Question

我正在研究mips编码并对某个特定问题有一个解决方案，我正在努力理解它并且已经根据我的知识定义了每一行，但是我并没有就如何解释他们主要工作的最后一行是fib（n-1）+ fib（n-2）如何返回？我看到在beq失败之后，并且bne失败t1变为0并且达到退出t1的值被存储到v0用于结果/表达式并且我们的（n）在被从堆栈中删除之前被重新加载但是我不是很看到fib（n-1）+ fib（n-2）？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);

新**答案/音译我的工作

＃

  #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 ***部分正确答案但不正确的文件音译 // ---------------------------

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

//

Answer 1

这两个版本都不会按照建议运行。我没有测试下面的代码;但这样的事情应该有效（没有错误检查（假设$ a0非负）;宽松的程序框架约定）：

        .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

这个C代码如何转换为MIPS指令？

1 个答案: