Why is the move constructor being called?

Question

Given my understanding of return value optimization, I am a confused as to why the move constructor is being called in the sample code below:

#include <vector>
#include <iostream>

class MyCustomType
{
public:
  MyCustomType()
  {
    std::cout << "Constructor called" << std::endl;
  }

  MyCustomType(const MyCustomType & inOther) : // copy constructor
    mData(inOther.mData)
  {
    std::cout << "Copy constructor called" << std::endl;
  }

  MyCustomType(MyCustomType && inOther) : // move constructor
    mData(std::move(inOther.mData))
  {
    std::cout << "Move constructor called" << std::endl;
  }

private:
  std::vector<int> mData;
};

MyCustomType getCustomType()
{
  MyCustomType _customType;
  return _customType;
}

int main()
{
  MyCustomType _t = getCustomType();
}

Output:

Constructor called
Move constructor called

I assumed that there would only be a single instance of MyCustomType constructed and assigned directly to _t.

For info, I am using the VC14 compiler.

Why did you assume that? Because of the C++ rules (which you would have misunderstood) or because of optimisations (which can be a bit unpredictable)? — BoBTFish, Mar 22 '17 at 07:54
@BoBTFish Both, I assumed the C++ standard permitted single-construction in such a case and hoped the compiler would be smart enough to notice it. — Harry, Mar 22 '17 at 07:58
If you only wanted a single instance constructed without any other constructor calls it would be `MyCustomType _t();` but in your post first a temporary instance is created `getCustomType()` and the assignment operator does not initiate a copy construction but a move construction. Which saves resources. C++ does not simplify your code that far, that it would convert those two statements. — Aeonos, Mar 22 '17 at 08:08
See also,http://stackoverflow.com/questions/17473753/c11-return-value-optimization-or-move . And if you compile in release mode, MSVC will likely eliminate the move as well. Note also that the move constructor normally should take a non-const argument — nos, Mar 22 '17 at 08:19
@Aeonos: You meant `MyCustomType _t{};` (as you have vexing parse: function declaration). — Jarod42, Mar 22 '17 at 08:42

Edgar Rokjān · Accepted Answer · 2017-03-22T08:34:28.593

1

In your example you assume that NRVO will be applied. But NRVO is just an optimization, which is not guaranteed to be used by the compiler ^.

I have tested your example with http://webcompiler.cloudapp.net/.

The default set of compiler options there is:

/EHsc /nologo /W4

In such case the output is similar with yours:

Constructor called

Move constructor called

But if appropriate optimization is enabled with, say, /O2 flag:

/O2 /EHsc /nologo /W4

Then the output is:

Constructor called

^ As @einpoklum has mentioned in the comments, rules have changed since C++17.

edited Mar 22 '17 at 08:34

answered Mar 22 '17 at 08:20

Edgar Rokjān

17,245
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1

Actually, in C++17 I think copy elision for return values will be [guaranteed](https://jonasdevlieghere.com/guaranteed-copy-elision/#guaranteedcopyelision) under reasonable conditions. – einpoklum Mar 22 '17 at 08:23
@einpoklum Good remark! Anyway, OP uses VC14 compiler :) – Edgar Rokjān Mar 22 '17 at 08:24
1

@einpoklum It's only guaranteed when returning a prvalue; at least not for the OP's case. – songyuanyao Mar 22 '17 at 08:25
@songyuanyao: Fair enough, but this answer says that NRVO is just an optimization, so I had to qualify that. – einpoklum Mar 22 '17 at 08:26
@einpoklum Yes. It might be more precise that the NRVO for OP's case is just an optimization. – songyuanyao Mar 22 '17 at 08:27
@einpoklum: prvalues are never *named*. Therefore, guaranteed elision never applies to named RVO. – Nicol Bolas Mar 22 '17 at 15:08

score 1 · Answer 2 · answered Mar 22 '17 at 08:34

1

Because expression getCustomType() is a prvalue(rvalue), and parameter can be more cv-qualified than argument.

In order to see "more" copy elision, you can try release compiling mode.

BTW, inOther.mData isn't moved, mData(std::move(inOther.mData)) will in fact call vector's copy constructor.

answered Mar 22 '17 at 08:34

felix

2,213
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Thanks, modified the move constructor to be non-const. – Harry Mar 22 '17 at 08:40

Why is the move constructor being called?

2 Answers2