如何廉价地将 C 样式数组分配给 std::vector? [英] How to cheaply assign C-style array to std::vector?

问题描述

目前我执行以下操作:

//float *c_array = new float[1024];void Foo::foo(float *c_array, size_t c_array_size) {//std::vectorcpp_array;cpp_array.assign(c_array, c_array + c_array_size);删除 [] c_array;}

我怎样才能优化这个分配?我不想执行元素复制，而只是交换指针.

解决方案

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.

解决方案

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.

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