我正在尝试使用MARS解决一个小任务问题，但我不断收到此错误。有什么我做错了吗？
我在linenum 11收到错误：这是lw $ t0,0
以下是我的程序代码。
.data
SOURCE: .word 0x0100
DEST: .word 0x0110
.text
lw $t0, 0
lw $t4 , -1
lw $t5 , 0
lw $t6, 20
lw $t7, 32
VERY_START:
beq $t6,$zero,EXIT
addi $t6,$t6,-1
lw $t7,32
la $t1,SOURCE
li $t2,1
li $s1,2
START:
and $t3, $t2 , $t1
beq $t4,-1,FIRST_LOOP
bne $t4,$t3,STORE
#bne is branch if not equal to.
add $t5 , $t5 , 1
addi $t7, $t7 , -1
beq $t7, $zero, VERY_START
# So we jump to the very start if we have 32 bits done.
sll $t2, $t2 , 1
j START
STORE:
sb $t5,DEST($t2)
#dest needs to be defined (It is implicit according to the question)
# after storing , we need to increment $t0 so that we can store the next element a byte away from this one. so
add $t0 , $t0 , 2
lw $t5,1
addi $t7, $t7 , -1
beq $t7, $zero, VERY_START
# So we jump to the very start if we have 32 bits done.
sll $t2, $t2 , 1
j START
FIRST_LOOP:
# we populate t4 here
move $t4,$t3
j START
EXIT:
#we find the number of counts by simply dividing the current t0 with 2
div $t0, $s1
mflo $t0
# we move the quotient to t0..
move $a0,$t0
li $v0, 1
syscall
答案 0 :(得分：1)
根据this reference，您尝试使用lw将文字加载到注册表中，这是不允许的：
RAM访问仅允许加载和存储指令
您有两种选择：
选项1
您可以从临时存储
加载到$t0
var1: .word 0 # declare storage for var1; initial value is 0
.text
lw $t0, var1 # load contents of RAM location into register $t0: $t0 = var1
...
选项2
您可以使用文字值
立即加载li
li $t0, 0 # $t0 = 0 ("load immediate")
您的代码中的多个位置都存在此问题。