函数调用示例

例题 求如下递归程序的MIPS汇编版本：

```

int fact (int n)

{

    if (n < 1) return (1);

          else return (n * fact(n – 1));

}

```

答：

```

fact:

addi  $sp, $sp, –8 # adjust stack for 2 items

sw    $ra, 4($sp)  # save the return address

sw    $a0, 0($sp)  # save the argument n

slti  $t0,$a0,1    # test for n < 1

  beq  $t0,$zero,L1  # if n >= 1, go to L1

addi  $v0,$zero,1 # return 1

  addi  $sp,$sp,8  # pop 2 items off stack

  jr    $ra        # return to caller

  lw $a0, 0($sp) # return from jal: restore argument n lw $ra, 4($sp) # restore the return address

addi $sp, $sp, 8 # adjust stack pointer to pop 2 items

mul $v0,$a0,$v0 # return n * fact (n – 1)

jr  $ra          # return to the caller

L1:

addi $a0,$a0,–1  # n >= 1: argument gets (n – 1)

  jal fact        # call fact with (n –1)

```