获取的内存位置。

于 2012-11-27T05:42:32.563 回答