Static method class instantiation

Question

How does a c++ line like this Foo t3 = Foo::construct(true); work when the default constructor is private? My assumption (which is obviously incorrect), is that the default constructor is called, followed my the assignment operator. That assumption must be incorrect because the default constructor is private, and cannot be called.

A concrete example:

class Foo {
private:
   Foo() {}
   bool bar;
public:
   Foo(bool t): bar(t) {}
   static Foo construct(bool t) {
      Foo temp; //calling private constructor;
      temp.bar = t;
      return temp;
   }
}

And a test method for instantiating this class looks like:

int main() {
   //Foo t1; //Not allowed, compile error, Foo() is private
   Foo t2(true); //Constructor, valid use
   Foo t3 = Foo::construct(true); //It works! Why?
   return 0;
}

What is really happening behind the scenes when t3 is instantiated?

`Foo t3 = Foo::construct(true);` makes a copy(symantically). What is wrong with that? — NathanOliver, Aug 26 '16 at 16:38
@NathanOliver I'm actually going unsure if the copy constructor would be auto generated for this case, would it? — πάντα ῥεῖ, Aug 26 '16 at 16:39
@πάνταῥεῖ per 12.8 *If the class definition does not explicitly declare a copy constructor, one is declared implicitly.* — NathanOliver, Aug 26 '16 at 16:40
@NathanOliver THX, as mentioned I was unsure if copy constructor and copy assignment operator would be auto generated for that case. — πάντα ῥεῖ, Aug 26 '16 at 16:42

Slava · Accepted Answer · 2016-08-26T17:00:24.617

3

Foo t3 = Foo::construct(true); //It works! Why?

Because this is not default initiliazation followed by assignment but copy initialization

1) when a named variable (automatic, static, or thread-local) of a non-reference type T is declared with the initializer consisting of an equals sign followed by an expression.

So according to this statement:

If T is a class type and the cv-unqualified version of the type of other is T or a class derived from T, the non-explicit constructors of T are examined and the best match is selected by overload resolution. The constructor is then called to initialize the object.

and this:

If other is an rvalue expression, move constructor will be selected by overload resolution and called during copy-initialization. There is no such term as move-initialization.

implicitly declared move constructor is used in your case.

edited Aug 26 '16 at 17:00

answered Aug 26 '16 at 16:40

Slava

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Thanks for the reference links! In my own words, a copy constructor is created by the compiler and used in the instantiation of t3. – SyntaxRules Aug 26 '16 at 16:52
Assuming c++11. Could move schematics be involved as well? – SyntaxRules Aug 26 '16 at 16:57
See updated answer, yes implicit move ctor is used in your case. – Slava Aug 26 '16 at 17:01

score 2 · Answer 2 · answered Aug 26 '16 at 16:39

2

Your line doesn't call the default constructor OR the copy-assignment operator.

It copy-constructs t3 from the temporary object returned from construct.

answered Aug 26 '16 at 16:39

Mark B

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score 0 · Answer 3 · answered Aug 26 '16 at 16:41

0

The static construct method belongs to the class and can access the private constructor - this is often used when implementing the factory pattern.

answered Aug 26 '16 at 16:41

grae22

104
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This may be useful: http://stackoverflow.com/questions/2062560/what-is-the-use-of-making-constructor-private-in-a-class – grae22 Aug 26 '16 at 16:46
The essence of the question is not if this pattern works. As you have proven, it does. But how t3 is constructed in my main method. What constructor is it using? – SyntaxRules Aug 26 '16 at 16:48

Static method class instantiation

3 Answers3