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。我需要答案,我想要的是直接把代码告诉我
// Read numbers into an array until a negative number is entered
// then print the numbers in reverse order
#include <stdio.h>
int numbers[1000];
int main(void) {
int i = 0;
while (i < 1000) {
int x;
scanf("%d", &x);
if (x < 0) {
break;
} else {
numbers[i] = x;
}
i++;
}
while (i > 0) {
i--;
printf("%d\n", numbers[i]);
}
}
# Read numbers into an array until a negative number is entered
# then print the numbers in reverse order
# i in register $t0
# registers $t1, $t2 & $t3 used to hold temporary results
main:
li $t0, 0 # i = 0
loop0:
bge $t0, 1000, end0 # while (i < 1000) {
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:
li $v0, 0 # return 0
jr $ra #
.data
numbers:
.space 4000
代码示例如下
# Read numbers into an array until a negative number is entered
# then print the numbers in reverse order
# i in register $t0
# registers $t1, $t2 & $t3 used to hold temporary results
main:
li $t0, 0 # i = 0
loop0:
bge $t0, 1000, output # while (i < 1000) {
li $v0, 5 # scanf("%d", &numbers[i]);
syscall #
bltz $v0,output
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 # }
output:
beqz $t0,end0
subi $t0,$t0,1
mul $t1, $t0, 4 # calculate &numbers[i]
la $t2, numbers #
add $t3, $t1, $t2 #
lw $a0, ($t3) # store entered number in array
li $v0,1
syscall
li $a0, '\n' # printf("%c", '\n');
li $v0, 11
syscall
j output
end0:
li $v0, 0 # return 0
jr $ra #
.data
numbers:
.space 4000
请随意使用任何一种gcc-misp交叉编译工具链生成.s汇编源码,源码一楼正解。