我如何故意丢弃 [[nodiscard]] return 值？

Question

说我有

[[nodiscard]] int foo ()
{
    return 0;
}

int main ()
{
    foo ();
}

然后

error: ignoring return value of ‘int foo()’, declared with attribute nodiscard [-Werror=unused-result]

但是如果

int x = foo ();

然后

error: unused variable ‘x’ [-Werror=unused-variable]

是否有一种清晰的方式告诉编译器"I want to discard this [[nodiscard]] value"？

Answer 1

Cast it to void:

[[nodiscard]] int foo ()
{
    return 0;
}

int main ()
{
    static_cast<void>(foo());
}

这基本上告诉编译器“是的，我知道我正在丢弃它，是的，我确定它。”

Answer 2

您还可以用另一个标签标记返回的 int：

[[nodiscard]] int foo ()
{
    return 0;
}

int main ()
{
    [[maybe_unused]] int i = foo ();
}

如果您有一些需要该值的仅调试代码，这可能会很有用。

Answer 3

WG14 nodiscard proposal 讨论了允许通过转换为 void 来使诊断静音的基本原理。它说转换为 void 是鼓励的（如果非规范的）沉默方式，它遵循现有实现对 __attribute__((warn_unused_result)):

的处理

The [[nodiscard]] attribute has extensive real-world use, being implemented by Clang and GCC as __attribute__((warn_unused_result)) , but was standardized under the name [[nodiscard]] by WG21. This proposal chose the identifier nodiscard because deviation from this name would create a needless incompatibility with C++.

The semantics of this attribute rely heavily on the notion of a use, the definition of which is left to implementation discretion. However, the non-normative guidance specified by WG21 is to encourage implementations to emit a warning diagnostic when a nodiscard function call is used in a potentially-evalulated discarded-value expression unless it is an explicit cast to void. This means that an implementation is not encouraged to perform dataflow analysis (like an initialized-but- unused local variable diagnostic would require). ...

C++ 方式是 static_cast<void>。

参见 C++ 标准草案 [[dcl.attr.nodiscard]p2:

[ Note: A nodiscard call is a function call expression that calls a function previously declared nodiscard, or whose return type is a possibly cv-qualified class or enumeration type marked nodiscard. Appearance of a nodiscard call as a potentially-evaluated discarded-value expression is discouraged unless explicitly cast to void. Implementations should issue a warning in such cases. This is typically because discarding the return value of a nodiscard call has surprising consequences. — end note]

这是一个注释，因此不规范，但基本上这是现有实现对 __attribute__((warn_unused_result)) 的处理方式。另外，请注意 nodiscard 的诊断也是非规范的，因此违反 nodiscard 的诊断不是格式错误，而是实施质量就像通过强制转换为 void 进行压制一样。

查看 clang document on nodiscard, warn_unused_result:

Clang supports the ability to diagnose when the results of a function call expression are discarded under suspicious circumstances. A diagnostic is generated when a function or its return type is marked with [[nodiscard]] (or __attribute__((warn_unused_result))) and the function call appears as a potentially-evaluated discarded-value expression that is not explicitly cast to void.

Answer 4

我使用了一个（空的）辅助函数 "discard"

template<typename T>
void discard(const T&) {}

[[nodiscard]] int foo ()
{
    return 0;
}

int main ()
{
    discard(foo());
}

故意丢弃 [[nodiscard]] 值。

Answer 5

With Boost:

#include <boost/core/ignore_unused.hpp>
int main ()
{
    boost::ignore_unused(foo ());
}

boost::ignore_unused 通过对 const 的引用获取其参数，因此参数必须是可以绑定到对 const 的引用的内容。我相当确定任何可以是函数 return 类型的东西（当然除了 void！）在这里应该没问题。

Answer 6

CppCoreGuidelines suggest 使用 std::ignore:

Never cast to (void) to ignore a [[nodiscard]] return value. If you deliberately want to discard such a result, first think hard about whether that is really a good idea (there is usually a good reason the author of the function or of the return type used [[nodiscard]] in the first place). If you still think it's appropriate and your code reviewer agrees, use std::ignore = to turn off the warning which is simple, portable, and easy to grep.

这与另一个答案中建议的 boost::ignore_unused 几乎相同，但超出了 std:: 范围。

但是，使用 std::ignore 有缺点：

像其他这样的辅助函数一样，它会实例化，花费编译时间，并被调用（在非优化调试中），花费运行时间
std::ignore 适用于 another purpose
std::ignore 甚至不能保证抑制警告

我如何故意丢弃 [[nodiscard]] return 值？

How can I intentionally discard a [[nodiscard]] return value?

c++

c++17

nodiscard