Question

我有一个C代码，我在MIPS中需要帮助。 C代码如下。

for (i=0; i<100; i++){
   a[i] = b[i] + c[i] - d[i];
}

我已将此转换为MIPS，但不知道将什么放入加载字的偏移量。

addi $t0, $zero, 0                #i = 0

for_loop:
   bgt $t0, 100, for_loop_done     #i <100
   addi $t0, $t0, 1               #i++ or i = i+1


   lw $t4, __($s0)              # load a in t4
   lw $t1, __($s1)              # load b in t1
   lw $t2, __($s2)              # load c in t2
   add $t4, $t2, $t1                 # add b with c and store in a
   lw $t3, __($s3)              # load d in t3
   sub $t4, $t3, $t4                 # sub contents of a from d
   sw $t4, __($s0)              # store contents of t4 into a

   j for_loop                     # go to start of loop

for_loop_done:

我们假设a，b，c，d分别在s0，s1，s2，s3，s4中。现在代码需要的是我们如何通过c代码中不断变化的“i”来偏移加载字和存储字。因为据我所知，加载词只使用静态值。

Answer 1

数组在内存中以最低的基址排列，并以4字节的偏移量索引到下一个元素。[图片来自Harris D. M.，Harris S. L. - 数字设计与计算机体系结构，第2版 - 2012]

 for_loop:
         bgt $t0, 100, for_loop_done     #i <100
         addi $t0, $t0, 1               #i++ or i = i+1


         lw $t4, 0($s0)              # load a in t4
         lw $t1, 4($s1)              # load b in t1
         lw $t2, 8($s2)              # load c in t2
         add $t4, $t2, $t1            # add b with c and store in a
         lw $t3, 12($s3)              # load d in t3
         sub $t4, $t3, $t4            # sub contents of a from d
         sw $t4, 0($s0)              # store contents of t4 into a

         j for_loop                   # go to start of loop

for_loop_done:

C到MIPS的转换。数组加载字偏移量？

1 个答案: