问题描述
我知道这就像打开潘多拉盒子，但它并没有停止打扰我.考虑一个简单的例子:
I know it's like opening the Pandora box but it doesn't stop bothering me. Consider a simple example:
#include <type_traits>
template <auto>
struct Foo: std::false_type { };
template <>
struct Foo<[](){return 1;}()>:std::true_type { };
int main() {
static_assert(Foo<1>::value);
}
我知道 lambdas 不能在未评估的上下文中声明，但显然这里不是这种情况.还有什么更奇怪的 clang 5.0.0(我猜它首先部分支持 constexpr lambda)编译它.
I know lambdas cannot be declared inside unevaluated context, but obviously this is not the case here. What is even more weird clang 5.0.0 (which, I guess, first partially supports constexpr lambda) does compile it.
这是编译器错误还是 C++17 允许这样做?
Is it a compiler bug or will C++17 allow this?
推荐答案
不，这是一个编译器错误.gcc 7.1 正确拒绝代码.
No, that is a compiler bug. gcc 7.1 correctly rejects the code.
[expr.prim.lambda]/2:
lambda 表达式是一个纯右值，其结果对象称为闭包对象.lambda 表达式不得出现在未求值的操作数、模板参数、别名声明、typedef 声明或函数声明中或函数体外的函数模板声明中和默认参数.
A lambda-expression is a prvalue whose result object is called the closure object. A lambda-expression shall not appear in an unevaluated operand, in a template-argument, in an alias-declaration, in a typedef declaration, or in the declaration of a function or function template outside its function body and default arguments.
从我标记为粗体的部分可以看出，lambda 表达式不能出现在模板参数列表中.
As you can see from the part that I marked as bold, a lambda expression cannot appear in a template argument list.
这在随后的注释中也有明确说明:
This is also made clear in a subsequent note:
[ 注意:目的是防止 lambda 出现在签名中.— 尾注 ]
[Note: The intention is to prevent lambdas from appearing in a signature. —end note]
如果我猜的话，我会说这个错误是因为从 C++17 开始，lambdas 是隐式的 constexpr，这使得它们在编译时表达式中被调用是有效的，比如模板参数.但实际上在模板参数中定义 lambda 仍然是非法的.
If I were to guess, I would say that the bug comes about because starting with C++17, lambdas are implicitly constexpr, which makes them valid to be called in compile time expressions, like template arguments. But actually defining a lambda in a template argument is still illegal.
请注意，此限制已在 C++20 中取消.:)
Note that this restriction has been lifted in C++20. :)
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