mips地址超出范围

Question

你好我是MIPS汇编语言的新手，我试着编写等价的B[8]=A[i-j]，其中变量f，g，h，i和j分配给寄存器$ s0，$ s1，$ s2，$分别为s3和$ s4。我假设数组A和B的基地址分别在寄存器$ s6和$ s7中。

我的代码

# Gets A[i-j]
sub $t1, $s3, $s4 
sll $t1, $t1, 2
add $t1, $s6, $t1

lw $t0, 0($t1)

# Set B[8] equal to above
addi $t2, $s0, 8
sll $t2, $t2, 2
add $t2, $s7, $t2

lw $t2, 0($t2)
sw $t2, 0($t0)

但是这会在0x0040000c处抛出运行时异常：地址超出范围0x00000000，有什么建议吗？

Answer 1

# Gets A[i-j]
sub $t1, $s3, $s4 
sll $t1, $t1, 2
add $t1, $s6, $t1

lw $t0, 0($t1)

# Set B[8] equal to above
addi $t2, $s0, 8              #this actually computes something like &B[f+8], not &B[8]
sll $t2, $t2, 2
add $t2, $s7, $t2

lw $t2, 0($t2)                #this loads the value B[f+8]
sw $t2, 0($t0)                #this stores B[f+8] to *(int *)A[i-j], which is probably not an address.

如果您希望代码符合您的描述，则必须：

# Gets A[i-j]
sub $t1, $s3, $s4 
sll $t1, $t1, 2
add $t1, $s6, $t1

lw $t0, 0($t1)

# Set B[8] equal to above
addi $t2, $zero, 8
sll $t2, $t2, 2
add $t2, $s7, $t2

##deleted##lw $t2, 0($t2) no need to load B[8] if you just want to write to it
sw $t0, 0($t2) #this stores A[i-j] to &B[8]