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可以借助解引用操作变通地对 std::unique_ptr 进行深拷贝 #32

@ltimaginea

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@ltimaginea

// std::unique_ptr<MyTime> mt3 = mt1; // error

课件这里,其实可以这样进行拷贝: std::unique_ptr<MyTime> mt3 = std::make_unique<MyTime>(*mt1);

虽然 std::unique_ptr 删除了 copy constructor 和 copy assignment operator ,但其实我们可以借助解引用操作变通地对 std::unique_ptr 进行拷贝。

deep copy 示例如下:

std::unique_ptr<std::string> up1(std::make_unique<std::string>("Good morning"));

// copy construct!
std::unique_ptr<std::string> up2(std::make_unique<std::string>(*up1));
// safe copy construct!
std::unique_ptr<std::string> up3(up1 ? std::make_unique<std::string>(*up1) : nullptr);
// copy assignment!
up2 = std::make_unique<std::string>(*up1);
// safe copy assignment!
up3 = up1 ? std::make_unique<std::string>(*up1) : nullptr;

其它的例证:

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