c++ - 运算符删除标准行为

Question

我无法理解为 void* == nullptr 调用 operator delete 需要做什么标准。

像这样:

void foo(void* ptr) // ptr == nullptr here
{
    delete ptr;
}

一方面，我们在标准中有如下声明:

ISO/IEC 14882:2011

5.3.5 删除[expr.delete]

1 ... The operand shall have a pointer to object type, or a class type having a single non-explicit conversion function (12.3.2) to a pointer to object type. The result has type void.78

78) This implies that an object cannot be deleted using a pointer of type void* because void is not an object type.

这使得删除 void* 的代码格式错误。另一方面，我们还有一个关于delete中nullptr的声明:

ISO/IEC 14882:2011

5.3.5 删除[expr.delete]

2 ... In the first alternative (delete object), the value of the operand of delete may be a null pointer value, a pointer to a non-array object created by a previous new-expression, or a pointer to a subobject (1.8) representing a base class of such an object (Clause 10). If not, the behavior is undefined. In the second alternative (delete array), the value of the operand of delete may be a null pointer value or a pointer value that resulted from a previous array new-expression.79 If not, the behavior is undefined.

在这种情况下需要做什么实现？

Answer 1

Null 和 void* 是两个不同的东西:

delete static_cast<int*>(nullptr); // deleting null pointer, of int*

您给定的代码可能格式错误，但它与指针的值(可能为空)无关，但它的类型(不能是 void*)。