c++ - lambda 函数/表达式是否支持 constexpr？

Question

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

在 gcc-4.6 中，我得到类似这样的信息，error: function needs to be constexpr。我尝试了将 constexpr 放在不同位置的多种组合，但没有成功。

constexpr 是否也支持 lambda 函数(无论是否指定了 return 类型)？正确的语法是什么？

有什么解决办法吗？

Answer 1

更新:从 C++17 开始，常量表达式中允许使用 lambda。

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根据 [expr.const]/(2.6)，Lambda 目前 (C++14) 不允许出现在常量表达式中，但它们一次会出现 N4487被接受(可以在工作草案 N4582 中找到):

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).