我需要将原始指针包装到 shared_ptr
中,以便将其传递给函数。该函数在返回后不保留对输入对象的任何引用。
{
MyClass i;
shared_ptr<MyClass> p(&i);
f(p);
// BAD: shared_ptr will delete i.
}
如何防止shared_ptr
删除引用的对象?
最佳答案
作为chris在评论中提到,写一个空的删除器:
#include <type_traits>
template <typename T>
struct empty_delete
{
empty_delete() /* noexcept */
{
}
template <typename U>
empty_delete(const empty_delete<U>&,
typename std::enable_if<
std::is_convertible<U*, T*>::value
>::type* = nullptr) /* noexcept */
{
}
void operator()(T* const) const /* noexcept */
{
// do nothing
}
};
使用示例:
#include <iostream>
#include <memory>
struct noisy
{
noisy() { std::cout << "alive" << std::endl; }
~noisy() { std::cout << "dead" << std::endl; }
noisy(const noisy&);
noisy& operator=(const noisy&);
};
template <typename T>
void take(T& yours)
{
std::cout << "Taking..." << std::endl;
{
auto mine = std::move(yours);
}
std::cout << "Took." << std::endl;
}
int main()
{
std::unique_ptr<noisy> a(new noisy());
std::shared_ptr<noisy> b(new noisy());
std::unique_ptr<noisy, empty_delete<noisy>> c(new noisy());
std::shared_ptr<noisy> d(new noisy(), empty_delete<noisy>());
take(a);
take(b);
take(c);
take(d);
}
输出:
alive
alive
alive
alive
Taking...
dead
Took.
Taking...
dead
Took.
Taking...
Took.
Taking...
Took.
当然,这个例子会泄漏内存。
关于c++ - 如何将原始指针包装到 shared_ptr 中并防止 shared_ptr 删除对象?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15534146/