Question

我需要在提示时检查他们是否输入快乐，悲伤或疯狂：输入一种情绪（快乐，悲伤，疯狂）到目前为止，这是我的代码 count = 0; while（count＆lt; 1）{

    cout << ", How are you feeling today? (happy,sad,mad)" << endl;

    cin >> mood;

    if (mood == sad){

        cout << "im sorry you're feeling sad" << endl;

            count++;
    }

    else if (mood == happy) {

        cout << "I'm glad you're feeling happy!" << endl;

            count++;
    }

    else if (mood == mad){
        cout << "Don't be angry, there's so many happy things in life!" << endl;
    }
    else{

        cout << " invalid input " << endl;

        count = 0;
    }

    }

但如果我处于快乐，悲伤或疯狂状态，那么它就会进入无效输入状态。我也尝试为它们声明字符串变量 string happy = happy; string sad = sad; string mad = mad;

Answer 1

两件事：

确保正确初始化字符串
```
string happy = "happy";
```
尝试使用string.compare()而不是==
```
mood.compare(happy);
```

有关详情：http://www.cplusplus.com/reference/string/string/compare/

C ++检查用户是否输入了有效答案

1 个答案: