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我的 MIPS 汇编代码有问题。它在 li $t4, $zero 行(实际代码的第 4 行)中显示语法错误。你能帮我解决吗?

    .data

    .globl funcall1
    .globl funcall2


    .text

funcall1:   
        lw $t1, 0($a0) # load the 1st argument - shows how many there are
        addiu $t2, $t1, -1 #t2 stores how many args
        addiu $t3, $t2, -4 #t3 stores how many args there will be apart from 4 main
        li $t4, $zero
        addiu $sp, $sp, -24 # make stack frame
        bgezal $t3, SETSTACK 
        addiu $sp, $sp, $t4 # make stack frame
        sw $ra, 16($sp) # save where to return

        # 3:0xAAAA:0:12:0:7 format of FET


        lw $v0, 4($a0) # load the op-comand or value into the return value
        lw $s0, $a0 # save FET for future reference
        jal EXPRESSION
        bgezal $t3, ADDMOREARGS

        lw $a0, $s1 # return the first value to a0

        jal $t0

        lw $ra, 16($sp) # restore $ra
        addiu $sp, $sp, 24  # restore $sp

        jr $ra # return

funcall2:                   # main function in file
                        # Arguments $a0 is address of input FET

        jr $ra           # return

SETSTACK:   
        addiu $sp, $sp, -24 # make stack 
        sw $ra, 16($sp) # save where to return
        mul $t4, $t3, 4 # this is how many extra bytes wil be needed
        div $t4, 8
        bne $hi, 0, ADDFOUR # if size not divisible by 8, add 4.
        mul $t4, $t4, -1

        lw $ra, 16($sp) # restore $ra
        addiu $sp,$sp,24  # restore $sp

        jr $ra # return

ADDFOUR:
        addiu $sp, $sp, -24 # make stack 
        sw $ra, 16($sp) # save where to return
        addiu $t4, $t4, 4 # t4 += 4
        lw $ra, 16($sp) # restore $ra
        addiu $sp,$sp,24  # restore $sp

        jr $ra # return 

EXPRESSION: 
        addiu $sp, $sp, -24 # make stack 
        sw $ra, 16($sp) # save where to return

        lw $t0,  4($s0) # save the operand
        lw $s1, 12($s0) # free a0 for addmoreargs, save the first value in s1 as well
        lw $a0, 12($s0) # save first value in a0
        lw $a1, 20($s0) # second value
        lw $a2, 28($s0) # third value
        lw $a3, 32($s0) # fourth value

        la $s9, 40($s0) # save the rest of address
        lw $s8, 40($s0) # save the rest of data
        la $s7, $a3     # save the next place for an argument
        addiu $s7, $s7, -8

        lw $ra, 16($sp) # restore $ra
        addiu $sp,$sp,24  # restore $sp

        jr $ra # return


ADDMOREARGS:

        sw $s7, 4($s8)
        lw $s8, 8($s8) # jump thorugh the data
        addiu $t3, $t3, -1
        bgezal $t3, ADDMOREARGS

PS我知道代码可能还有更多问题,我只是开始调试......

4

2 回答 2

1

$zero是一个寄存器,你不能使用liwhich is load immediate。使用li $t4, 0move $t4, $zero伪指令。

于 2014-04-22T12:45:11.393 回答
1

的目的li加载I调解$zero是寄存器,而不是立即数。您正在寻找的可能是move $t4,$zero.

您的代码中还有其他类似的问题。例如addiu $sp, $sp, $t4(这似乎是不必要的,因为你应该在 中输入零$t4)。

于 2014-04-22T12:45:38.827 回答