如何写一个通用的可变参数lambda，抛弃它的参数？ [英] How to write a generic variadic lambda that discards its parameters?

问题描述

我想写一个lambda，通过通用引用获取任意数量的参数，并完全忽略它们。显而易见的方法是使用可变参数通用参数包的语法，并省略参数名：

I want to write a lambda that takes an arbitrary number of arguments by universal reference and ignores them entirely. The obvious method would be to use the syntax for a variadic universal parameter pack and omit the parameter name:

auto my_lambda = [](auto&&...) { return 42; };

这很好（使用gcc 4.9.2），直到尝试传递非三维可复制对象：

This works fine (with gcc 4.9.2) until I try to pass a non trivially-copyable object:

struct S { S() {} S(S const&) {} };
my_lambda("meow", 42, S{});
^ error: cannot pass objects of non-trivially-copyable type 'struct S' through '...'

发生了什么事？我的代码是不成形的，还是这是gcc中的一个错误？

What's going on? Is my code ill-formed, or is this a bug in gcc?

在这两种情况下，最好的解决方法是什么？我发现命名参数工程，但后来我遇到一个未使用的参数警告：

In either case, what's the best workaround? I found that naming the parameter works, but then I ran into an unused-parameter warning:

auto my_lambda = [](auto&&... unused) { return 42; };
^ error: unused parameter 'unused#0' [-Werror=unused-parameter]
^ error: unused parameter 'unused#1' [-Werror=unused-parameter]
^ error: unused parameter 'unused#2' [-Werror=unused-parameter]

模板参数包中未使用的参数警告？

How do you suppress an unused-parameter warning on a template parameter pack?

推荐答案

这是一个解析错误在GCC（你自己报告！ auto&&&<... 在语法上不清楚，可以解析为 auto&&& $ c>或参数包声明（技术上，问题是 ... 是 parameter-declaration-clause em> abstract-declarator ）;标准说它要解析为后者;

It's a parsing bug in GCC (which you yourself reported!). auto&&... is grammatically ambiguous and can be parsed as either the equivalent of auto&&, ... or a parameter pack declaration (technically, the question is whether ... is part of the parameter-declaration-clause or the abstract-declarator); the standard says it's to be parsed as the latter; GCC parses it as the former.

命名包解析解析歧义：

auto my_lambda = [](auto&&... unused) { return 42; };

要禁止警告，可以应用 __ attribute __（（__ unused __））（或，如@Luc Danton建议， [[gnu :: unused]] ）：

To suppress the warning, one could apply __attribute__((__unused__)) (or, as @Luc Danton suggested, [[gnu::unused]]):

auto my_lambda = [](auto&&... unused __attribute__((__unused__))) { return 42; };

或使用 sizeof ... p>

or use sizeof...

auto my_lambda = [](auto&&... unused) { (void) sizeof...(unused); return 42; };

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