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C++ 面试常问问题(一)

C++ 面试常问问题(一)

作者: 恋恋风辰 | 来源:发表于2022-03-03 11:52 被阅读0次

    这篇文章讲解C++ 面试常问的几个问题。本文通过demo讲解初始化列表,继承,字符串等常问问题。看下边这个例子

    初始化列表

    //基类
    class Base
    {
    public:
        Base() : m_nbase(0), m_nbase2(m_nbase + 100) {}
        Base(int n) : m_nbase(n), m_nbase2(m_nbase + 100)
        {
            cout << "this is Base construct " << endl;
        }
        ~Base()
        {
            cout << "this is Base destruct " << endl;
        }
    
       virtual void printData()
        {
            cout << "this is Base printData " << endl;
            cout << "data is " << m_nbase << endl;
            // cout << "base2 data is " << m_nbase2 << endl;
        }
    
        void printData2()
        {
            cout << "base2 data is " << m_nbase2 << endl;
        }
    
        int SizeOf(char p[])
        {
            return sizeof(p);
        }
    
        int SizeOf2(char *p)
        {
            return sizeof(p);
        }
    
    private:
        int m_nbase;
        int m_nbase2;
    };
    

    实现了一个类Base,类的构造函数采用了初始化列表,初始化列表按顺序初始化初始类成员。
    接下来在main函数里调用如下

    Base b1(1);
    b1.printData();
    b1.printData2();
    

    1 问Base的初始化列表是否会报错?
    回答:
    不会有问题,因为初始化列表按顺序初始化类成员。所以会分别输出m_nbase的值为1, m_nbase2的值为101

    继承问题

    继承问题常问到的是基类和子类的关系,继承先构造基类再构造子类。析构时先析构子类,再析构基类。
    我们实现一个Derive类继承自Base

    class Derive : public Base
    {
    public:
        Derive(int n) : Base(n), m_nderive(n) 
        { cout << "this is Derive construct " << endl; }
    
         ~Derive()
        {
            cout << "this is Derive destruct " << endl;
        }
    
        void printData()
        {
            cout << "this is Derive printData" << endl;
        }
    
    private:
        int m_nderive;
    };
    

    Derive类继承了Base类,接下来我们在主函数中调用如下

    Derive d1(2);
    d1.printData();
    cout << " ................." << endl;
    

    1 问 上面程序输出什么
    答 输出如下

    this is Base construct 
    this is Derive construct 
    this is Derive printData
    this is Derive destruct 
    this is Base destruct 
    

    因为构造时,先构造基类再构造子类,析构时先析构子类再析构基类
    如果在主函数中调用如下

    Base *b1 = new Derive(2);
    b1->printData();
    delete b1;
    

    2 问上面程序输出什么?
    答输出如下

    this is Base construct 
    this is Derive construct 
    this is Derive printData
    this is Base destruct 
    

    因为构造时同样先构造基类,再构造子类。
    由于创建子类对象返回基类指针,调用虚函数printData会触发多态机制,调用了子类的printData。
    析构时由于析构函数不是虚函数,所以不会调用子类的析构函数!!!
    所以只会输出Base的析构函数。
    3 问如何解决多态情况下子类无法析构问题?
    答 将基类和子类的析构函数都设置为虚析构函数即可,少一个都不行!!!
    修改如下

    class Base
    {
    public:
        Base() : m_nbase(0), m_nbase2(m_nbase + 100) {}
        Base(int n) : m_nbase(n), m_nbase2(m_nbase + 100)
        {
            cout << "this is Base construct " << endl;
        }
        ~Base()
        {
            cout << "this is Base destruct " << endl;
        }
    
       virtual void printData()
        {
            cout << "this is Base printData " << endl;
            cout << "data is " << m_nbase << endl;
            // cout << "base2 data is " << m_nbase2 << endl;
        }
        //....省略
    };
    
    class Derive : public Base
    {
    public:
        Derive(int n) : Base(n), m_nderive(n) 
        { cout << "this is Derive construct " << endl; }
    
        virtual  ~Derive()
        {
            cout << "this is Derive destruct " << endl;
        }
    
        void printData()
        {
            cout << "this is Derive printData" << endl;
        }
    
    private:
        int m_nderive;
    };
    

    此时再调用

    Base *b1 = new Derive(2);
    b1->printData();
    delete b1;
    

    程序输出

    this is Base construct 
    this is Derive construct 
    this is Derive printData
    this is Derive destruct 
    this is Base destruct 
    

    字符串sizeof

    字符串的sizeof也是常问的问题

    char carry[100] = "Hello World";
    cout << "sizeof(carry)  " << sizeof(carry) << endl;
    char *pstr = "Hello World";
    cout << "sizeof(pstr)   " << sizeof(pstr) << endl;
    cout << "sizeof(*pstr)  " << sizeof(*pstr) << endl;
    Base b1;
    cout << "b1.SizeOf(carry)  " << b1.SizeOf(carry) << endl;
    cout << "b1.SizeOf2(carry) " << b1.SizeOf2(carry) << endl;
    

    1 问上边程序输出什么
    答 输出如下

    sizeof(carry)  100
    sizeof(pstr)  4
    sizeof(*pstr)  1
    b1.SizeOf(carry) 4
    b1.SizeOf2(carry) 4
    

    carry是个数组,所以sizeof(carry)是数组的大小为100
    pstr是个char指针,所以sizeof(pstr)是指针大小为4字节,当然在64位机器为8
    我的机器上输出的是8,这个根据机器不同而不同,记住是指针大小就行。
    pstr指向字符串首地址,也是第一个字符的地址,pstr为第一个字符
    sizeof(
    pstr)为一个字符的大小,所以为1
    b1.SizeOf以及b1.SizeOf2内部调用的都是sizeof()一个char指针,所以都为4,
    我的机器上输出为8

    bool,float和0比较

    bool和0比较

    bool bval = true;
    if(!bval){
        //...
        //bval为0的逻辑
    }else{
        //...
    }
    

    float和0比较

        float f1 = 0.1;
        if (f1 <= FLT_EPSILON && f1 >= -FLT_EPSILON)
        {
            cout << "this is float 0" << endl;
        }
        else
        {
            cout << "this is not float 0" << endl;
        }
    
        double d1 = 0.0;
        if (d1 <= DBL_EPSILON && d1 >= -DBL_EPSILON)
        {
            cout << "this is double 0" << endl;
        }
        else
        {
            cout << "this is not double 0" << endl;
        }
    

    总结

    本文介绍了一些面试常见问题
    源码链接
    https://gitee.com/secondtonone1/cpplearn
    想系统学习更多C++知识,可点击下方链接。
    C++基础

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