constexpr是否支持lambda函数/表达式？ [英] Is constexpr supported with lambda functions / expressions?

问题描述

  struct Test 
 {
 static const int value = []（） - > int {return 0; }（）; 
}; 
  

使用gcc-4.6，我得到类似于错误：function需要constexpr 。我尝试了多种组合在 constexpr 在各个地方，但没有运气。

是 constexpr 也支持lambda函数（不管 return 是否指定类型）？

解决方案

> Lambdas目前（C ++ 14）不允许在[expr.const] /（2.6）的常量表达式中，但他们会一次 N4487 （可在工作草案N4582中找到）：

此建议建议在常量
表达式中允许使用lambda表达式，删除现有的限制。作者建议
确定某个λ-expression s和对某些闭包
对象的操作允许出现在常量表达式中。在这样做时，
我们还建议，如果
每个数据成员的类型是字面类型，则将闭包类型视为文字类型;并且如果
在lambda声明符中省略了 constexpr 说明符，那么
生成的函数调用操作符 constexpr 如果它满足
需求的一个 constexpr 函数（类似于
constexpr 对隐式定义的
构造函数和赋值操作符函数已经发生的推断）。

struct Test
{
  static const int value = []() -> int { return 0; } ();
};

With gcc-4.6 I get something like, error: function needs to be constexpr. I have tried multiple combinations of putting constexpr at various places, but no luck.

Is constexpr supported for lambda functions as well (irrespective of return type specified or not) ? What is the correct syntax ?

Any work around possible ?

解决方案

Lambdas are currently (C++14) not allowed in constant expressions as per [expr.const]/(2.6), but they will once N4487 is accepted (which can be found in the working draft N4582):

This proposal suggests allowing lambda-expressions in constant expressions, removing an existing restriction. The authors propose that certain lambda-expressions and operations on certain closure objects be allowed to appear within constant expressions. In doing so, we also propose that a closure type be considered a literal type if the type of each of its data-members is a literal type; and, that if the constexpr specifier is omitted within the lambda-declarator, that the generated function call operator be constexpr if it would satisfy the requirements of a constexpr function (similar to the constexpr inference that already occurs for implicitly defined constructors and the assignment operator functions).

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