我应该将编译器生成的构造函数标记为constexpr吗？

问题描述

两者之间有区别吗？

X() = default;

和

constexpr X() = default;

在常量表达式中默认构造类很好，所以这两个示例之间有区别吗？我应该在另一个上使用吗？

Default-constructing the class within constant expressions work fine, so is there a difference between these two examples? Should I use one over the other?

推荐答案

因为隐式构造函数实际上是 constexpr 在您的情况下…

Since the implicit constructor is actually constexpr in your case…

[C ++ 11：7.1.5 / 3]： constexpr 函数的定义应满足以下约束：

[C++11: 7.1.5/3]: The definition of a constexpr function shall satisfy the following constraints:

• it不得为虚拟（10.3）；


• 其返回类型应为文字类型；


• 其每个参数类型均应为文字类型；


• 其功能主体应为 =删除， =默认值 ，或者仅包含
  
  
  
  • null语句的 b $ b
  • static_assert-declarations
  
  
  • typedef 声明和 alias-declarations ，它们未定义类或枚举，
  
  
  • 使用声明，
  
  
  • using指令，
  
  
  • 和一个返回语句；
  
  
  • it shall not be virtual (10.3);
  • its return type shall be a literal type;
  • each of its parameter types shall be a literal type;
  • its function-body shall be = delete, = default, or a compound-statement that contains only
    • null statements,
    • static_assert-declarations
    • typedef declarations and alias-declarations that do not define classes or enumerations,
    • using-declarations,
    • using-directives,
    • and exactly one return statement;

    …声明实际上是等效的：

    … the declarations are actually equivalent:

    
    
    • 该函数被隐式认为是 constexpr 如果隐式声明为，
    
    
    • 它被隐式认为具有相同的 exception-specification ，就好像它已被隐式声明（15.4），
    
    
    • 对于复制构造函数，move构造函数，复制赋值运算符或move赋值运算符而言，它应具有
    
    
    • it is implicitly considered to be constexpr if the implicit declaration would be,
    • it is implicitly considered to have the same exception-specification as if it had been implicitly declared (15.4), and
    • in the case of a copy constructor, move constructor, copy assignment operator, or move assignment operator, it shall have the same parameter type as if it had been implicitly declared.

    所以要么—

    So do either — it doesn't matter.

    在一般情况下，如果您确实希望构造函数为 constexpr ，最好保留关键字，这样如果不符合条件，则至少会出现编译器错误；忽略它，您可能会得到一个非 constexpr 构造函数而没有意识到。

    In the general case, if you definitely want a constructor to be constexpr, though, it may be wise to leave the keyword in so that you at least get a compiler error if it does not meet the criteria; leaving it out, you may get a non-constexpr constructor without realising it.

    这篇关于我应该将编译器生成的构造函数标记为constexpr吗？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！