#include <iostream>using namespace std;void fibonacci(int n, int &f0, int &f1) { if (n == 0) { f0 = 0; f1 = 1; } else { fibonacci(n - 1, f1, f0); f1 += f0; }}int main() { int n; cout << "Please input the number of terms of the Fibonacci sequence: "; cin >> n; cout << "The first " << n << " terms of the Fibonacci sequence are: "; int f0 = 0, f1 = 1; for (int i = 0; i < n; i++) { cout << f0 << " "; fibonacci(i, f0, f1); } cout << endl; return 0;} 翻译成mips汇编程序

下面是翻译成MIPS汇编程序的代码：

.data
output: .asciiz "The first %d terms of the Fibonacci sequence are: \n"
        .align 2
f0:     .word   0
f1:     .word   1
        .text
        .globl main
main:
        li $v0, 4             # print the output message
        la $a0, output
        syscall
        li $v0, 5             # read in the value of n
        syscall
        move $s0, $v0         # save the value of n in register s0
        la $s1, f0            # load the address of f0 into register s1
        la $s2, f1            # load the address of f1 into register s2
        addi $s3, $zero, 0    # initialize the counter to 0
        j fib                 # jump to the fib subroutine
fib:
        beq $s3, $s0, done    # if the counter reaches n, exit the subroutine
        lw $t0, 0($s1)        # load the value of f0 into register t0
        lw $t1, 0($s2)        # load the value of f1 into register t1
        addi $sp, $sp, -8     # allocate space on the stack for the two arguments
        sw $s2, 4($sp)        # store the address of f1 on the stack
        sw $s1, 0($sp)        # store the address of f0 on the stack
        addi $s0, $s0, -1     # decrement the value of n
        jal fib               # recursive call to fib(n-1)
        addi $s0, $s0, 1      # increment the value of n
        lw $s1, 0($sp)        # restore the address of f0 from the stack
        lw $s2, 4($sp)        # restore the address of f1 from the stack
        addi $sp, $sp, 8      # deallocate space on the stack
        add $t2, $t0, $t1     # calculate the next term in the sequence
        sw $t1, 0($s1)        # store the value of f1 in f0
        sw $t2, 0($s2)        # store the value of f2 in f1
        li $v0, 1             # print the current term
        move $a0, $t0
        syscall
        li $v0, 4             # print a space
        la $a0, " "
        syscall
        addi $s3, $s3, 1      # increment the counter
        j fib                 # jump to the beginning of the subroutine
done:
        li $v0, 10            # exit the program
        syscall

这个程序使用了递归的方式来计算斐波那契数列的每一项，并将它们打印出来。在每一次递归调用中，程序使用栈来保存f0和f1两个变量的地址，并将n的值减1传递给递归函数。在递归函数返回时，程序恢复f0和f1的地址，并将它们的值更新为下一个斐波那契数。最后，程序打印出计算出的斐波那契数列的每一项。