c++ - 在 MIPS 上调用两次的子程序

Question

我在将此 C++ 程序转换为 MIPS 程序集时遇到问题:

教授给出的原始 C++ 代码示例

float vect1[64], vect2[64];
int num1,num2;
float average;
int numag;
void main() {
    int i, count;
    float sum;
    float fnum;
    read_vect_float(vect1,num1);
    read_vect_float(vect2,num2);
    sum=0.0;
    for(i=0;i<num1;i++)
        sum=sum+vect1[i];
     fnum=(float)num;
     sum=sum/fnum;
    average=sum;

    cout<< "The average is: " << average;
    cout << "\n";

    count=0;
    for(i=0;i<num2;i++)
       if(vect2[i]>average)
          count=count+1;
    numag=count;
    cout << "In vector 2 there are ";
    cout << numag;
    cout << " elements major of ";
    cout << average;
    cout << "\n";
}

void read_vect_float(float v[],int &n) {
    int i;
    float x;

    cout << "Number of elements: ";
    cin >> n;
    for(i=0;i<n;i++) {
        cout=<< "Element: " ;
        cin >> x;
        vect[i]=x;
    }
}

C++ 代码由我审核，您可以运行

#include <iostream>

using namespace std;

void read_vect_float(float v[],int &n);

float vect[64],vect1[64], vect2[64];
int num,num1,num2;
float average;
int numag;

int main() {
    int i, count;
    float sum;
    float fnum;
    read_vect_float(vect1,num1);
    read_vect_float(vect2,num2);
    sum=0.0;
    for(i=0;i<num1;i++)
        sum=sum+vect1[i];
     fnum=(float)num;
     sum=sum/fnum;
    average=sum;

    cout<< "The average is: " << average;
    cout << "\n";

    count=0;
    for(i=0;i<num2;i++)
       if(vect2[i]>average)
          count=count+1;
    numag=count;
    cout << "In vector 2 there are ";
    cout << numag;
    cout << " elements major of ";
    cout << average;
    cout << "\n";
    return 0;
}

void read_vect_float(float v[],int &n) {
    int i;
    float x;

    cout << "Number of elements: ";
    cin >> n;
    for(i=0;i<n;i++) {
        cout << "Element: " ;
        cin >> x;
        vect[i]=x;
    }
}

程序的目的是读取两个 float 数组的输入元素，计算存储项的平均值，并找出第二个数组中有多少元素大于第一个数组。

这是我的 MIPS 汇编代码:

.data
vect: .float
.space 256
vect1: .float
.space 256
vect2: .float
.space 256
num1: .word
.space 4
num2: .word
.space 4
average: .float
.space 4
numag: .word
.space 4
i: .word
.space 4
count: .word
.space 4
sum: .float
.space 4
fnum: .float
.space 4
str1: .asciiz "The average is: "
.align 2
str2: .asciiz "\n"
.align 2
str3: .asciiz "In vector 2 there are "
.align 2
str4: .asciiz " element major of "
.align 2
x: .float
.space 4
str5: .asciiz "Number of elements: "
.align 2
n: .word
.space 4
str6: .asciiz "Element: "
.align 2
.text
.globl main
main:
# read_vect_float(vect1,num1);
lwc1 $f12,vect1
lw $a0,num1
jal read_vect_float
# read_vect_float(vect2,num2);
lwc1 $f13,vect2
lw $a1,num2
jal read_vect_float
# sum=0.0;
mtc1 $zero,$f0
cvt.s.w $f0,$f0
# for(i=0;i<num1;i++)
li $t0,0
for3:
# sum=sum+vect1[i];
sll $t0,$t0,2
lwc1 $f1,vect1($t0)
add.s $f2,$f2,$f1

addi $t0,$t0,1
blt $t0,$t1,for3
# fnum=(float)num;
mtc1 $t1,$f3
cvt.s.w $f3,$f3
# sum=sum/fnum;
div.s $f2,$f2,$f3
# average=sum;
swc1 $f2,average
# cout<< "The average is: " << average;
li $v0,4
la $a0,str1
syscall

li $v0,2
lwc1 $f12,average
syscall
# cout << "\n";
li $v0,4
la $a0,str2
syscall
# count=0;
li $t2,0
sw $t2,count
# for(i=0;i<num2;i++)
li $t0,0
for2:
# if(vect2[i]>average)
#ble $f3,$f2,next
c.le.s $f2,$f3
bc1t next
# count=count+1;
addi $t2,$t2,1
sll $t0,$t0,2
sw $t2,numag
addi $t1,$t1,1
blt $t0,$t1,for2
next:
# cout << "In vector 2 there are ";
li $v0,4
la $a0,str3
syscall
# cout << numag;
li $v0,1
lw $a0,numag
syscall
# cout << " elements major of ";
li $v0,4
la $a0,str4
syscall
# cout << average;
li $v0,2
lwc1 $f12,average
syscall
# cout << "\n";
li $v0,4
la $a0,str2
syscall

li $v0,10
syscall

read_vect_float:
# cout << "Number of elements: ";
li $v0,4
la $a0,str5
syscall
# cin >> n;
li $v0,5
syscall
sw $v0,n
lw $t5,n
# for(i=0;i<n;i++)
li $t0,0
for:
# cout=<< "Element: " ;
li $v0,4
la $a0,str6
syscall
# cin >> x;
li $v0,6
syscall
swc1 $f0,x
# vect[i]=x;
sll $t0,$t0,2
swc1 $f0,vect($t0)

addi $t0,$t0,1
blt $t0,$t5,for

jr $ra

但是当我在 MARS MIPS 模拟器上运行代码时，结果并不像预期的那样，我希望得到您的一些建议，以了解问题出在哪里。

我认为问题出在被调用两次的子例程中，并且在读取第二个数组的元素并覆盖它们之前不会保存第一个数组的元素。

我觉得要解决这个问题需要把这个元素保存在一个栈中，作为支撑，但是我不知道如何做好。

Answer 1

我设法以这种方式解决了这个练习:

.data
vect1: .float
.space 256
vect2: .float
.space 256
num1: .word
.space 4
num2: .word
.space 4
average: .float
.space 4
numag: .word
.space 4
i: .word
.space 4
count: .word
.space 4
sum: .float
.space 4
fnum: .float
.space 4
str1: .asciiz "The average is: "
.align 2
str2: .asciiz "\n"
.align 2
str3: .asciiz "In vector 2 there are "
.align 2
str4: .asciiz " elements major of "
.align 2
x: .float
.space 4
str5: .asciiz "Number of elements: "
.align 2
str6: .asciiz "Element: "
.align 2
.text
.globl main
main:
# read_vect1_float(vect1,num1);
lwc1 $f12,vect1
lw $a0,num1
jal read_vect1_float
# leggi_vet2_float(vect2,num2);
lwc1 $f13,vect2
lw $a1,num2
jal read_vect2_float
# sum=0.0;
mtc1 $zero,$f1
cvt.s.w $f1,$f1
# for(i=0;i<num1;i++)
li $t0,0
for:
# sum=sum+vect1[i];
move $t3,$t0
sll $t3,$t3,2
lwc1 $f2,vect1($t3)
add.s $f1,$f1,$f2

addi $t0,$t0,1
blt $t0,$t1,for
# fnum=(float)num1;
mtc1 $t1,$f3
cvt.s.w $f3,$f3
# sum=sum/fnum;
div.s $f1,$f1,$f3
# average=sum;
swc1 $f1,average
# cout<< "The average is: " << average;
li $v0,4
la $a0,str1
syscall

li $v0,2
lwc1 $f12,average
syscall
# cout << "\n";
li $v0,4
la $a0,str2
syscall
# count=0;
li $t4,0
sw $t4,count
# for(i=0;i<num2;i++)
li $t0,0
for2:
move $t3,$t0
sll $t3,$t3,2
lwc1 $f2,vect2($t3)
# if(vect2[i]>average)
c.le.s $f2,$f1
bc1t next
# count=count+1;
addi $t4,$t4,1

addi $t0,$t0,1
blt $t0,$t2,for2
next:
# numag=count;
sw $t4,numag
# cout << "In vector 2 there are ";
li $v0,4
la $a0,str3
syscall
# cout << numag;
li $v0,1
lw $a0,numag
syscall
# cout << " elements major of "
li $v0,4
la $a0,str4
syscall
# cout << average;
li $v0,2
lwc1 $f12,average
syscall
# cout << "\n";
li $v0,4
la $a0,str2
syscall

li $v0,10
syscall

read_vect1_float:
# cout << "Number of elements: ";
li $v0,4
la $a0,str5
syscall
# cin >> num1;
li $v0,5
syscall
sw $v0,num1
lw $t1,num1
# for(i=0;i<num1;i++)
li $t0,0
forsub1:
# cout=<< "Element: " ;
li $v0,4
la $a0,str6
syscall
# cin >> x;
li $v0,6
syscall
swc1 $f0,x
# vect1[i]=x;
move $t4,$t0
sll $t4,$t4,2
swc1 $f0,vect1($t4)

addi $t0,$t0,1
blt $t0,$t1,forsub1

jr $ra

read_vect2_float:
# cout << "Number of elements: ";
li $v0,4
la $a0,str5
syscall
# cin >> num2;
li $v0,5
syscall
sw $v0,num2
lw $t2,num2
# for(i=0;i<num2;i++)
li $t0,0
forsub2:
# cout=<< "Element: " ;
li $v0,4
la $a0,str6
syscall
# cin >> x;
li $v0,6
syscall
swc1 $f0,x
# vect2[i]=x;
move $t4,$t0
sll $t4,$t4,2
swc1 $f0,vect2($t4)

addi $t0,$t0,1
blt $t0,$t2,forsub2

jr $ra

我想知道固定的练习是否顺利，或者您可以做一些改进。