C语言问题，编写一个将C语言 变成 MIPS 汇编程序

望各位帮忙看看，现在就想要答案，蟹蟹各位潜水专家，感谢解答，感谢回答，感谢解答。编写一个 MIPS 汇编程序，相当于这个 C 程序：在原有的基础上完成这个程序。在# PUT YOUR CODE HERE下面写内容.我需要答案，不要告诉我需要自己下什么东西然后就可以得到，我想要的是直接把代码告诉我

// Read 10 numbers into an array
// print 0 if they are in non-decreasing order
// print 1 otherwise
#include <stdio.h>
int main(void) {
    int i;
    int numbers[10] = { 0 };
    i = 0;
    while (i < 10) {
        scanf("%d", &numbers[i]);
        i++;
    }
    int swapped = 0;
    i = 1;
    while (i < 10) {
        int x = numbers[i];
        int y = numbers[i - 1];
        if (x < y) {
            swapped = 1;
        }
        i++;
    }
    printf("%d\n", swapped);
}
 

# Read 10 numbers into an array
# print 0 if they are in non-decreasing order
# print 1 otherwise
# i in register $t0
main:
    li   $t0, 0         # i = 0
loop0:
    bge  $t0, 10, end0  # while (i < 10) {
    li   $v0, 5         #   scanf("%d", &numbers[i]);
    syscall             #
    mul  $t1, $t0, 4    #   calculate &numbers[i]
    la   $t2, numbers   #
    add  $t3, $t1, $t2  #
    sw   $v0, ($t3)     #   store entered number in array
    addi $t0, $t0, 1    #   i++;
    j    loop0          # }
end0:
 
    # PUT YOUR CODE HERE
 
    li   $a0, 42        # printf("%d", 42)
    li   $v0, 1         #
    syscall
    li   $a0, '\n'      # printf("%c", '\n');
    li   $v0, 11
    syscall
    jr   $ra
.data
numbers:
    .word 0, 0, 0, 0, 0, 0, 0, 0, 0, 0  # int numbers[10] = {0};
 

代码如下
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# Read 10 numbers into an array
# then print the numbers which are
# larger than the last number read.
# i in register $t0
# registers $t1, $t2 & $t3 used to hold temporary results
main:
    li   $t0, 0         # i = 0
loop0:
    bge  $t0, 10, end0  # while (i < 10) {
    li   $v0, 5         #   scanf("%d", &numbers[i]);
    syscall             #
    mul  $t1, $t0, 4    #   calculate &numbers[i]
    la   $t2, numbers   #
    add  $t3, $t1, $t2  #
    sw   $v0, ($t3)     #   store entered number in array
    #addi $t4,$v0,0      # store last number
    addi $t0, $t0, 1    #   i++;
    j    loop0          # }
end0:
     #load number[0]
    li $t5,0#flag = 0
    la   $t2, numbers   #
    lw   $a1, ($t2) 
    li   $t0, 1         # i = 1
    j loop1
change:
    li $t5,1
    j out
loop1:
    bge  $t0, 10, out  # while (i < 10) {
    mul  $t1, $t0, 4    #   calculate &numbers[i]
    la   $t2, numbers   #
    add  $t3, $t1, $t2  #
    lw   $a0, ($t3)     #   load numbers[i] into $a0
    bgt $a1,$a0,change
    addi $a1,$a0,0
    addi $t0, $t0, 1    #   i++
    #blt $a0,$t4,loop1
    #li   $v0, 1         #   printf("%d", numbers[i])
     j    loop1          # }
 out: 
    addi $a0,$t5,0    #load flag
    li   $v0, 1
    syscall
    li $a0,'\n'
    li   $v0, 11
    syscall
    
   
end1:
    jr   $ra              # return
.data
numbers:
    .word 0, 0, 0, 0, 0, 0, 0, 0, 0, 0  # int numbers[10] = {0};