Question

我对我的教授在研讨会上所做的一项练习表示怀疑。它基本上是从C到MIPS的翻译。 C代码如下：

void subr1(char *a, int v[]) {
   int i;
   char c[12];
   for (i=11; i >= 0; i--) {
     v[i] = subr2(c[i], *a);
  }
}

我教授的翻译如下（我写了关于每条指令的作用以及我不理解的内容的评论）：

addiu $sp,$sp,-28 #Activation block
sw $ra, 24($sp) #Space for $ra (4 bytes)
sw $s0, 20($sp) #Space for *a (32-bit address = 4 bytes)
sw $s1, 16($sp) #Space for c (Q1: Shouldn't this be 12 bytes? c is an array of 12 chars = 12*1Byte=12Bytes)
sw $s2 12($sp) #Space for i (4 bytes)
(Q2: Why is he stopping at 12? Doesn't he need to save space for int v[]?)

move $s0,$a0 #Save 'a' in stack
move $s1,$a1 #Save 'c' in stack
li $s2, 11 #i=11

For:
blt $s2, $zero, End #if i<0, then go to 'End'
addu $t0, $s2, $sp (Q3: I'm not sure, but I think he's trying to access v[i] by summing $sp and 'i'. Can you do that, without initialising any register at the beginning like we did for $s0 and $s1?)
lb $a0, 0($t0) #$a0 = v[i] (Q4: Isn't this wrong? It should access c[i], not v[i])
lb $a1, 0($s0) #$a1 = *a
jal sub2 #Jump to sub2
sll $t1, $s2, 2 #$t1 = 4*i
addu $t1,$t1,$sp #$t1 = 4*i + $sp (By this, I guess he's accessing v[i])
sw $v0,0($t1) #v[i] = $v0 (returned value from sub2)
addiu $s2, $s2, -1 #i--
b For

end:
#Close the activation block
lw $s2,12($sp)
lw $s1,16($sp)
lw $s0,20($sp)
lw $ra,24($sp)
addiu $sp,$sp,28
jr $ra #Jump to register address

有人能帮帮我吗？提前谢谢。

使用堆栈将C中的循环与数组转换为MIPS

0 个答案: