将数组的原始指针转换为unique_ptr [英] Cast raw pointer of array to unique_ptr

问题描述

我正在使用黑盒框架(cdg)，该框架用值填充uint32_t的数组. 呼叫看起来像这样:

I am working against a blackbox framework (cdg), which fills an array of uint32_t with values. The call looks like that:

std::size_t dataCount = 100;
uint32_t* data = new uint32_t[dataCount];
cdg.generate(data);

不幸的是，该框架未使用模板，因此我必须传递uint32_t*.为了摆脱原始指针，我想将其包装"到std::unique_ptr<uint32_t>中.因此，这是一个数组，我想我必须使用 std::unique_ptr<uint32_t[]>.有没有一种方法可以将原始指针转换为unique_ptr，或者我应该做类似的事情:

Unfortunately, the framework doesn't use templates, so I have to pass in a uint32_t*. To get rid of the raw pointer I want to "wrap" it into a std::unique_ptr<uint32_t>. Thus it is an array I think I have to use a std::unique_ptr<uint32_t[]>. Is there a way to convert the raw pointer into a unique_ptr or should I do something like that:

const std::size_t dataCount = 100;
std::unique_ptr<uint32_t[]> data = std::make_unique<uint_32_t[]>(dataCount);
cdg.generate(data.get());

推荐答案

智能指针非常有用，但最好考虑使用std::vector，因为您注意到它是一个数组.

Smart pointers are great and all, but perhaps consider using std::vector, since you note that it is an array.

您可以使用获取指向数据的指针，因为它是C数组(对于您的旧版接口). data() .

You can get a pointer to the data as it were a C array (for your legacy interface) with data().

std::vector<uint32_t> nums(100);   // creates a vector of size 100
uint32_t *ptr = nums.data();
cdg.generate(ptr);                 // passes uint32_t* as needed

如果您需要确保唯一的所有权，那么unique_ptr是正确的选择.但这听起来像是您确实只想要一些可以清理数组的东西，但仍然提供了C方法的原始指针.

If you need to ensure unique ownership, unique_ptr is the right choice. But it sounded like you really just wanted something that will clean up your array, but still gives a raw pointer for C methods.

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