使用add和shift的乘法：從Java轉換為MIPS

Question

我必須在MIPS中編寫一個使用add和shift方法將兩個數相乘的程序。 在嘗試了很多次之后，我得到了一個我認為應該工作的程序，但事實並非如此，然后我用Java編寫它，代碼在Java中工作。 然后我嘗試將它從Java轉換為MIPS（通常我更容易從高級語言的代碼轉換為低級語言），並且在翻譯它之后，仍然無法正常工作。 這是我寫的代碼，如果有人發現它們有任何問題或者知道如何修復它們，請告訴我。

謝謝，

在Java中：

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中：

# 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

最后在srl上復制錯誤：

srl $t1, $t1, 1

Answer 2

除了@Joop Eggen的更正之外，您還必須考慮延遲分支是否到位。 如果您使用的MIPS具有延遲分支，則應相應地修改您的程序。 最簡單的方法是在跳轉/分支之后添加一條nop指令（在兩個beq之后和j之后）。

除此之外，在代碼的末尾，您將結果（$ s2）添加到$ a0而不是將其移動到那里。

所以，總結一下：

• 考慮延遲分支，即在beq和j之后添加nop
• 將srl $t0, $t0更改為srl $t1, $t1, 1
• 更改add $a0, $a0 ，$ s2以add $a0, $0, $s2

Answer 3

一個使用add和shift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