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我正在研究 mips 编码并有一个给定问题的解决方案，我正在努力理解它，并尽我所知定义了每一行，但我并没有了解它们主要如何工作的几行最后一行 fib(n-1) + fib(n-2) 是如何返回的？我看到在 beq 失败并且 bne 失败后 t1 变为 0 并达到退出 t1 的值存储到 v0 以用于结果/表达式，并且我们的 (n) 在从堆栈中删除之前被重新加载但我不完全看到 fib(n-1) + fib(n-2) 了吗？帮助？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);

新**答案/音译我工作

# 
  #fib function
  loop:
  addi $sp, $sp, -4                 #creating item on stack -> int n given caller value
  sw $ra, 0($sp)                       #saving address to stack
  addi $t0, $zero, $zero               #temp 0 is given value of 0
  beq 0($sp), $t0, exit                #if equal return 0  (if (n == 0)
  addi $t1, $zero, 1                   #temp1 gets value of 1
  beq 0($sp), $t1, exit                #if equal return 0 (else if(n==0)
  lw $t2, 0($sp)                       #storing n to temp 2 
  sub $t2, $t2, 1                      #n - 1
  lw $t3, 0($sp)                       #storing n to temp 3
  sub $t3, $t3, 2 #n-2
  add $t4, $t2, $t3                    # (fib(n-1) + fib( n-1) 
  sw 0($sp), $s4                       #storing n's new value back to its original       location
  bne 0($sp), $zero, loop              #jump to loop function with new value of n
  exit: jr $ra                         #return value of register address to caller

//OLD***fibanocci 部分正确但音译不正确//---------------------------

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

//

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1 回答 1

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这两个版本都不会按建议工作。我没有测试下面的代码；但是这样的事情应该可以工作（没有错误检查（假设 $a0 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
于 2013-10-17T20:13:05.057 回答