Nullable< T&gt ;：和重载运算符，布尔？ &布尔 [英] Nullable<T>: and overloaded operators, bool? & bool

问题描述

1. 为什么为布尔定义了运算符&，而没有为运算符&&定义？


2. 这到底如何1）布尔？ &布尔？和2）布尔？和布尔工作？

在Nullable上还有其他有趣的运算符语义吗？任何泛型T的重载运算符？

Any other "interesting" operator semantics on Nullable? Any overloaded operators for generic T?

推荐答案

Nullable< T> 是解除运算符。这意味着什么：如果T有运算符，T？

Operators on Nullable<T> are "lifted" operators. What this means is: if T has the operator, T? will have the "lifted" counterpart.

&&和||并不是与&和-例如，它们不能被重载-根据ECMA规范14.2.2运算符重载：

&& and || aren't really operators in the same sense as & and | - for example, they can't be overloaded - from the ECMA spec 14.2.2 Operator overloading:

可重载的二进制运算符为：
+-* /％& | ^<< >> ==！=>< > =< =只有上面列出的运算符可以重载
。特别是，不可能
重载成员访问权限，
方法调用或=，&&，||，
??，？：，选中，未选中，新

The overloadable binary operators are: + - * / % & | ^ << >> == != > < >= <= Only the operators listed above can be overloaded. In particular, it is not possible to overload member access, method invocation, or the =, &&, ||, ??, ?:, checked, unchecked, new, typeof, as, and is operators.

同样，根据ECMA规范，14.2.7解除了运算符，解除运算符的是：

Likewise, from the ECMA spec, 14.2.7 Lifted operators, the lifted operators are:

对于一元运算符
+ ++--！ 〜

For the unary operators + ++ - -- ! ~

对于二进制运算符
+-* /％& | ^<< >>

For the binary operators + - * / % & | ^ << >>

对于相等运算符
==！=

For the equality operators == !=

对于关系运算符< >< => =

For the relational operators < > <= >=

因此，基本上，短路运算符未定义为提升运算符。

So basically, the short-circuiting operators aren't defined as lifted operators.

• 提升运算符：编译器提供 Nullable< T> ，基于T的运算符-例如： int +运算符被解除在 int？上定义为：

• Lifted operator: a compiler provided operator on Nullable<T>, based on the operators of T - for example: the int "+" operator gets "lifted" onto int?, defined as:

（int？x，int？y）=>（x。 HasValue& y.HasValue）吗？ （x.Value + y.Value）：（int？）null;

(int? x, int? y) => (x.HasValue && y.HasValue) ? (x.Value + y.Value) : (int?) null;

运算符重载：为给定类型提供自定义运算符实现的操作；例如十进制和 DateTime 提供各种运算符重载

Operator overloading: the act of providing a custom operator implementation for a given type; for example decimal and DateTime provide various operator overloads

短路：& 和 || 的正常行为（在许多语言中，包括C ++和C＃）-即第二个操作数可能不被评估-即

Short-circuiting: the normal behavior of && and || (in many languages, including C++ and C#) - i.e. the second operand might not be evaluated - i.e.

（expression1，expression2）=> expression1（）吗？ expression2（）：false;

(expression1, expression2) => expression1() ? expression2() : false;

或者也许是一个更简单的示例：

Or perhaps a simpler example:

bool someFlag = Method1() && Method2();

如果 Method1（）返回false，则 Method2（）不会执行（因为编译器已经知道总体答案是错误的）。如果 Method2（）有副作用，这一点很重要，因为保存到数据库...

if Method1() returns false, then Method2() isn't executed (since the compiler already knows that the overall answer is false). This is important if Method2() has side-effects, since as saving to the database...

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