从移位运算符，位运算符和sizeof运算符产生的结果的值类别是什么？ [英] What is the value category of result yielded from shift operators, bit-wise operators, and sizeof operator?

问题描述

Shift操作符：< >>
位运算符： code>〜，& ， ^ ， |
sizeof运算符： sizeof（）

Shift operators: << >> bit-wise operators: ~, &, ^, | sizeof operator: sizeof()

标准（n3797），我只能确认〜产生prvalue（5.3.1 / 2），但不是上面的其他。

Per the C++ standard (n3797), I can only confirm that ~ yields prvalue (5.3.1/2), but not the others above.

推荐答案

至于我可以告诉结果是 prvalues ，但这只是一个投机。这似乎在类似于中所涵盖的操作数的值类别中的指定方式。未指定时使用C ++运算符？和在应用间接时，标准是否要求指针变量的左值到值的转换？ / a>。

As far as I can tell the results are prvalues but that is just a speculative. This seems to be under-specified in a similar way to the value categories of operands which is covered in What is the value category of the operands of C++ operators when unspecified? and Does the standard mandate an lvalue-to-rvalue conversion of the pointer variable when applying indirection?.

我们可以从 3.10 部分看到Lvalues和Rvalues下面的注释：

We can see from section 3.10 Lvalues and rvalues has the following note:

第5节中每个内置操作符的讨论表明它产生的值的
类别，

但我们可以看到 code>仅在少数情况下明确说明了值类别。在这个特定问题的情况下，仅仅明确地说明& 和〜的结果的值类别 code>。

but as we can see section 5 only spells out the value category explicitly in a few cases. In the case of this particular question only spells out explicitly the value category of the result for & and ~.

即使不足，我们可以找到指向一致方向的线索。我们可以提出一个参数，说明>> ，<< ， ^ 和 | 转换为 prvalues 。这不会决定结果的值类别，但它似乎排除了结果 xvalue 根据 5 段落 7 其中包含以下注释：

Even though it is underspecified we can find clues that point us in a consistent direction. We can make an argument that the operands for >>, <<, ^ and | are converted to prvalues. This does not dictate the value category of the result but it does seem to exclude the result being an xvalue as per section 5 paragraph 7 which has the following note:

如果是xvalue，表达式是：

An expression is an xvalue if it is:

- 调用
函数的结果，无论是隐式还是显式，返回类型是对象类型的
值，

— the result of calling a function, whether implicitly or explicitly, whose return type is an rvalue reference to object type,

- 转换为
对象类型的右值引用，

— a cast to an rvalue reference to object type,

- 类成员访问表达式指定
非静态数据成员非引用类型，其中对象
表达式是xvalue，或

— a class member access expression designating a non-static data member of non-reference type in which the object expression is an xvalue, or

- a $ *
中的指向成员表达式第一个操作数是xvalue，第二个操作数是一个
指向数据成员的指针。

— a .* pointer-to-member expression in which the first operand is an xvalue and the second operand is a pointer to data member.

一般来说，这个规则的效果是
命名的右值引用被视为左值，未命名的右值
引用对象被视为xvalues;

In general, the effect of this rule is that named rvalue references are treated as lvalues and unnamed rvalue references to objects are treated as xvalues; rvalue references to functions are treated as lvalues whether named or not.

我没有看到任何合理的论据，结果可能是一个左值，所以基本上留给我们一个 prvalue 。

I don't see any reasonable argument that the result could be an lvalue so that basically leaves us with a prvalue.

因此细节如下：

〜和& 5.3.1 一元运算符段 2 ：

每个下列一元运算符的结果是一个prvalue。

The result of each of the following unary operators is a prvalue.

5.8 段落 部分中说明的部分中包含的整体促销：

The shift operators require the integral promotions which is covered in section 5.8 Shift operators paragraph 1 which says:

操作数应为整数或无范围的枚举类型，并执行整数提升。

The operands shall be of integral or unscoped enumeration type and integral promotions are performed.

，我们可以看到积分促销活动需要 4.5 整体促销部分中的 prvalues 开头为：

and we can see that the integral promotions require prvalues from section 4.5 Integral promotions which says in every paragraph starts with:

一个价值[...]

A prvalue of [...]

^ 和 | 需要通常的算术转换说：

Both ^ and | require the usual arithmetic conversions and both say:

运算符仅适用于整数或无范围的枚举操作数

operator applies only to integral or unscoped enumeration operands

b $ b

，因此常用算术转换的最后一句适用于：

，将对两个操作数执行积分促销（4.5）。 ⁵⁹

Luc Danton在回答经验确定C ++ 11表达式的值类别？。该方法使用以下代码：

Luc Danton has an empirical approach to determining the value category of an expression in his answer to Empirically determine value category of C++11 expression?. The approach uses the following code:

template<typename T>
struct value_category {
    // Or can be an integral or enum value
    static constexpr auto value = "prvalue";
};

template<typename T>
struct value_category<T&> {
    static constexpr auto value = "lvalue";
};

template<typename T>
struct value_category<T&&> {
    static constexpr auto value = "xvalue";
};

// Double parens for ensuring we inspect an expression,
// not an entity
#define VALUE_CATEGORY(expr) value_category<decltype((expr))>::value

，答案概述逻辑如下：

一个左值表达式导致一个左值引用类型，
中的x值，右值引用类型，以及只是类型中的prvalue。

an lvalue expression results in an lvalue reference type, an xvalue in an rvalue reference type, and a prvalue in just the type.

以下示例都会生成 prvalue （ see it live

int x = 10, y = 2 ;
int &xr = x ;
int &yr = y ;

std::cout << VALUE_CATEGORY( x << y ) << std::endl ;
std::cout << VALUE_CATEGORY( 10 << 2 ) << std::endl ;
std::cout << VALUE_CATEGORY( xr << yr ) << std::endl ;

std::cout << VALUE_CATEGORY( x | y ) << std::endl ;
std::cout << VALUE_CATEGORY( 10 | 2 ) << std::endl ;

std::cout << VALUE_CATEGORY( x ^ y ) << std::endl ;
std::cout << VALUE_CATEGORY( 10 ^ 2 ) << std::endl ;

std::cout << VALUE_CATEGORY( sizeof( int ) ) << std::endl ;
std::cout << VALUE_CATEGORY( sizeof( x ) ) << std::endl ;

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