Question

我在c中编写了一个简单的计算器，用户输入两个数字，然后选择要应用的操作。（Mul，Add，Div，Sub）程序可以正常工作，但它会跳过它应该采取的部分操作数的用户输入。我做错了什么

#include<stdio.h>
#include<sstream>
#include<iostream>
#include<conio.h>
#include<string.h>

using std::cout;
using std::cin;

using namespace std;

int main() {

char Operative[100];

int a;
int b;
int c;


printf("Enter First Number\n");
scanf("%d",&a);

printf("Enter First Number\n");
scanf("%d",&b);

printf("\nPlease Enter Operation(M,A,D,S)");
gets(Operative);

//getline(cin,Operative);

if (Operative == "M")
{
    c = a*b;
    printf("Multiplication value is %d",c);

}

else if (Operative == "A")
{
    c = a+b;
    printf("Addition value is %d",c);

}

else if (Operative == "D")
{
    c = a/b;
    printf("Division value is %d",c);

}

else if (Operative == "S")
{
    c = a-b;
    printf("\nSubtraction value is %d",c);

}

}

Answer 1

使用strcmp比较字符串值。 ==比较指针：

if (strcmp (Operative,  "M") == 0) ...

Answer 2

您必须先调用cin.ignore()（scanf之后，gets之前），因为缓冲区中仍有\0。

Answer 3

您也可以使用switch声明。

 switch(operation){
        case 'P':
            result = a+b;
            printf("%d + %d = %d", a,b,result);
            .
            .
            .

Answer 4

如果您希望使用int main()和printf替换，只需在scanf中对此进行编码即可。

do {

        cout << "Option 1 <<endl;
        cout << "Option 2 <<endl;
        cout << "Option 3 <<endl;

        cin >> x;

        switch (x) {
            case 1:
                //Your operation here
                break;
            case 2:
                //Your operation here
                 break;
            case 3:
                //Your operation here
                break;
            case 4:
                exit(0);
                break;
            default:
                cout << "Input a valid option" <<endl;
    }

    } while(x);

程序跳过用户输入

4 个答案: