explicit运算符

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  引用StackOverFlow中的例子:

struct Foo {
  int value; // Foo stores an int, called value
  Foo(int v):value(v) {}; // Foo can be constructed (created) from an int
  explicit Foo(double d):value(d) {}; // Foo can be constructed (created) from a double
  // ^^^ note that keyword.  It says that it can only be EXPLICITLY created from a double
};


按照规定，只有一个参数的构造函数也定义了一个隐式转换，将该构造函数对应数据类型的数据转换为该类对象
在此构造函数前加了explicit后，用来抑制隐式转换
explicit只对构造函数起作用

Foo i = 0.0;

This fails to compile. I said that the constructor for Foo(double) is explicit, and the above only chooses to call constructors that are non-explicit. On the other hand:

Foo i(0.0);
Foo i = Foo(0.0);

these both are willing to call explicit constructors, and work fine.

explicit运算符

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原文地址：http://www.cnblogs.com/yzz0110/p/7668032.html