为什么`decltype(static_cast< T>(...))`并不总是`T`? [英] Why is `decltype(static_cast<T>(...))` not always `T`?

问题描述

对于以下代码，除最后一个断言外，所有消息均通过:

For the following code, all but the last assertion passes:

template<typename T>
constexpr void assert_static_cast_identity() {
    using T_cast = decltype(static_cast<T>(std::declval<T>()));
    static_assert(std::is_same_v<T_cast, T>);
}

int main() {
    assert_static_cast_identity<int>();
    assert_static_cast_identity<int&>();
    assert_static_cast_identity<int&&>();
    // assert_static_cast_identity<int(int)>(); // illegal cast
    assert_static_cast_identity<int (&)(int)>();
    assert_static_cast_identity<int (&&)(int)>(); // static assert fails
}

为什么最后一个断言失败，并且 static_cast< T> 并不总是返回 T ?

Why is this last assertion failing, and static_cast<T> not always returning a T?

推荐答案

这是在 static_cast 的定义中硬编码的:

This is hard-coded in the definition of static_cast:

[expr.static.cast] (强调我的意思)

1 表达式 static_cast< T>(v)是结果将表达式 v 转换为 T 的过程.如果 T 是左值引用类型或函数类型的右值引用，结果是左值；如果 T 是对对象类型的右值引用，则结果为一个xvalue;否则，结果为prvalue. static_cast 操作员不得抛弃常数.

1 The result of the expression static_­cast<T>(v) is the result of converting the expression v to type T. If T is an lvalue reference type or an rvalue reference to function type, the result is an lvalue; if T is an rvalue reference to object type, the result is an xvalue; otherwise, the result is a prvalue. The static_­cast operator shall not cast away constness.

decltype 尊重其操作数的值类别，并为左值表达式生成左值引用.

decltype respects the value category of its operand, and produces an lvalue reference for lvalue expressions.

推理可能是由于函数名称本身始终是左值，因此函数类型的右值不能在野外"出现.因此，强制转换为这种类型可能毫无意义.

The reasoning may be due to function names themselves always being lvalues, and so an rvalue of a function type cannot appear "in the wild". As such, casting to that type probably makes little sense.

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