Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
159 views
in Technique[技术] by (71.8m points)

c++ - What is an int() Called?

It's been rehashed over and over that primitive types don't have constructors. For example this _bar is not initialized to 0 when I call Foo():

class Foo{
    int _bar;
};

So obviously int() is not a constructor. But what is it's name?

In this example I would say i is: (constructed? initialized? fooed?)

for(int i{}; i < 13; ++i)

Loki Astari mentions here that the technique has some sort of name.

EDIT in response to Mike Seymour:

#include <iostream>

using namespace std;

class Foo{
    int _bar;
public:
    void printBar(){ cout << _bar << endl; }
};

int main()
{
    Foo foo;

    foo.printBar();

    Foo().printBar();

    return 0;
}

Running this code on Visual Studio 2013 yields:

3382592
3382592

Interestingly on gcc 4.8.1 yields:

134514651
0

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

It's been rehashed over and over that primitive types don't have constructors.

That's right.

For example this bar is not initialized to 0 when I call Foo()

Yes it is. Foo() specifies value-initialisation which, for class like this with no user-provided constructor, means it's zero-initialised before initialising its members. So _bar ends up with the value zero. (Although, as noted in the comments, one popular compiler doesn't correctly value-initialise such classes.)

It would not be initialised if you were to use default-initialisation instead. You can't do that with a temporary; but a declared variable Foo f; or an object by new F will be default-initialised. Default-initialisation of primitive types does nothing, leaving them with an indeterminate value.

It would also not be initialised if the class had a user-provided default constructor, and that constructor didn't specifically initialise _bar. Again, it would be default-initialised, with no effect.

So obviously int() is not a constructor. But what is it's name?

As an expression, it's a value-initialised temporary of type int.

Syntactically, it's a special case of an "explicit type conversion (functional notation)"; but it would be rather confusing to use that term for anything other than a type conversion.

In this example I would say i is: (constructed? initialized? fooed?)

Initialised. List-initialised (with an empty list), value-initialised, or zero-initialised, if you want to be more specific.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...