explicit 关键字

c++中explicit关键字的含义和用法

explicit关键字用来修饰类的构造函数，表明该构造函数是显式的，既然有\显式\那么必然就有\隐式\，那么什么是显示而什么又是隐式的呢？

如果c++类的构造函数有一个参数，那么在编译的时候就会有一个缺省的转换操作：将该构造函数对应数据类型的数据转换为该类对象，如下面所示： class MyClass {

public:

MyClass( int num ); } ....

MyClass obj = 10; //ok,convert int to MyClass

在上面的代码中编译器自动将整型转换为MyClass类对象，实际上等同于下面的操作： MyClass temp（10）； MyClass obj = temp；

上面的所有的操作即是所谓的\隐式转换\。

如果要避免这种自动转换的功能，我们该怎么做呢？嘿嘿这就是关键字explicit的作用了，将类的构造函数声明为\显示\，也就是在声明构造函数的时候 前面添加上explicit即可，这样就可以防止这种自动的转换操作，如果我们修改上面的MyClass类的构造函数为显示的，那么下面的代码就不能够编 译通过了，如下所示： class MyClass {

public:

explicit MyClass( int num ); }

MyClass obj = 10; { public:

explicit isbn_missmatch(const std::string &s):std:logic_error(s){}

isbn_mismatch(const std::string &s,const std::string &lhs,const std::string &rhs): std::logic_error(s),left(lhs),right(rhs){} const std::string left,right;

virtual ~isbn_mismatch() throw(){}

//err,can't non-explict convert

class isbn_mismatch:public std::logic_error

};

Sales_item& operator+(const Sales_item &lhs,const Sales_item rhs) {

if(!lhs.same_isbn(rhs))

throw isbn_mismatch(\

Sales_item ret(lhs); ret+rhs; return ret; }

Sales_item item1,item2,sum; while(cin>>item1>>item2) {

try{

{

} }

用于用户自定义类型的构造函数，指定它是默认的构造函数，不可用于转换构造函数。因为构造函数有三种：1拷贝构造函数2转换构造函数3一般的构造函数（我自己的术语^_^）

另：如果一个类或结构存在多个构造函数时，explicit 修饰的那个构造函数就是默认的 class isbn_mismatch:public std::logic_error { public:

explicit isbn_missmatch(const std::string &s):std:logic_error(s){}

isbn_mismatch(const std::string &s,const std::string &lhs,const std::string &rhs): std::logic_error(s),left(lhs),right(rhs){} const std::string left,right;

virtual ~isbn_mismatch() throw(){} };

cerr<

}catch(const isbn_mismatch &e)

Sales_item& operator+(const Sales_item &lhs,const Sales_item rhs) {

if(!lhs.same_isbn(rhs))

throw isbn_mismatch(\Sales_item ret(lhs); ret+rhs; return ret; }

Sales_item item1,item2,sum; while(cin>>item1>>item2) {

try{

sun=item1+item2;

}catch(const isbn_mismatch &e) {

cerr<

这个 《ANSI/ISO C++ Professional Programmer's Handbook 》是这样说的

explicit Constructors

A constructor that takes a single argument is, by default, an implicit conversion operator, which converts its argument to an object of its class (see also Chapter 3, \class string {

private: int size; int capacity; char *buff; public: string();

string(int size); // constructor and implicit conversion operator

string(const char *); // constructor and implicit conversion operator ~string(); };

Class string has three constructors: a default constructor, a constructor that takes int, and a constructor that constructs a string from const char *. The second constructor is used to create an empty string object with an initial preallocated buffer at the specified size. However, in the case of class string, the automatic conversion is dubious. Converting an int into a string object doesn't make sense, although this is exactly what this constructor does. Consider the following: int main() {

string s = \int ns = 0;

s = 1; // 1 oops, programmer intended to write ns = 1, }

In the expression s= 1;, the programmer simply mistyped the name of the variable ns, typing s instead. Normally, the compiler detects the incompatible types and issues an error message. However, before ruling it out, the compiler first searches for a user-defined conversion that allows this expression; indeed, it finds the constructor that takes int. Consequently, the compiler interprets the expression s= 1; as if the programmer had written s = string(1); You might encounter a similar problem when calling a function that takes a string argument. The following example can either be a cryptic coding style or simply a programmer's typographical error. However, due to the implicit conversion constructor of class string, it will pass unnoticed: int f(string s); int main() {

f(1); // without a an explicit constructor, //this call is expanded into: f ( string(1) );

//was that intentional or merely a programmer's typo? }

'In order to avoid such implicit conversions, a constructor that takes one argument needs to be declared explicit: