为什么使用 std::auto_ptr<> 是错误的?使用标准容器?
为什么在标准容器中使用 std::auto_ptr<> 是错误的?
Why is it wrong to use std::auto_ptr<> with standard containers?
推荐答案
C++ 标准规定 STL 元素必须是可复制构造的"和可分配的".换句话说,一个元素必须能够被分配或复制,并且这两个元素在逻辑上是独立的.std::auto_ptr 不满足此要求.
The C++ Standard says that an STL element must be "copy-constructible" and "assignable." In other words, an element must be able to be assigned or copied and the two elements are logically independent. std::auto_ptr does not fulfill this requirement.
以这段代码为例:
class X
{
};
std::vector<std::auto_ptr<X> > vecX;
vecX.push_back(new X);
std::auto_ptr<X> pX = vecX[0]; // vecX[0] is assigned NULL.
要克服此限制,您应该使用 std::unique_ptr, std::shared_ptr 或 std::weak_ptr 智能指针或 boost 等价物,如果您没有 C++11.这里是这些智能指针的 boost 库文档.
To overcome this limitation, you should use the std::unique_ptr, std::shared_ptr or std::weak_ptr smart pointers or the boost equivalents if you don't have C++11. Here is the boost library documentation for these smart pointers.
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