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
// swap any pair of numbers which are out of order
// then print the array
#include <stdio.h>
int main(void) {
int i;
int numbers[10] = { 0 };
i = 0;
while (i < 10) {
scanf("%d", &numbers[i]);
i++;
}
i = 1;
while (i < 10) {
int x = numbers[i];
int y = numbers[i - 1];
if (x < y) {
numbers[i] = y;
numbers[i - 1] = x;
}
i++;
}
i = 0;
while (i < 10) {
printf("%d\n", numbers[i]);
i++;
}
}
# read 10 numbers into an array
# swap any pairs of of number which are out of order
# then print the 10 numbers
# i in register $t0,
# registers $t1 - $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 $t0, $t0, 1 # i++;
j loop0 # }
end0:
# PUT YOUR CODE HERE
li $t0, 0 # i = 0
loop1:
bge $t0, 10, end1 # 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
li $v0, 1 # printf("%d", numbers[i])
syscall
li $a0, '\n' # printf("%c", '\n');
li $v0, 11
syscall
addi $t0, $t0, 1 # i++
j loop1 # }
end1:
li $v0, 0
jr $ra # return 0
.data
numbers:
.word 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 # int numbers[10] = {0};