MIPS提取地址未在字边界上对齐，已使用.align 4，仍然无法执行 [英] MIPS fetch address not aligned on word boundary, used .align 4, still no go
问题描述
有什么想法吗?为什么我得到: 运行时异常位于0x00400020:提取地址未在字边界0x00000007上对齐 问题行是:lw $ s1,0($ a1)#copy arg2 =数组大小
any ideas? Why I am getting: Runtime exception at 0x00400020: fetch address not aligned on word boundary 0x00000007 Problem line is: lw $s1,0($a1) #copy arg2 = size of array
.data
.align 4 #added this, didnt work
size: .word 7
.align 4 #added this, didnt work
search: .word 30
.align 4 #added this,didnt work
array: .word 10,20,30,40,50,60,70
.align 4
.text
main:
la $a0,array #$a0 = address of array
lw $a1,size #a1 = size of array
lw $a2,search #$a2 = search key
COUNT:
lw $s0,0($a0) #copy arg1 = address array
addi $s1,$zero,7
lw $s1,0($a1) #copy arg2 = size of array
lw $s2,0($a2) #copy arg3 = search key (n)
addi $s2,$zero,30
COUNTLOOP:
add $v0,$zero,$zero #v0 = res
add $t0,$zero,$zero #$t0 = init i to 0
slt $t1,$t0,$s1 #check if i > size of array
beq $t1,$zero,DONECOUNT #i is n so end
sll $t2,$s0,2 #$t2 = get off set for a[i]
lw $t3,0($t2) #$t3 = get value of a[i]
bne $t3,$s2,CLOOPBTM #check if a[i] == seach key
addi $v0,$v0,1 #if above then increment res
CLOOPBTM:
addi $t0,$t0,1
j COUNTLOOP
DONECOUNT:
推荐答案
代码的问题是，您使用的不是存储大小的地址，而是大小本身:
The problem with the code is, that you're not using the address where the size is stored but the size itself:
在这里将地址加载到A0中，将大小(7)加载到A1中:
Here you load the address into A0 and the size (7) into A1:
la $a0,array
lw $a1,size #a1 = size of array
在这里加载数组中存储的第一个单词(将加载10).这不是您想要的.
Here you load the first word stored at your array (that will load a 10). This is not what you've intended.
lw $s0,0($a0) #copy arg1 = address array
addi $s1,$zero,7
在这里加载存储在位置0x000007的第一个单词. (您的尺码).这 也可能不是故意的，并且会因为地址未对齐而导致异常:
Here you load the first word stored at location 0x000007. (your size). This is probably also not intended and will cause an exception because the address is not aligned:
lw $s1,0($a1) #copy arg2 = size of array
以此类推.
在我看来，您对LW指令的功能有误解.它将存储单元读取到寄存器中.您希望在循环的序幕中制作一个寄存器的副本.
It seems to me, that you have a misunderstanding what the LW instruction does. It reads a memory location into a register. What you want in the prolog of your loop is to make copies of a register.
为此，如果汇编程序支持，可以使用move伪指令.否则，使用OR指令复制寄存器，如下所示:
To do so you can use the move pseudo instruction if your assembler supports it. Otherwise use the OR instruction to copy registers like this:
COUNT:
or $s0, $a0, $a0 #copy arg1 = address array
addi $s1, $zero,7
or $s1, $a1, $a1 #copy arg2 = size of array
or $s2, $a2, $a2 #copy arg3 = search key (n)
addi $s2, $zero,30
COUNTLOOP:
...
有关线性搜索循环的完整示例，请尝试以下操作(未经测试，并期望汇编程序在乎延迟槽)
for a complete example of a linear search loop try this (untested and expects that the assembler cares about the delay slots)
main:
la $a0,array # $a0 = address of array
lw $a1,size # $a1 = size of array
lw $a2,search # $a2 = search key
beq $a1, $zero, NOTFOUND # handle the size==0 case..
or $v0, $zero, $zero # init counter to zero
LOOP:
lw $s0, 0($a0) # load element
beq $s0, $a2, FOUND # branch if key found:
addiu $a0, $a0, 4 # increment array pointer
addiu $v0, $v0, 1 # increment loop counter
bne $v0, $a1, LOOP # repeat until we've processed the array.
NOTFOUND:
# --------------------------------------
# if you reach this, key does not exist:
# --------------------------------------
li $v0, -1 # load a -1 to signal key not found.
jr $lr # return to caller
FOUND:
# -----------------------------------------
# v0 now contains the position of the key.
# -----------------------------------------
jr $lr
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