C++0x 仍然可以使用全局运算符 new 显式分配吗？
维基百科指出：
使用运算符 new 可以使类型无法分配：
struct NonNewable { void *operator new(std::size_t) = 删除; };这种类型的对象只能作为堆栈对象或其他类型的成员进行分配。如果没有不可移植的技巧，它不能直接进行堆分配。 （由于放置 new 是在用户分配的内存上调用构造函数的唯一方法，并且如上所述已禁止这种使用，因此无法正确构造该对象。）
删除运算符 new 类似于在当前 C++ 中将其设为私有，但不是这样t 显式使用全局运算符 new，这避免了特定于类的查找，仍然有效 C++0x？
NonNewable *p = ::new NonNewable();
// neither non-portable nor trickery, though perhaps not widely known
我是否遗漏了草稿中的某些内容？
需要明确的是，这是有效的 C++03 并且工作正常：
struct NonNewable {
private:
void *operator new(std::size_t); // not defined
};
int main() {
// ignore the leaks, it's just an example
void *mem = operator new(sizeof(NonNewable));
NonNewable *p = ::new(mem) NonNewable();
p = ::new NonNewable();
return 0;
}
Wikipedia states:
A type can be made impossible to allocate with operator new:
struct NonNewable { void *operator new(std::size_t) = delete; };An object of this type can only ever be allocated as a stack object or as a member of another type. It cannot be directly heap-allocated without non-portable trickery. (Since placement new is the only way to call a constructor on user-allocated memory and this use has been forbidden as above, the object cannot be properly constructed.)
Deleting operator new is similar to making it private in current C++, but isn't explicitly using global operator new, which avoids class-specific lookup, still valid C++0x?
NonNewable *p = ::new NonNewable();
// neither non-portable nor trickery, though perhaps not widely known
Have I missed something in the draft?
To be clear, this is valid C++03 and works fine:
struct NonNewable {
private:
void *operator new(std::size_t); // not defined
};
int main() {
// ignore the leaks, it's just an example
void *mem = operator new(sizeof(NonNewable));
NonNewable *p = ::new(mem) NonNewable();
p = ::new NonNewable();
return 0;
}
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我相信你是对的，维基百科是错的。 C++0x 草案标准将“删除的函数”(8.4p10) 描述为不能以任何方式使用的函数（否则程序格式错误）。与普通函数不同，它们在范围或名称查找中不发挥作用。关于新表述的相关段落保持不变：
所以是的，表达式
::new NonNewable[或::new(mem) NonNewable] 将选择::operator new的重载，忽略函数NonNewable::operator new，并且不会使程序格式错误。I believe you are right and wikipedia is wrong. The C++0x draft standard describes "deleted functions" (8.4p10) as functions which may not be used in any way (or else the program is ill-formed). They play no part in scope or name lookup different from normal functions. And the relevant paragraphs concerning new expressions have remained the same:
So yes, the expression
::new NonNewable[or::new(mem) NonNewable] would choose an overload of::operator new, ignoring the functionNonNewable::operator new, and would not make the program ill-formed.