C ++ 11 std :: copy内部函数

Question

I am writing a new cross-platform toolbox and would like to be up-to-date with the C++11 extensions. I ran into a problem, where std::copy doesn't behave as I would expect. If anyone has an explanation, I would very much appreciate this.

I completely understand the errors I get and a work-around is to use old-style memcpy, but if there is a solution, where I can use std::copy, I would appreciate this.

The error I get is:

error: no matching function for call to 'begin(const float*&)'

and I'am using gcc 4.7.2-5 with C++11 enabled.

There errors I get for the std::copy within the set function

#include <vector>
#include <array>
#include <algorithm>
#include <cstring>

class A {
public:
  float data[3];
  void set(const float c[3]) {
    std::copy(std::begin(c),std::end(c), std::begin(data));
  }
};

int main() {
  const float a[3] = {1,2,3};
  float b[] = {4,5,6};
  std::copy(std::begin(a),std::end(a), std::begin(b));
  return 0;
}

Answer 1

This function parameter

const float c[3]

is actually a pointer to float , or const float* . It is confusing, and is really used to document that the argument should be an array, but any float* could be passed to it. Inside the function, c is a pointer and that is why std::begin doesn't work.

If you want to pass a reference to an array to a function, you need a different syntax:

void set(const float (&c)[3]);

Alternatively, you can use a user-defined array type, std::array<float, 3> .

Note that std::begin and std::end live in the <iterator> header. Also note that you can use the old-school approach:

std::copy(c, c + 3, &data[0]);

Answer 2

const float c[3] as a function parameter is longhand for const float* c due to C legacy compatibility.

Try const float (&c)[3] or std::array<float,3> const& c or the like.

I like template<class T>using Type=T; Type<float[3]> const& c template<class T>using Type=T; Type<float[3]> const& c myself.