隐式将null指针强制转换为bool [英] Implicit cast of null pointer to bool

问题描述

假设我有Foo* foo = nullptr;

如果我正在检查foo是否为nullptr，我可以写吗

If I'm checking whether or not foo is nullptr, am I permitted to write

if (!foo)

或者我应该写

if (foo == nullptr)

推荐答案

请参阅此标准参考(加粗强调):

See this standard reference (bold emphasis mine):

C ++ 11§4.12布尔转换

算术，无作用域枚举，指针或指向的指针的prvalue 成员类型可以转换为bool类型的prvalue. 零值， 空指针值，或将空成员指针值转换为 错误的;任何其他值都将转换为true.类型的prvalue std :: nullptr_t可以转换为bool类型的prvalue；这 结果值为假.

A prvalue of arithmetic, unscoped enumeration, pointer, or pointer to member type can be converted to a prvalue of type bool. A zero value, null pointer value, or null member pointer value is converted to false; any other value is converted to true. A prvalue of type std::nullptr_t can be converted to a prvalue of type bool; the resulting value is false.

中间一句很重要:它告诉您可以将空指针值(foo = nullptr) 隐式转换为本身具有bool类型的false.因此if (!foo)是定义明确的.

The middle sentence is relevant: it is telling you that the null pointer value (foo = nullptr) can be implicitly cast to false which itself has type bool. Therefore if (!foo) is well-defined.

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