Question

所以是的另一个家庭作业问题。但是我真的以为我这次钉了它..无论如何我得到了4个错误，对我来说似乎没有任何意义O_O或者也许这只是我的盲目眼光或我对代码的可怜借口但是是的。我收到了这些错误

mp5.s:70: error: invalid combination of opcode and operands
mp5.s:77: error: invalid combination of opcode and operands
mp5.s:83: error: invalid combination of opcode and operands
mp5.s:90: error: invalid combination of opcode and operands

此次赋值是使汇编代码进行矩阵乘法。我将粘贴我的代码并突出显示第70,77,83和90行

我可能会粘贴一大堆代码，但我需要知道的是为什么我会收到这些错误。它应该是同一类型，我不知道它为什么不起作用......

extern  printf

segment .data

test1   db  'M','i','s','m','a','t','c','h',0
test2   db  'h','e','l','l','o','!',0
sout    db  "%s", 10, 0

segment .text

    global  matMul


matMul: 
    enter   12,0

    mov eax,    0
    mov [ebp-04], eax   ;i=0
    mov [ebp-08], eax   ;j=0
    mov [ebp-12], eax   ;k=0

    mov eax,    [ebp+24]
    mov ebx,    [ebp+28]
    cmp eax,    ebx
    je  start

    ;; When The Matrices can't b multiplied It prints out "Mismatch" then exits
    push    test1
    push    sout
    call    printf
    jmp done

;;; start with I loop
start:  
iloop:
    mov eax,    [ebp-04]
    mov ebx,    [ebp+20]
    cmp eax,    ebx ;while i < xrow keep looping
    je  done

    inc eax
    mov [ebp-04], eax   ;increments i++
jloop:  
    ;; j loop starts here
    mov eax,    [ebp-8]
    mov ebx,    [ebp+32]
    cmp eax,    ebx ;while j < ycol keep looping
    je  resetJ

    inc eax
    mov eax,    [ebp-08] ;increments j++
kloop:  
    ;; k loop starts here
    mov eax,    [ebp-12]
    mov ebx,    [ebp+24]
    cmp eax,    ebx ;while k < xcol keep loopinbg
    je  resetK

    ;; operation starts
    mov esi,    [ebp+8] ;address of the matrix 1
    mov edi,    [ebp+12] ;address of matrix 2

    mov eax,    [ebp-04]
    mov ebx,    [ebp-12]
    mov ecx,    [ebp+24]
    **mul   ecx,    eax ;xcol * i**
    add ecx,    ebx ;xcol * i + k
    add esi,    ecx ;esi + shift    now esi have the location for x[col*i+k]

    mov eax,    [ebp-12]
    mov ebx,    [ebp-08]
    mov ecx,    [ebp+32]
    **mul   ecx,    eax ;ycol * k**
    add ecx,    ebx ;ycol * k + j
    add edi,    ecx ;edi + shift    now esi have the location for y[col*k + j]

    mov edx,    [esi]   ;edx <- have the value we need
    mov eax,    [edi]   ;eax <- have the 2nd value we need
    **mul   edx,    eax ;x[xcol * i + k] * y[ycol * 1 + k] = edx**

    mov esi,    [ebp+16] ;esi now have the first location of the 3rd matrix

    mov eax,    [ebp-04]
    mov ebx,    [ebp-08]
    mov ecx,    [ebp+28]
    **mul   ecx,    eax ;yrow * i**
    add ecx,    ebx ;yrow * i + j
    add esi,    ecx ;esi + shift    now esi have the location for z[yrow * i + j]
    mov eax,    [esi]   ;eax <- [esi]

    add eax,    edx ;z[yrow * i + j] + x[xcol * i + k] * y[ycol * k + j] = eax;
    mov [esi],  edx
    ;; operation end

    mov eax,    [ebp-12]
    inc eax
    mov eax,    [ebp-12]    ;increment k++


    jmp kloop       ;k loop
    jmp jloop       ;j loop 
    jmp iloop       ;i loop back to start


resetJ:
    mov eax,    0
    mov [ebp-08], eax
    jmp iloop

resetK:
    mov eax,    0
    mov [ebp-12], eax
    jmp jloop

done:   
    leave
    ret



testing:
    push    test2
    push    sout
    call    printf
    jmp done

第70,77,83和90行是里面** **我不知道为什么，但我试图让它们粗体字体所以它是显而易见的:(我不知道它为什么不无论如何，这整个代码基本上都是我的C矩阵乘法程序的重写

我将在此处粘贴

void mm(int *x, int *y, int *z, int xrow, int xcol, int yrow, int ycol){
  if(xcol == yrow){
    int i,j,k;

    for(i = 0; i < xrow; i++){
      for(j = 0; j < ycol; j++){
    for(k = 0; k < xcol; k++){
      z[yrow * i + j] = z[yrow * i + j] + x[xcol * i + k] * y[ycol * k + j];
    }
      }
    }

  }

  else{
    printf("Size Mismatch. Can't Multiply.\n");
  }
}

Answer 1

MUL，无符号乘法，具有以下形式（英特尔语法）：

MUL r/m8
    AX ← AL∗ r/m8
MUL r/m16 
    DX:AX ← AX ∗ r/m16
MUL r/m32
    EDX:EAX ← EAX ∗ r/m32
MUL r/m64
    RDX:RAX ← RAX ∗ r/m64

Answer 2

我编码80x86汇编程序已经有一段时间了，但是MUL操作码不只占用一个操作数，而隐式使用累加器（EAX）作为术语之一，而EAX + EDX存储结果？（当然，当我在汇编时进行编码时，它只是普通的老式AX和DX：对我们来说，没有这些奇特的32或64位的东西！）

我为什么会遇到这些错误？（汇编x86）矩阵乘法

2 个答案:

我为什么会遇到这些错误？ （汇编x86）矩阵乘法

2 个答案:

我为什么会遇到这些错误？（汇编x86）矩阵乘法