如何转换 std::vector<vector>作废* [英] how to convert std::vector<vector> to void*
问题描述
我想知道如何将 std::vector 转换为 void*，例如:
I am wondering how can I convert std::vector<vector> to void*, for example:
std::vector<ColorData> pixels_ (w*h, background_color);
现在我想将 pixels_ 转换为 void* 以便我可以 memcpy pixels_.
Now I want to convert pixels_ to void* so that I can memcpy the pixels_.
memcpy ((void*)destinationBuffer->pixels_, (void*)sourceBuffer->pixels_, \
sizeof(ColorData)*destinationBuffer->width_*destinationBuffer->height_);
但是当我运行此代码时，我收到一个错误消息:
But when I run this code, I get an error which says:
从类型‘std::vector
invalid cast from type ‘std::vector<image_tools::ColorData>’ to type ‘void*’
如何将 std::vector 转换为 void*?
推荐答案
要将向量转换为 void* 类型，有两个选项:
To transform your vector into a void* type, there are two options:
- Pre C++11:
(void*)&pixels_[0](推荐使用!empty()检查) - 自 C++11 起:
static_cast
- Pre C++11:
(void*)&pixels_[0](!empty()check is recommended) - Since C++11:
static_cast<void*>(pixels_.data())
但是，如果您想复制元素，请直接使用 STL 函数:
However, if you want to copy the elements, go straight with STL functions:
它们是类型安全的，您的代码看起来更简洁，而且大多数时候性能都相当.此外，与 std::memcpy 不同的是，它还支持非平凡的可复制类型(例如 std::shared_ptr).
They are type safe, your code looks much cleaner and the performance will be on par most of the times. Also, unlike std::memcpy, it also supports non-trivially copyable types (for example std::shared_ptr).
记住:
我们应该忘记小效率，比如大约 97% 的时间:过早优化是万恶之源.然而，我们不应该在关键的 3% 中放弃我们的机会
We should forget about small efficiencies, say about 97% of the time: premature optimization is the root of all evil. Yet we should not pass up our opportunities in that critical 3%
更新:从 C++17 开始，你也可以使用 std::data，它适用于任何使用连续内存存储的容器(例如 std::vector、std::string、std::array).
Update: since C++17 you can also use std::data, which will work for any container using contiguous memory storage (e.g. std::vector, std::string, std::array).
ColorData* buffer = std::data(pixels_);
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