This is a case where you should really measure.
And I'm looking at the OP's copy assignment operator and seeing inefficiency:
A& operator=(A const& other)
{A temp = other; std::swap(*this, temp); return *this;}
What if *this and other have the same s?
It seems to me that a smarter copy assignment could avoid making a trip to the heap if s == other.s. All it would have to do is the copy:
A& operator=(A const& other)
{
if (this != &other)
{
if (s != other.s)
{
delete [] p;
p = nullptr;
s = 0;
p = new int[other.s];
s = other.s;
}
std::copy(other.p, other.p + s, this->p);
}
return *this;
}
If you don't need strong exception safety, only basic exception safety on copy assignment (like std::string, std::vector, etc.), then there is a potential performance improvement with the above. How much? Measure.
I've coded this class three ways:
Design 1:
Use the above copy assignment operator and the OP's move assignment operator #1.
Design 2:
Use the above copy assignment operator and the OP's move assignment operator #2.
Design 3:
DeadMG's copy assignment operator for both copy and move assignment.
Here is the code I used to test:
#include <cstddef>
#include <algorithm>
#include <chrono>
#include <iostream>
struct A
{
std::size_t s;
int* p;
A(std::size_t s) : s(s), p(new int[s]){}
~A(){delete [] p;}
A(A const& other) : s(other.s), p(new int[other.s])
{std::copy(other.p, other.p + s, this->p);}
A(A&& other) : s(other.s), p(other.p)
{other.s = 0; other.p = nullptr;}
void swap(A& other)
{std::swap(s, other.s); std::swap(p, other.p);}
#if DESIGN != 3
A& operator=(A const& other)
{
if (this != &other)
{
if (s != other.s)
{
delete [] p;
p = nullptr;
s = 0;
p = new int[other.s];
s = other.s;
}
std::copy(other.p, other.p + s, this->p);
}
return *this;
}
#endif
#if DESIGN == 1
// Move assignment operator #1
A& operator=(A&& other)
{
swap(other);
return *this;
}
#elif DESIGN == 2
// Move assignment operator #2
A& operator=(A&& other)
{
delete [] p;
s = other.s;
p = other.p;
other.s = 0;
other.p = nullptr;
return *this;
}
#elif DESIGN == 3
A& operator=(A other)
{
swap(other);
return *this;
}
#endif
};
int main()
{
typedef std::chrono::high_resolution_clock Clock;
typedef std::chrono::duration<float, std::nano> NS;
A a1(10);
A a2(10);
auto t0 = Clock::now();
a2 = a1;
auto t1 = Clock::now();
std::cout << "copy takes " << NS(t1-t0).count() << "ns\n";
t0 = Clock::now();
a2 = std::move(a1);
t1 = Clock::now();
std::cout << "move takes " << NS(t1-t0).count() << "ns\n";
}
Here is the output I got:
$ clang++ -std=c++11 -stdlib=libc++ -O3 -DDESIGN=1 test.cpp
$ a.out
copy takes 55ns
move takes 44ns
$ a.out
copy takes 56ns
move takes 24ns
$ a.out
copy takes 53ns
move takes 25ns
$ clang++ -std=c++11 -stdlib=libc++ -O3 -DDESIGN=2 test.cpp
$ a.out
copy takes 74ns
move takes 538ns
$ a.out
copy takes 59ns
move takes 491ns
$ a.out
copy takes 61ns
move takes 510ns
$ clang++ -std=c++11 -stdlib=libc++ -O3 -DDESIGN=3 test.cpp
$ a.out
copy takes 666ns
move takes 304ns
$ a.out
copy takes 603ns
move takes 446ns
$ a.out
copy takes 619ns
move takes 317ns
DESIGN 1 looks pretty good to me.
Caveat: If the class has resources that need to be deallocated "quickly", such as mutex lock ownership or file open-state ownership, the design-2 move assignment operator could be better from a correctness point of view. But when the resource is simply memory, it is often advantageous to delay deallocating it as long as possible (as in the OP's use case).
Caveat 2: If you have other use cases you know to be important, measure them. You might come to different conclusions than I have here.
Note: I value performance over "DRY". All of the code here is going to be encapsulated within one class (struct A). Make struct A as good as you can. And if you do a sufficiently high quality job, then your clients of struct A (which may be yourself) won't be tempted to "RIA" (Reinvent It Again). I much prefer to repeat a little code within one class, rather than repeat the implementation of entire classes over and over again.
