Question

如何将以下代码翻译成MIPS汇编语言？

# include < stdio .h >
int fibRecursive ( int n )
{
   int answer ;
   if ( n < 2 ) {
   answer = n ;
   }
   else {
     answer = fibRecursive ( n - 1 ) + fibRecursive ( n - 2 ) ;
   }
   return answer ;
}

int main ( int argc , char * argv [] )
{
   int input = 10;
   int result = fibRecursive ( input ) ;
   printf ( " The %d - th Fibonacci number is % d .\ n " , input , result ) ;
   return 0;
}

感谢您的帮助。我遇到了问题：

answer = fibRecursive ( n - 1 ) + fibRecursive ( n - 2 ) ;

这是我到目前为止的MIPS：

fibRecursive:
addi $sp, $sp, -4
sw $ra, 0($sp)
addi $s0, $zero, 0
slti $t0, $a0, 2
beq $t0, $zero, ELSE
add $s0, $a0, $zero
j EXIT

ELSE:
addi $sp, $sp, -4
sw $a0, 0($sp)
addi $a0, $a0, -1
jal fibRecursive

EXIT:
add $v0, $zero, $s0

Answer 1

我会运行一个编译器（如GCC），其标志设置为发出汇编代码。（例如，对于GCC，您可以使用-S选项）

Answer 2



// Swapped to MIPs syntax :)
main:
  jal function
  ...

function:
  // if code here
  // return value here
  jal function
  // gather answer into reg.
  jal function
  // add away
  // return value here
  ...

在那里，您现在看到如何在main中调用两次递归函数。清理最终案例留给用户练习。

Answer 3

编译为MIPS我得到了这个 .file 1“c2mips.c”

 # -G value = 8, Cpu = 3000, ISA = 1
 # GNU C version cygnus-2.7.2-970404 (mips-mips-ecoff) compiled by GNU C version cygnus-2.7.2-970404.
 # options passed:  -msoft-float
 # options enabled:  -fpeephole -ffunction-cse -fkeep-static-consts
 # -fpcc-struct-return -fcommon -fverbose-asm -fgnu-linker -msoft-float
 # -meb -mcpu=3000

gcc2_compiled.:
__gnu_compiled_c:
    .text
    .align  2
    .globl  fibRecursive
    .ent    fibRecursive
fibRecursive:
    .frame  $fp,40,$31      # vars= 8, regs= 3/0, args= 16, extra= 0
    .mask   0xc0010000,-8
    .fmask  0x00000000,0
    subu    $sp,$sp,40
    sw  $31,32($sp)
    sw  $fp,28($sp)
    sw  $16,24($sp)
    move    $fp,$sp
    sw  $4,40($fp)
    lw  $2,40($fp)
    slt $3,$2,2
    beq $3,$0,$L2
    lw  $2,40($fp)
    sw  $2,16($fp)
    j   $L3
$L2:
    lw  $3,40($fp)
    addu    $2,$3,-1
    move    $4,$2
    jal fibRecursive
    move    $16,$2
    lw  $3,40($fp)
    addu    $2,$3,-2
    move    $4,$2
    jal fibRecursive
    addu    $3,$16,$2
    sw  $3,16($fp)
$L3:
    lw  $3,16($fp)
    move    $2,$3
    j   $L1
$L1:
    move    $sp,$fp         # sp not trusted here
    lw  $31,32($sp)
    lw  $fp,28($sp)
    lw  $16,24($sp)
    addu    $sp,$sp,40
    j   $31
    .end    fibRecursive
    .rdata
    .align  2
$LC0:
    .ascii  " The %d - th Fibonacci number is % d . n \000"
    .text
    .align  2
    .globl  main
    .ent    main
main:
    .frame  $fp,32,$31      # vars= 8, regs= 2/0, args= 16, extra= 0
    .mask   0xc0000000,-4
    .fmask  0x00000000,0
    subu    $sp,$sp,32
    sw  $31,28($sp)
    sw  $fp,24($sp)
    move    $fp,$sp
    sw  $4,32($fp)
    sw  $5,36($fp)
    jal __main
    li  $2,10           # 0x0000000a
    sw  $2,16($fp)
    lw  $4,16($fp)
    jal fibRecursive
    sw  $2,20($fp)
    la  $4,$LC0
    lw  $5,16($fp)
    lw  $6,20($fp)
    jal printf
    move    $2,$0
    j   $L4
$L4:
    move    $sp,$fp         # sp not trusted here
    lw  $31,28($sp)
    lw  $fp,24($sp)
    addu    $sp,$sp,32
    j   $31
    .end    main

C代码转换为MIPS

3 个答案: