此C代码如何转换为MIPS指令？

Question

我正在研究mips编码并为给定的问题提供解决方案，并且我正在努力理解它，并据我所知定义了每行，但是我并没有就它们的主要工作方式选择两行fib（n-1）+ fib（n-2）的最后一行如何返回？ 我看到在beq失败之后，并且bne失败，t1变为0并达到了退出状态，将t1的值存储到v0以进行结果/表达式，并且在从堆栈中删除之前重新加载了（n），但我不太清楚看到fib（n-1）+ fib（n-2）吗？ 救命？ 谢谢！

C code:
int fib(int n){
if (n==0)
return 0;
else if (n == 1)
return 1;
else
fib(n-1) + fib(n–2);

我研究过的新**答案/音译

＃

compare:
addi $sp, $sp, –4 #add immediate adjusts stack for one more item
sw   $ra, 0($sp)  #saves return address on stack of our new item

add $s0, $a0, $0  #add, stores argument 0 + (0) to s0
add $s1, $a1, $0  #add, stores argument 0 + (0) to s1

jal sub           #jump and link to subtract

addi $t1, $0, 1   #add immediate, temp 1 = add 0 + 1
beq $v0, $0, exit #branch on equal, if value in 0 is equal to zero go to -> exit 
slt $t2, $0, $v0  #set less than, if 0 < value at 0  then temp2 equals 1 else 0
bne $t2, $0, exit #branch on not equal, if temp2 not equal to zero go to -> exit
addi $t1, $0, $0  #add immediate, temp1 = 0 + 0

exit:
add $v0, $t1, $0  #add value at 0 = t1 + 0
lw $ra, 0($sp)    #loads register address from stack 0()
addi $sp, $sp, 4  #add immediate, deletes stack pointer pops it off stack
jr $ra            #jump register, return to caller from return address

sub:
sub $v0, $a0, $a1 #subtract, value at 0 = argument 1 - argument 2
jr $ra            #jump register, return to caller from return address

// OLD ***部分正确的答案，但对斐波那契的音译不正确// ---------------------------

 compare: addi $sp, $sp, –4 #add immediate adjusts stack for one more item sw $ra, 0($sp) #saves return address on stack of our new item add $s0, $a0, $0 #add, stores argument 0 + (0) to s0 add $s1, $a1, $0 #add, stores argument 0 + (0) to s1 jal sub #jump and link to subtract addi $t1, $0, 1 #add immediate, temp 1 = add 0 + 1 beq $v0, $0, exit #branch on equal, if value in 0 is equal to zero go to -> exit slt $t2, $0, $v0 #set less than, if 0 < value at 0 then temp2 equals 1 else 0 bne $t2, $0, exit #branch on not equal, if temp2 not equal to zero go to -> exit addi $t1, $0, $0 #add immediate, temp1 = 0 + 0 exit: add $v0, $t1, $0 #add value at 0 = t1 + 0 lw $ra, 0($sp) #loads register address from stack 0() addi $sp, $sp, 4 #add immediate, deletes stack pointer pops it off stack jr $ra #jump register, return to caller from return address sub: sub $v0, $a0, $a1 #subtract, value at 0 = argument 1 - argument 2 jr $ra #jump register, return to caller from return address 

//

Answer 1

两种版本均无法按照建议工作。 我还没有测试下面的代码； 但是这样的事情应该可以工作（不进行错误检查（假定$ a0为非负数；宽松的过程框架约定）：

        .text
fib:  addi $sp, $sp, -24
      sw $ra, 16($sp)
      sw $a0, 20(sp)         # recursive calls will overwrite original $a0
      sw $s0. 0($sp)         # holds fib(n-1)
      # end prologue

      slti $t0, $a0, 4      # fib(i) = i for i = 1, 2, 3; fib(0) = 0 by C code
      beq $t0, $zero, L1
      addi $v0, $a0, 0      # see prior comment (assumes $a0 non-negative integer)
      j exit

      # fib(n) = fib(n-1) + fib(n-2)
L1:   addi $a0, $a0, -1
      jal fib
      addi $s0, $v0, 0       # $s0 = fib(n-1)
      addi $a0, $a0, -1
      jal fib                # upon return, $v0 holds fib(n-2)
      add $v0, $v0, $s0

exit: # unwind stack and return
      lw $s0, 0($sp)
      lw $a0, 20($sp)
      lw $ra, 16($sp)
      addi $sp, $sp, 24
      jr $ra