mips地址超出范围

Question

您好，我是MIPS汇编语言的新手，我试图编写相当于B[8]=A[i-j]的变量f，g，h，i和j分配给寄存器$ s0，$ s1，$ s2， $ 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，这将引发一个运行时异常在0x0040000c：地址超出范围00000000，有什么建议？

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. 

如果您希望您的代码以符合你的描述，你必须：

# 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]