且构网

分享程序员开发的那些事...
且构网 - 分享程序员编程开发的那些事

如何廉价地将 C 样式数组分配给 std::vector?

更新时间:2023-09-06 23:25:34

当前的 std::vector 不提供任何功能或接口来获取先前分配的存储的所有权.据推测,意外传递堆栈地址太容易了,导致问题多于解决的问题.

如果您想避免复制到向量中,您要么需要在整个调用链中使用向量,要么始终使用 float[] 以 C 方式进行.你不能混合它们.你可以保证 &vec[0] 将等同于 C 数组,但完全连续,所以在整个程序中使用向量可能是可行的.>

Currently I do the following:

// float *c_array = new float[1024];

void Foo::foo(float *c_array, size_t c_array_size) {
  //std::vector<float> cpp_array;

  cpp_array.assign(c_array, c_array + c_array_size);
  delete [] c_array;
}

How can I optimize this assigning? I would like not to perform elementwise copy but just swap pointers.

The current std::vector doesn't provide any capability or interface to take ownership of previously allocated storage. Presumably it would be too easy to pass a stack address in by accident, allowing more problems than it solved.

If you want to avoid copying into a vector, you'll either need to use vectors through your entire call chain, or do it the C way with float[] the entire time. You can't mix them. You can guaranteed that &vec[0] will be equivalent to the C-array though, fully contiguous, so using vector in the whole program may be feasible.