Question

我是C ++的新手，我得到了一项任务，将 b-c d转换为c ++表达式这是我的代码行

#include<iostream>
#include<conio.h>
#include<stdio.h>
using namespace std;
int main ()
{
	int a,b,c,d,e,f ,sum;
	cout<<"enter value of A";
	cin>>a;
	cout<<"enter value of b";
	cin>>b;
	cout<<"Your Answer is";
	cin>>a*b=e;
	cout<<"enter value of c";
	cin>>a;
	cout<<"enter value of d";
	cin>>b;
	cout<<"Your Answer is";
	cin>>c*d=f;
	cin>>e-f;
	sum=e-f;
    getch();
}

您的帮助将不胜感激。

此致

Answer 1

#include<iostream>
#include<conio.h>
#include<stdio.h>
using namespace std;
int main ()
{
    int a,b,c,d,e,f ,sum;
    cout<<"enter value of A";
    cin>>a;
    cout<<"enter value of b";
    cin>>b;
    e=a*b;
    cout<<"Your Answer is"<<e<<endl;

    cout<<"enter value of c";
    cin>>a;
    cout<<"enter value of d";
    cin>>b;
    f=c*d;
    cout<<"Your Answer is"<<f<<end;

    sum=e-f;
    getch();
}

你使用cin语句来获取用户的输入，评估不是在cin语句中完成的，如果它适合你，请考虑这段代码。

Answer 2

我已经修复了你想要做的一些事情，假设你使用e和f作为临时变量来计算ab-cd，并将sum作为最终变量。

            #include<iostream>
            #include<conio.h>
            #include<stdio.h>
            using namespace std;
            int main ()
            {
                int a,b,c,d,e,f ,sum;
                cout<<"enter value of A";
                cin>>a;
                cout<<"enter value of b";
                cin>>b;
                cout<<"Your Answer is";
                e = a*b;
                cout<<"enter value of c";
                cin>>c;
                cout<<"enter value of d";
                cin>>d;
                cout<<"Your Answer is";
                f = c*d;
                sum=e-f;
                cout << sum;
            }

如何在c ++表达式中转换* b-c * d

2 个答案: