initializing static class variables

[Neil Zanella]

Hello,

When an "std::map<int, int> foobar;" statement appears in a C++ program
followed by a statement of the form "foobar[123];" where nothing has ever
been inserted at position 123 into map foobar, the C++ standard states
that a new element is created at position 123 and is initialized to
zero. That is, in this case, the pair (123, 0) is inserted into the
map.

I am interested in knowing what the standard says should happen in the
following case. Does the standard mandate that struct Foo's pointers be
initialized to zero as well, upon reference to hello[110], or does this
program lead to undefined behavior.

Thank you for your feedback and explanations,

Neil

#include <iostream>
#include <map>

struct Foo {
int *x;
int *y;
};

typedef std::map<int, Foo> Hello;

int main() {
Hello hello;
//initialize entiry structure, hence also its pointers, to zero
//(I believe this is according to the ANSI/ISO C++ standard too)
std::cout << hello[110].x << std::endl;
//hence dereferencing zero memory address causes segfault
//std::cout << *hello[110].x << std::endl;
//hence same goes for assigning
*hello[110].x = 112;
}

 

[Dave Townsend]

I believe the STL standard says an object will be constructed
with the default constructor in this situation. Since you don't have
one, the values of x and y will be undefined (garbage) when you
insert a new key

dave

 

[John Harrison]

I believe the STL standard says an object will be constructed
with the default constructor in this situation. Since you don't have
one, the values of x and y will be undefined (garbage) when you
insert a new key

dave

I don't believe that is correct, the standard says that hello[110] is
equivalvent to

(*((insert(make_pair(100,Foo()))).first)).second;

he crucial bit is Foo() which is value initialisation not default
initialisation. For value initialisation of Foo, x and y will be zero
initialised.

However I think value iniitalisation is a fairly new concept in the
C++ standard and I wouldn't be surprised if a compiler got this wrong.

Also I'm not completely certain of this, and I'm really just posting to
put the alternative view. The releveant sections of the standard are 8.5,
23.3.1.2, 5.2.3 for anyone who want's to check this out. Maybe one of the
gurus could clear this up.

Either way the safe thing to do is to add a default constructor to Foo.

john

Hello,

When an "std::map<int, int> foobar;" statement appears in a C++ program
followed by a statement of the form "foobar[123];" where nothing has
ever
been inserted at position 123 into map foobar, the C++ standard states
that a new element is created at position 123 and is initialized to
zero. That is, in this case, the pair (123, 0) is inserted into the
map.

I am interested in knowing what the standard says should happen in the
following case. Does the standard mandate that struct Foo's pointers be
initialized to zero as well, upon reference to hello[110], or does this
program lead to undefined behavior.

Thank you for your feedback and explanations,

Neil

#include <iostream>
#include <map>

struct Foo {
int *x;
int *y;
};

typedef std::map<int, Foo> Hello;

int main() {
Hello hello;
//initialize entiry structure, hence also its pointers, to zero
//(I believe this is according to the ANSI/ISO C++ standard too)
std::cout << hello[110].x << std::endl;
//hence dereferencing zero memory address causes segfault
//std::cout << *hello[110].x << std::endl;
//hence same goes for assigning
*hello[110].x = 112;
}

 

[Dave Townsend]

John,

thanks for point out this subtlety. Can you post an example which
demonstrates the difference, I've heard mention of these two types
of initialization but never got on to what the difference was.

thanks, dave

I believe the STL standard says an object will be constructed
with the default constructor in this situation. Since you don't have
one, the values of x and y will be undefined (garbage) when you
insert a new key

dave

I don't believe that is correct, the standard says that hello[110] is
equivalvent to

(*((insert(make_pair(100,Foo()))).first)).second;

he crucial bit is Foo() which is value initialisation not default
initialisation. For value initialisation of Foo, x and y will be zero
initialised.

However I think value iniitalisation is a fairly new concept in the
C++ standard and I wouldn't be surprised if a compiler got this wrong.

Also I'm not completely certain of this, and I'm really just posting to
put the alternative view. The releveant sections of the standard are 8.5,
23.3.1.2, 5.2.3 for anyone who want's to check this out. Maybe one of the
gurus could clear this up.

Either way the safe thing to do is to add a default constructor to Foo.

john

Hello,

When an "std::map<int, int> foobar;" statement appears in a C++ program
followed by a statement of the form "foobar[123];" where nothing has
ever
been inserted at position 123 into map foobar, the C++ standard states
that a new element is created at position 123 and is initialized to
zero. That is, in this case, the pair (123, 0) is inserted into the
map.

I am interested in knowing what the standard says should happen in the
following case. Does the standard mandate that struct Foo's pointers be
initialized to zero as well, upon reference to hello[110], or does this
program lead to undefined behavior.

Thank you for your feedback and explanations,

Neil

#include <iostream>
#include <map>

struct Foo {
int *x;
int *y;
};

typedef std::map<int, Foo> Hello;

int main() {
Hello hello;
//initialize entiry structure, hence also its pointers, to zero
//(I believe this is according to the ANSI/ISO C++ standard too)
std::cout << hello[110].x << std::endl;
//hence dereferencing zero memory address causes segfault
//std::cout << *hello[110].x << std::endl;
//hence same goes for assigning
*hello[110].x = 112;
}

 

[John Harrison]

John,

thanks for point out this subtlety. Can you post an example which
demonstrates the difference, I've heard mention of these two types
of initialization but never got on to what the difference was.

thanks, dave

Well I don't want to sound more certain than I am, but here goes.

8.5 para 7 of the standard says

An object whose initialiser is an empty set of parentheses i.e., (), shall
be value-initialized.

So I reckon in this code

class Bar
{
Bar() : foo() {}
Bar(int) {}
private:
Foo foo;
};

the first constructor value initialises foo, and the second default
initialises it.

Ditto

Foo* foo = new Foo(); // value initialisation
Foo* foo2 = new Foo; // default initialisation

john

 

[Neil Zanella]

Interesting, you must all be referring to the new C++ standard
available
from ANSI (at USD 18.0 for the electronic version). It also seems like
the
older 1998 standard has been discontinued.

http://webstore.ansi.org/ansidocstore/product.asp?sku=INCITS/ISO/IEC+14882-2003

So what is the difference between default initialization
and value initialization? Is this something introduced
in the 2003 standard?

Thanks,

Neil

class Foo {
public:
Foo() { }
};

class Bar {
public:
Bar(): foo() { }
Bar(int) { }
private:
Foo foo;
};

int main() {
Foo *foo = new Foo();
Foo *foo2 = new Foo;
Foo foo3;
Foo foo4();
}

 

[John Harrison]

Interesting, you must all be referring to the new C++ standard
available
from ANSI (at USD 18.0 for the electronic version). It also seems like
the
older 1998 standard has been discontinued.

http://webstore.ansi.org/ansidocstore/product.asp?sku=INCITS/ISO/IEC+14882-2003

So what is the difference between default initialization
and value initialization? Is this something introduced
in the 2003 standard?

Thanks,

Neil

I believe it was introduced in 2003 standard but I can't be sure since I
don't have the 1998 standard available.

I guess the difference is best illustrated with an example

struct Foo
{
int x;
};

The standard says (8.5 para 5) that if Foo is default initialised then its
default constructor is called. This is obviously an implicit default
constructor and 12.1 para 7 and 12.6.2 para 4 together say that x is not
initialised in this case (i.e. it has an undefined value).

Hoever 8.5 para 5 says that if Foo is value initialised then because Foo
does not have a user-declared constructor each of its members is value
initialised. And when a scalar type is value initalsed it is in fact zero
initialised.

So

#include <iostream>
using namespace std;

struct Foo
{
int x;
};

int main()
{
int* a = new int;
cout << *a << '\n';
int* b = new int();
cout << *b << '\n';
Foo* c = new Foo;
cout << c->x << '\n';
Foo* d = new Foo();
cout << d->x << '\n';
}

Output with MSVC++ 7.1

-842150451
0
-842150451
0

But I'm still learning this stuff myself, so watch out I could easily be
wrong.

john

 

[Mark]

I believe it was introduced in 2003 standard but I can't be sure since I
don't have the 1998 standard available.

I guess the difference is best illustrated with an example

struct Foo
{
int x;
};

The standard says (8.5 para 5) that if Foo is default initialised then its
default constructor is called. This is obviously an implicit default
constructor and 12.1 para 7 and 12.6.2 para 4 together say that x is not
initialised in this case (i.e. it has an undefined value).

Hoever 8.5 para 5 says that if Foo is value initialised then because Foo
does not have a user-declared constructor each of its members is value
initialised. And when a scalar type is value initalsed it is in fact zero
initialised.
[....]

But I'm still learning this stuff myself, so watch out I could easily be
wrong.

john

Default initilization when viewed from the 1998 and 03 version of the
standard seem contradictory.

The 1998 standard for C++ (ISO/IEC 14882-1998) says:

"An object whose initializer is an empty set of parentheses, i.e., (),
shall be default-initialized."

A little before that, it says:

"To default-initialize an object of type T means:
** if T is a non-POD class type (clause 9), the default constructor
for T is called (and the initialization is ill-formed if T has no
accessible default constructor);
** if T is an array type, each element is default-initialized;
** otherwise, the storage for the object is zero-initialized."

(Section 8.5, Initializers)

So given

struct data_t
{
int Idx;
int Jdx;
int Kdx;
} data;

data_t type is a POD class type (plain old data), so with empty
parentheses, it should be zero-initialized.

data_t() is not exactly a constructor call. It denotes a temporary
object of type data_t, constructed with default initialization.

So data(data_t()) initializes the object named "data" with a
temporary, default-initialized object of its type, and for this type
default initialization is zero-initialization.

Note that some compilers may not do this right. In any event, it
seems to me the 03 version of the standard refers to this as value
initilization.

Mark

 

[John Harrison]

I believe it was introduced in 2003 standard but I can't be sure since I
don't have the 1998 standard available.

I guess the difference is best illustrated with an example

struct Foo
{
int x;
};

The standard says (8.5 para 5) that if Foo is default initialised then
its
default constructor is called. This is obviously an implicit default
constructor and 12.1 para 7 and 12.6.2 para 4 together say that x is not
initialised in this case (i.e. it has an undefined value).

Hoever 8.5 para 5 says that if Foo is value initialised then because Foo
does not have a user-declared constructor each of its members is value
initialised. And when a scalar type is value initalsed it is in fact
zero
initialised.
[....]

But I'm still learning this stuff myself, so watch out I could easily be
wrong.

john

Default initilization when viewed from the 1998 and 03 version of the
standard seem contradictory.

The 1998 standard for C++ (ISO/IEC 14882-1998) says:

"An object whose initializer is an empty set of parentheses, i.e., (),
shall be default-initialized."

That's directly contradicted by the 2003 standard which says value
initialised.

A little before that, it says:

"To default-initialize an object of type T means:
** if T is a non-POD class type (clause 9), the default constructor
for T is called (and the initialization is ill-formed if T has no
accessible default constructor);
** if T is an array type, each element is default-initialized;
** otherwise, the storage for the object is zero-initialized."

(Section 8.5, Initializers)

So given

struct data_t
{
int Idx;
int Jdx;
int Kdx;
} data;

data_t type is a POD class type (plain old data), so with empty
parentheses, it should be zero-initialized.

Assuming you are talking about the 1998 standard I don't think so. Look up
the definition of compiler generated default constructor. I think you will
see that it says Idx, Jdx and Kdx remain uninitialised not zero
initialised.

data_t() is not exactly a constructor call. It denotes a temporary
object of type data_t, constructed with default initialization.

So data(data_t()) initializes the object named "data" with a
temporary, default-initialized object of its type, and for this type
default initialization is zero-initialization.

Note that some compilers may not do this right. In any event, it
seems to me the 03 version of the standard refers to this as value
initilization.

And that's the crucial difference, because value initialisation does not
use the compiler generated default constructor.

john

 

[Mark]

On Sun, 01 Aug 2004 07:36:46 +0100, "John Harrison"

[...]

Assuming you are talking about the 1998 standard I don't think so. Look up
the definition of compiler generated default constructor. I think you will
see that it says Idx, Jdx and Kdx remain uninitialised not zero
initialised.

If I run test program below through Visual Studio 7.1. Then run said
program through Borland C++ Builder 4. The end result. Builder 4 did
zero initialize the struct while Visual Studio didn't.


# include<cstdio>

class test
{
private:
struct data_t
{
int mem1;
double mem2;
char mem3;
double mem4[50];
} data;
public:
test()
:data(data_t()) //zero initialises data
{
printf( "mem1 = %d\n", data.mem1 );
printf( "mem2 = %f\n", data.mem2 );
}
};

int main(int argc, char* argv[])
{
test mytest;
}

Mark

 

[John Harrison]

On Sun, 01 Aug 2004 07:36:46 +0100, "John Harrison"

[...]

Assuming you are talking about the 1998 standard I don't think so. Look
up
the definition of compiler generated default constructor. I think you
will
see that it says Idx, Jdx and Kdx remain uninitialised not zero
initialised.

If I run test program below through Visual Studio 7.1. Then run said
program through Borland C++ Builder 4. The end result. Builder 4 did
zero initialize the struct while Visual Studio didn't.

Strange, my copy of VC++ 7.1 does zero initialise the struct. I thought
VC++ 7.1 was one of the few compilers to get this right.

john

 

[Neil Zanella]

Given the contradictions making the 1998 and 2003 C++ standards clash
on this
particular issue I would expect that some older C++ programs would
malfunction
or even segfault when compiled with a newer compiler and then run by
the user.

In particular, several books seem to emphasize that programmers should
not document language features in their code. In this case, it seems
it
could be hard to spot where a particular program is relying on some
feature
from the 1998 standard that makes it behave differently in the 2003
standard.

In any case, I have not come across any books that discuss the
differences
between default initialization and value initialization, not even
older books.
Does anyone know of any C++ books (i.e. not the standard itself) that
actually
mention this issue? If so I would be interested in knowing which ones,
as it
seems to me that default versus value initialization is somewhat of an
advanced topic (not having seen it in any C++ book).

I wonder how many people actually rely on it (perhaps not even knowing
it).

So, elaborating on the example posted by John Harrison, what does the
2003 standard say about the following example? In particular, is it
possible to
default initialize instances of Bar and FooBar in the same way, and
how
does the 2003 standard say it should be done if possible to do so?

#include <iostream>
using namespace std;

struct Foo
{
int x;
};

struct Bar
{
Bar(): x(10) { }
int x;
}

struct FooBar
{
Bar(int x): x(x) { }
int x;
}

int main()
{
int* a = new int;
cout << *a << '\n';
int* b = new int();
cout << *b << '\n';
Foo* c = new Foo;
cout << c->x << '\n';
Foo* d = new Foo();
cout << d->x << '\n';
}

Thanks,

Neil