【C++】: 隐式构造函数和显示构造函数的实验和讨论

1. 先放一段实验代码及编译结果。

#include <iostream>

class Stock1
{
public:
    Stock1(std::string str)
    {
        name = str;
    }

private:
    std::string name;
};

class Stock2
{
public:
    explicit Stock2(std::string str)
    {
        name = str;
    }

private:
    std::string name;
};

class Price
{
public:
    Price(int n)
    {
        value = n;
    }

private:
    int value;
};

int main()
{
    Stock1 nvidia = Stock1("nvidia");
    Stock1 intel("intel");
    std::string str1 = "amd";
    Stock1 amd = str1;

    Stock2 samsung = Stock2("samsung");
    Stock2 ibm("ibm");
    std::string str2 = "micro";
    Stock2 micro = str2;    // compile error
    micro = "micro";    // compile error, operator=() need implementation
    Stock2 tsmc = "tsmc";   // compile error

    Stock1 arm = nvidia;    // default copy constructor, shallow copy

    Price price = 1;

    return 0;
}

 编译结果如下：

practice.cpp: In function ‘int main()’:
practice.cpp:49:20: error: conversion from ‘std::__cxx11::string {aka std::__cxx11::basic_string<char>}’ to non-scalar type ‘Stock2’ requested
     Stock2 micro = str2;    // compile error
                    ^~~~
practice.cpp:50:13: error: no match for ‘operator=’ (operand types are ‘Stock2’ and ‘const char [6]’)
     micro = "micro";    // compile error
             ^~~~~~~
practice.cpp:15:7: note: candidate: Stock2& Stock2::operator=(const Stock2&)
 class Stock2
       ^~~~~~
practice.cpp:15:7: note:   no known conversion for argument 1 from ‘const char [6]’ to ‘const Stock2&’
practice.cpp:15:7: note: candidate: Stock2& Stock2::operator=(Stock2&&)
practice.cpp:15:7: note:   no known conversion for argument 1 from ‘const char [6]’ to ‘Stock2&&’
practice.cpp:51:19: error: conversion from ‘const char [5]’ to non-scalar type ‘Stock2’ requested
     Stock2 tsmc = "tsmc";   // compile error

2. 对结果进行讨论

代码中一共设置了3处错误：

(1) line49

class Stock2的构造函数前面加上了限定符(specifier) explicit，要求进行显示构造。

(2) line50

编译错误中提示需要实现函数operator=，我们的类中没有这样的成员函数。

(3) line51

很多人容易混淆std::string和字符串常量。也就是说“tsmc”并不是std::string，而是const char[]。所以在类型上不匹配，但是我们通过间接的方式，std::string str1 = "amd"; Stock1 amd = str1; 就能够匹配到Stock1的构造函数。

相关资料：

https://en.cppreference.com/w/cpp/language/explicit

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---------- 欢迎讨论和指正 ----------