尝试分配变量时发生MIPS程序集分析错误

Question

我的MIPS程序集代码有问题。它显示了li$t4$zero行（实际代码的第4行）中的语法错误。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

附言：我知道代码可能还有一些问题，我只是开始调试。。。

Answer 1

li的目的是L加载I即时消息。$zero是一个寄存器，而不是立即数。你要找的可能是move $t4,$zero。

您的代码中还有其他类似的问题。例如addiu $sp, $sp, $t4（这似乎是不必要的，因为你应该在$t4中放零）。

Answer 2

$zero是一个寄存器，不能将其与li（即load immediate）一起使用。使用li $t4, 0或move $t4, $zero伪指令。