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How can I check if a move constructor is being generated implicitly?

I have several classes for which I wish to check whether a default move constructor is being generated. Is there a way to check this (be it a compile-time assertion, or parsing the generated object files, or something else)?


Motivational example:

class MyStruct : public ComplicatedBaseClass {
    std::vector<std::string> foo; // possibly huge
    ComplicatedSubObject bar;
};

If any member of any base or member of either Complicated...Object classes cannot be moved, MyStruct will not have its implicit move constructor generated, and may thus fail to optimize away the work of copying foo, when a move could be done, even though foo is movable.


I wish to avoid:

  1. tediously checking the conditions for implicit move ctor generation,
  2. explicitly and recursively defaulting the special member functions of all affected classes, their bases, and their members—just to make sure a move constructor is available.

I have already tried the following and they do not work:

  1. use std::move explicitly—this will invoke the copy constructor if no move constructor is available.
  2. use std::is_move_constructible—this will succeed when there is a copy constructor accepting const Type&, which is generated by default (as long as the move constructor is not explicitly deleted, at least).
  3. use nm -C to check the presence of move constructor (see below). However, an alternative approach is viable (see answer).

I tried looking at the generated symbols of a trivial class like this:

#include <utility>

struct MyStruct {
    MyStruct(int x) : x(x) {}
    //MyStruct(const MyStruct& rhs) : x(rhs.x) {}
    //MyStruct(MyStruct&& rhs) : x(rhs.x) {}
    int x;
};

int main() {
    MyStruct s1(4);
    MyStruct s2(s1);
    MyStruct s3(std::move(s1));
    return s1.x + s2.x + s3.x; // Make sure nothing is optimized away
}

The generated symbols looks like this:

$ CXXFLAGS="-std=gnu++11 -O0" make -B x; ./x; echo $?; nm -C x | grep MyStruct | cut -d' ' -f3,4,5
g++ -std=gnu++11 -O0    x.cc   -o x
12
.pdata$_ZN8MyStructC1Ei
.pdata$_ZSt4moveIR8MyStructEONSt16remove_referenceIT_E4typeEOS3_
.text$_ZN8MyStructC1Ei
.text$_ZSt4moveIR8MyStructEONSt16remove_referenceIT_E4typeEOS3_
.xdata$_ZN8MyStructC1Ei
.xdata$_ZSt4moveIR8MyStructEONSt16remove_referenceIT_E4typeEOS3_
MyStruct::MyStruct(int)
std::remove_reference<MyStruct&>::type&&

The output is the same when I explicitly default the copy and move constructors (no symbols).

With my own copy and move constructors, the output looks like this:

$ vim x.cc; CXXFLAGS="-std=gnu++11 -O0" make -B x; ./x; echo $?; nm -C x | grep MyStruct | cut -d' ' -f3,4,5
g++ -std=gnu++11 -O0    x.cc   -o x
12
.pdata$_ZN8MyStructC1Ei
.pdata$_ZN8MyStructC1EOKS_
.pdata$_ZN8MyStructC1ERKS_
.pdata$_ZSt4moveIR8MyStructEONSt16remove_referenceIT_E4typeEOS3_
.text$_ZN8MyStructC1Ei
.text$_ZN8MyStructC1EOKS_
.text$_ZN8MyStructC1ERKS_
.text$_ZSt4moveIR8MyStructEONSt16remove_referenceIT_E4typeEOS3_
.xdata$_ZN8MyStructC1Ei
.xdata$_ZN8MyStructC1EOKS_
.xdata$_ZN8MyStructC1ERKS_
.xdata$_ZSt4moveIR8MyStructEONSt16remove_referenceIT_E4typeEOS3_
MyStruct::MyStruct(int)
MyStruct::MyStruct(MyStruct&&)
MyStruct::MyStruct(MyStruct const&)
std::remove_reference<MyStruct&>::type&& std::move<MyStruct&>(MyStruct&)

So it appears this approach also doesn't work.


However if the target class has a member with explicit move constructor, the implicitly generated move constructor will be visible for the target class. I.e. with this code:

#include <utility>

struct Foobar {
    Foobar() = default;
    Foobar(const Foobar&) = default;
    Foobar(Foobar&&) {}
};

struct MyStruct {
    MyStruct(int x) : x(x) {}
    int x;
    Foobar f;
};
int main() {
    MyStruct s1(4);
    MyStruct s2(s1);
    MyStruct s3(std::move(s1));
    return s1.x + s2.x + s3.x; // Make sure nothing is optimized away
}

I will get the symbol for MyStruct's move constructor, but not the copy constructor, as it appears to be fully implicit. I presume the compiler generates a trivial inlined move constructor if it can, and a non-trivial one if it must call other non-trivial move constructors. This still doesn't help me with my quest though.

like image 231
Irfy Avatar asked Nov 26 '15 13:11

Irfy


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Are move constructors automatically generated?

If a copy constructor, copy-assignment operator, move constructor, move-assignment operator, or destructor is explicitly declared, then: No move constructor is automatically generated. No move-assignment operator is automatically generated.

What does implicit move constructor do?

Implicitly-defined move constructor For non-union class types (class and struct), the move constructor performs full member-wise move of the object's bases and non-static members, in their initialization order, using direct initialization with an xvalue argument.

When copy constructor is implicitly called?

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Why should move constructor be Noexcept?

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1 Answers

Declare the special member functions you want to exist in MyStruct, but don't default the ones you want to check. Suppose you care about the move functions and also want to make sure that the move constructor is noexcept:

struct MyStruct {
    MyStruct() = default;
    MyStruct(const MyStruct&) = default;
    MyStruct(MyStruct&&) noexcept; // no = default; here
    MyStruct& operator=(const MyStruct&) = default;
    MyStruct& operator=(MyStruct&&); // or here
};

Then explicitly default them, outside the class definition:

inline MyStruct::MyStruct(MyStruct&&) noexcept = default;
inline MyStruct& MyStruct::operator=(MyStruct&&) = default;

This triggers a compile-time error if the defaulted function would be implicitly defined as deleted.

like image 161
T.C. Avatar answered Oct 18 '22 15:10

T.C.