使用添加和移位的乘法：從Java翻譯到MIPS

Question

我必須在MIPS中編寫一個程序，使用add和shift方法將兩個數字相乘。經過許多次嘗試後，我得到了一個我認爲應該可以工作的程序，但是它沒有，然後我用Java編寫了它，而代碼用Java工作。然後我嘗試將它從Java轉換爲MIPS（通常，它更容易讓我從高級語言的代碼翻譯成低級語言），並且在翻譯它之後，它仍然不起作用。這是我寫的代碼，如果有人發現他們有任何問題或知道如何解決問題，請告訴我。

感謝，

在Java：

static int prod = 0; 

public static int mult (int mcand, int mier) 
{ 
    while (mier != 0) 
    { 
     int rem = mier % 2; 

     if (rem != 0) 
     { 
      prod += mcand; 
     } 

     mcand = mcand << 1; 
     mier = mier >> 1; 
    } 

    return prod; 
} 

在MIPS：

# A program that multiplies two integers using the add and shift method 

.data # Data declaration section 

.text 

main: # Start of code section 

    li $s0, 72 # the multiplicand 
    li $t0, 72 # the multiplicand in a temporary register 
    li $s1, 4 # the multiplier 
    li $t1, 4 # the multiplier in a temporary register 
    li $s2, 0 # the product 
    li $s3, 2 # the number 2 in a register for dividing by 2 always for checking if even or odd 

LOOP: # first we check if the multiplier is zero or not 

    beq $t1, $zero, END 

    div $t1, $s3 
    mfhi $t3 # remainder is now in register $t3 

    beq $t3, $zero, CONTINUE # if the current digit is 0, then no need to add just go the shifting part 

    add $s2, $s2, $t0 # the adding of the multiplicand to the product 

CONTINUE: # to do the shifting after either adding or not the multiplicand to the product 
    sll $t0, $t0, 1 
    srl $t0, $t0, 1 

    j LOOP # to jump back to the start of the loop 

END:  
    add $a0, $a0, $s2 
    li $v0, 1 
    syscall 

# END OF PROGRAM 

---

Answer 1

除了@Joop Eggen的更正之外，您還必須考慮延遲分支是否到位。 如果你正在使用的MIPS有延遲分支，你應該相應地修改你的程序。最簡單的方法是在跳轉/分支之後添加nop指令（在兩個beq之後和j之後）。

除此之外，在代碼結束時，您將結果（$ s2）添加到$ a0，而不是將結果移動到那裏。

因此，要總結：

• 考慮延遲分支，即在BEQ的和j
• 變化srl $t0, $t0，1至srl $t1, $t1, 1
• 變化add $a0, $a0，$ S2添加nop到add $a0, $0, $s2

Answer 2

上SRL複製錯誤在最後：

srl $t1, $t1, 1 

Answer 3

使用添加和移位方法將兩個整數相乘的程序：

.data # Data declaration section 

.text 

main: # Start of code section 

    li $s0, 72 # the multiplicand 
    li $t0, 72 # the multiplicand in a temporary register 
    li $s1, 4 # the multiplier 
    li $t1, 4 # the multiplier in a temporary register 
    li $s2, 0 # the product 
    li $s3, 2 # the number 2 in a register for dividing by 2 always for checking if even or odd 

LOOP: # first we check if the multiplier is zero or not 

    beq nop $t1, $zero, END 

    div $t1, $s3 
    mfhi $t3 # remainder is now in register $t3 

    beq nop $t3, $zero, CONTINUE # if the current digit is 0, then no need to add just go the shifting part 

    add $s2, $s2, $t0 # the adding of the multiplicand to the product 

CONTINUE: # to do the shifting after either adding or not the multiplicand to the product 
    sll $t0, $t0, 1 
    srl $t1, $t1, 1 

    j nop LOOP # to jump back to the start of the loop 

END:  
    add $a0, $0, $s2 
    li $v0, 1 
    syscall