Thread: constructor for POD types?

Thanks CornedBee,


From yout post #13 about quoted standard, it does not cover your point (2) below about for POD it is not initialized.

However, can not test. I am not sure whether debugging into assembly code could help or we can see the final source code (when instantise T to int, what code looks like) to verify?

Originally Posted by CornedBee

1) This covers the behaviour for default-initialization.
2) This covers the behaviour of the statement "T t;". Note that it does not cause default-initialization for primitives, and thus doesn't conflict with #1.

regards,
George

George, why are you trying to test every single statement in the C/C++ standard? That seems pretty counter-productive to me.

>> does any Spec covers T t does not make zero-initialization for POD type?
anon's post mentions that area. Please refer to it as the standard, not the "Spec". People may not know what you mean by Spec, and there is one C++ standard that everybody can refer to.

Also, if you haven't already, please spend the $15 USD or whatever it is to get a PDF copy of the C++ standard so you can look these things up on your own. (However, if you do that please don't ask clarification questions on what the standard means, it might drive people crazy.)

Originally Posted by CornedBee

You can't, because the default constructor of PODs does nothing.

I'm confused about the terminology - Stroustrup in TC++PL (section 10.4.2, page 244) says "Built-in types also have default constructors (Section 6.2.8)." So, is int() (for example), which does zero initialization, a default constructor, or not?

Thanks anon,

Originally Posted by anon

The standard (draft) says a little further: "Otherwise, if no initializer is specified for an object, the object and its subobjects, if any, have an indeterminate initial value."

That should cover T t; if T is for example int or some other POD. This should be quite common knowledge in C++, and I don't see why you wouldn't trust your compiler on this one.

Not sure whether you mean for POD, it is zero-initialized?


Thanks CornedBee,


Sorry that I am lost in the context of discussion. Do you mean for POD types, it is zero-initialized?

Originally Posted by CornedBee

It is. As I said, my earlier wording was inaccurate.

regards,
George

>> Not sure whether you mean for POD, it is zero-initialized?
No, that's not what anon means.

>> Do you mean for POD types, it is zero-initialized?
No, that's not what CornedBee means.

>> Sorry that I am lost in the context of discussion.
Please give more context for your question if you truly want it answered. Many different cases has been covered in this thread. If you have a question about one of the cases, find where it is mentioned and ask if you need clarification. I think your answers are all here.

Thanks Daved,


I should say sorry to make this thread a little in-convenient to read. Too many people and cluttered.

Let me repeat and simplify my question, suppose we are writing a template (template function or template class), for statement like,

(suppose T is type parameter of a template, and T is a POD type)

1. T t;
2. T();

In both cases, for the initialization of new object, will be zero-initialized, default-initialized or not initialized by standard?

Thanks and let me know if my question is not clear and I can enhance it. :-)

Originally Posted by Daved

>> Not sure whether you mean for POD, it is zero-initialized?
No, that's not what anon means.

>> Do you mean for POD types, it is zero-initialized?
No, that's not what CornedBee means.

>> Sorry that I am lost in the context of discussion.
Please give more context for your question if you truly want it answered. Many different cases has been covered in this thread. If you have a question about one of the cases, find where it is mentioned and ask if you need clarification. I think your answers are all here.

regards,
George

That question has been answered in this thread. To summarize:

(suppose T is type parameter of a template, and T is a POD type)
1. T t; will leave t uninitialized.
2. T(); will zero-initialize.

Code:

#include <iostream>

using namespace std;

template <typename T>
struct Test
{
	Test()
	:	m_Zero( T() ) {}

	T	m_Zero;
};

int main()
{
	int i;
	int j = int();
	Test<int> test;

	cout << "i = " << i << endl
	     << "j = " << j << endl
	     << "test.m_Zero = " << test.m_Zero << endl;

	return 0;
}

Output is this:

Code:

i = 4206628
j = 0
test.m_Zero = 0

Thanks for your great comments and summary, Daved!


English is not my native language, so sometimes confused.

Originally Posted by Daved

That question has been answered in this thread. To summarize:

(suppose T is type parameter of a template, and T is a POD type)
1. T t; will leave t uninitialized.
2. T(); will zero-initialize.

Thanks cpjust,


Your reply is great, conforming to Daved's comments in post #25.

Originally Posted by cpjust

Code:

#include <iostream>

using namespace std;

template <typename T>
struct Test
{
	Test()
	:	m_Zero( T() ) {}

	T	m_Zero;
};

int main()
{
	int i;
	int j = int();
	Test<int> test;

	cout << "i = " << i << endl
	     << "j = " << j << endl
	     << "test.m_Zero = " << test.m_Zero << endl;

	return 0;
}

Output is this:

Code:

i = 4206628
j = 0
test.m_Zero = 0

regards,
George