std::copy for dynamically allocated pointers

Question

This is probably a very newbie question. The class I'm working on (let's call it MyClass) contains the following variables

 int nbins;
 double * bins; // = new double[nbins+i]

Since I'm implementing a copy constructor

 MyClass( const MyClass& source )

I need to copy source.bins. I was going to use std::copy from algorithm, but all the examples I've found show how to copy static array like double bins[] = ... or standard data structure like std::vector. I can't find any example in which std::copy is used on pointers like double * bins = new double[nbins+i]. So, my question is: is legal to use std::copy on pointers, like for example

double * bins = new double[nbins+1]; //is this necessary?
std::copy( std::begin( source.bins ), std::end(source.bins), std::begin(bins) );

or should I rather rely on the C memcpy

double * bins = new double[nbins+1]; //again, is this necessary?
memcpy( source.bins, bins, (nbins+1) /* *sizeof(double) ?*/ );

So, my questions are basically two:

1) Does std::end(bins) know where is the end of bins (=new double[nbins+1])?

2) Do I need to allocate the memory for the destination before doing the copy? Or the copy (or memcpy) takes care of that?

Answer 1

1) Does std::end(bins) know where is the end of bins (=new double[nbins+1])?

No, std::begin and std::end require that the array be of fixed size.

2) Do I need to allocate the memory for the destination before doing the copy? Or the copy (or memcpy) takes care of that?

You need to allocate the memory. Both copy and memcpy assume that you provide them with a block of memory that can be written to.

So yes, you can use std::copy, but without std::begin and std::end:

double * bins = new double[nbins+1]; //is this necessary?
std::copy( source.bins , source.bins + N, bins );

where N is the length or the array pointed at by source.bins. Obviously you have to make sure you are not accessing either array out of bounds.