Question

制作一个允许用户输入输入的程序，以便他们可以获得简单数学科目的帮助。我以为我已经成功了，如果他们进入了一个不在那里的主题，那么就会说（不）见下面的其他声明。知道为什么当我运行它时，它仍然包含输入主题正确时的else语句？ http://prntscr.com/ey3qyh

search = "triangle";

pos = sentence.find(search);
if (pos != string::npos)
    // source http://www.mathopenref.com/triangle.html
    cout << "A closed figure consisting of three line segments linked end-to-end. A 3 - sided polygon." << endl;

search = "square";
pos = sentence.find(search);
if (pos != string::npos)
    // source http://www.mathopenref.com/square.html
    cout << "A 4-sided regular polygon with all sides equal and all internal angles 90°" << endl;

else
    cout << "Sorry, it seems I have no information on this topic." << endl;

case 2:
    break;
default:
    cout << "Invalid input" << endl;
}

Answer 1

这是您的计划所做的事情：

寻找＆＃34;三角形＆＃34;
如果＆＃34;三角形＆＃34;找到，打印三角形的定义。
寻找＆＃34; square＆＃34;
如果＆＃34; square＆＃34;找到了，打印出正方形的定义。否则，打印道歉。

因为＆＃34;三角形＆＃34;找到了＆＃34; square＆＃34;没有找到，它打印出三角形的定义和道歉。换句话说，计算机正在完全按照你的要求去做 - 问题是什么？

Answer 2

您至少有两个选择：

嵌套选项，以便仅在前一个选项不起作用时评估每个后续选项：

search = "triangle";
pos = sentence.find(search);
if (pos != string::npos){
    // link to info about triangle
    cout << "...info about triangle..." << endl;
}else{
    search = "square";
    pos = sentence.find(search);
    if (pos != string::npos){
        // link to info about square
        cout << "...info about square..." << endl;
    }else{
        search = "pentagon";
        pos = sentence.find(search);
        if (pos != string::npos){
            // link to info about pentagon
            cout << "...info about pentagon..." << endl;
        }else{
            cout << "Sorry, it seems I have no information on this topic." << endl;
        }
    }
}

或者，如果您可以将if条件的所有代码放入if语句中，则可以使用if语句：

if (sentence.find("triangle") != string::npos){
    // link to info about triangle
    cout << "...info about triangle..." << endl;
}else if (sentence.find("square") != string::npos){
    // link to info about square
    cout << "...info about square..." << endl;
}else if (sentence.find("pentagon") != string::npos){
    // link to info about pentagon
    cout << "...info about pentagon..." << endl;
}else{
    cout << "Sorry, it seems I have no information on this topic." << endl;
}

您使用哪一个取决于您是否可以将if条件的所有代码都放入条件本身。

C ++忽略else语句

2 个答案: