Tuesday, May 26, 2015

Digest for comp.lang.c++@googlegroups.com - 25 updates in 6 topics

Victor Bazarov <v.bazarov@comcast.invalid>: May 26 07:47AM -0400

On 5/22/2015 7:41 PM, Öö Tiib wrote:
> it is interger zero or std::nullptr_t.
 
>> I think those GNU folks want to be clever, and often overdo it.
 
> They likely picked std::nullptr_t.
 
Either way, if it's 'nullptr_t', it should pick foo(int*), if it's
'int', it should pick 'foo(int)', in both cases the other one requires a
conversion of lower rank, if my reading is correct. If that is so, then
it still should not be ambiguous.
 
Now, the bad thing about picking 'nullptr_t' instead of 'int' would be
that the code that used to lead to calling 'foo(int)' could now be
calling 'foo(int*)' *silently*.
 
Unless I am mistaken in reading the relevant chapter 4 and chapter 13
parts... In that case, please enlighten me.
 
V
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Luca Risolia <luca.risolia@linux-projects.org>: May 26 05:26PM +0200

On 26/05/2015 13:47, Victor Bazarov wrote:
> Either way, if it's 'nullptr_t', it should pick foo(int*)
 
std::nullptr_t is defined as decltype(nullptr). If NULL is of type
std::nullptr_t, then the first choice for foo(NULL) should be
foo(std::nullptr_t), while both foo(int*) and foo(int) should be a
second choice.
Victor Bazarov <v.bazarov@comcast.invalid>: May 26 11:56AM -0400

On 5/26/2015 11:26 AM, Luca Risolia wrote:
> std::nullptr_t, then the first choice for foo(NULL) should be
> foo(std::nullptr_t), while both foo(int*) and foo(int) should be a
> second choice.
 
What if there is no foo(std::nullptr_t)? Does conversion from nullptr_t
to int* has the same rank as conversion from nullptr_t to int?
 
V
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Luca Risolia <luca.risolia@linux-projects.org>: May 26 07:41PM +0200

Il 26/05/2015 17:56, Victor Bazarov ha scritto:
>> second choice.
 
> What if there is no foo(std::nullptr_t)? Does conversion from nullptr_t
> to int* has the same rank as conversion from nullptr_t to int?
 
std::nullptr_t cannot be converted to int, but can be implicitly
converted to int* (and to any raw pointer types). So I was imprecise in
the second part of my statement: foo(int) is not an option at all if
NULL is of type std::nullptr_t. To answer your question then, if there's
no foo(std::nullptr_t), then the only valid option would be foo(int*).
Ergo in some C++11 implementations (like mine), which complain about
ambiguities in the foo(NULL) call, NULL is not of type std::nullptr_t.
Victor Bazarov <v.bazarov@comcast.invalid>: May 26 03:50PM -0400

On 5/26/2015 1:41 PM, Luca Risolia wrote:
> no foo(std::nullptr_t), then the only valid option would be foo(int*).
> Ergo in some C++11 implementations (like mine), which complain about
> ambiguities in the foo(NULL) call, NULL is not of type std::nullptr_t.
 
If we go back to the earlier post of yours:
On 5/22/2015 4:54 PM, Luca Risolia wrote:
> [..] The above code does not compile on my C++
> implementation because the overload is really ambiguous.
 
Doesn't what you've just written mean that the your C++ implementation
is defective? I see no other explanation why it would deem the call
'foo(NULL)' ambiguous.
 
V
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Luca Risolia <luca.risolia@linux-projects.org>: May 26 10:45PM +0200

Il 26/05/2015 21:50, Victor Bazarov ha scritto:
> > implementation because the overload is really ambiguous.
 
> Doesn't what you've just written mean that the your C++ implementation
> is defective?
 
No (see below).
 
> I see no other explanation why it would deem the call
> 'foo(NULL)' ambiguous.
 
One conforming C++11 implementation is allowed define NULL as:
 
#define NULL 0L
 
in which case foo(NULL) is ambiguous if you only have the foo(int) and
foo(int*) overloads defined, because conversion from long to int and 0L
to int* are considered equally good.
Glen Stark <mail@glenstark.net>: May 26 07:30PM

Hi Everyone.
 
Consider the following:
 
class A; // some non trivial class.
class B; // another non trivial class.
 
template<typename T> class Foo
{
public:
// some public stuff usint T.
private:
const int m_code; // integer needed for legacy stuff.
// some private stuff which uses T.
}
 
I would like to do something like:
 
Foo<A> f; -> instantiating Foo<A> results in m_code == 5
Foo<B> g; -> instantiating Foo<B> reusults in m_code == 10
 
In fact, then the m_code could be a constexpr, which might open up
further refactorings.
 
 
I figure if I experiment and read around I'll be able to come up with a
solution, but I'm doubtful I'll come up with the most elegant solution,
and I figure one of you could maybe tell me what that is?
 
Thanks for your help, hope my question was clearly phrased.
 
Glen
Ian Collins <ian-news@hotmail.com>: May 27 07:52AM +1200

Glen Stark wrote:
> Foo<B> g; -> instantiating Foo<B> reusults in m_code == 10
 
> In fact, then the m_code could be a constexpr, which might open up
> further refactorings.
 
You could make m_code a static member and specialise:
 
template<> const int Foo<A>::m_code = 5;
template<> const int Foo<B>::m_code = 10;
 
--
Ian Collins
Victor Bazarov <v.bazarov@comcast.invalid>: May 26 03:57PM -0400

On 5/26/2015 3:43 PM, Stefan Ram wrote:
> Glen Stark <mail@glenstark.net> writes:
>> I would like to do something like:
 
> I am not an expert for this topic, here is a wild guess:
 
Truly wild...
 
> template<> struct C< A >: common< A >{ const int c = 5; };
> template<> struct C< B >: common< B >{ const int c = 10; };
 
> int main(){ ::std::cout << C< A >{}.c << '\n'; }
 
There is no need to use inheritance, IMHO. You need some kind of type
traits stuff, specialized for A and B, and use it in Foo:
 
class A; // some non trivial class.
class B; // another non trivial class.
 
template<typename T> struct traits_used_by_Foo {
};
 
template<> struct traits_used_by_Foo<A> {
enum { code = 5 };
};
 
template<> struct traits_used_by_Foo<B> {
enum { code = 10 };
};
 
template<typename T> class Foo
{
public:
// some public stuff usint T.
private:
const int m_code = traits_used_by_Foo<T>::code;
// some private stuff which uses T.
};
 
V
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bleachbot <bleachbot@httrack.com>: May 26 02:33PM +0200

ram@zedat.fu-berlin.de (Stefan Ram): May 26 07:43PM

>I would like to do something like:
 
I am not an expert for this topic, here is a wild guess:
 
#include <iostream>
#include <ostream>
 
class A;
class B;
 
template< typename T >struct common {};
template< typename T >struct C : common< T >{ const int m_code; };
template<> struct C< A >: common< A >{ const int c = 5; };
template<> struct C< B >: common< B >{ const int c = 10; };
 
int main(){ ::std::cout << C< A >{}.c << '\n'; }
 
.
Doug Mika <dougmmika@gmail.com>: May 26 09:37AM -0700

By definition: A function object that combines two function objects is called an adaptive function object. Can anyone provide me with an example of an adaptive function object or a link to an example? Or is it simply that there are two operator() functions in a class that is an adaptive function object?
Victor Bazarov <v.bazarov@comcast.invalid>: May 26 01:04PM -0400

On 5/26/2015 12:37 PM, Doug Mika wrote:
> example of an adaptive function object or a link to an example? Or
> is it simply that there are two operator() functions in a class that
> is an adaptive function object?
 
Here it is:
 
http://lmgtfy.com/?q=%22adaptive+function+object%22+C%2B%2B
 
V
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legalize+jeeves@mail.xmission.com (Richard): May 26 05:11PM

[Please do not mail me a copy of your followup]
 
Doug Mika <dougmmika@gmail.com> spake the secret code
 
>By definition: A function object that combines two function objects is
>called an adaptive function object. Can anyone provide me with an
>example of an adaptive function object or a link to an example?
 
#include <algorithm>
#include <functional>
#include <iostream>
#include <iterator>
#include <vector>
 
template <
typename T,
typename Less = std::less<T>,
typename More = std::greater<T>
 
class range_comparison
{
public:
range_comparison(T min_bound, T max_bound)
: m_min_bound(min_bound),
m_max_bound(max_bound)
{
}
bool operator()(T val)
{
return !lesser(val, m_min_bound) && !greater(val, m_max_bound);
 
}
 
private:
Less lesser;
More greater;
T m_min_bound;
T m_max_bound;
};
 
int main()
{
std::vector<int> vals{1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
 
std::copy_if(std::begin(vals), std::end(vals),
std::ostream_iterator<int>(std::cout, "\n"),
range_comparison<int>(3, 6));
}
 
gives
 
3
4
5
6
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fl <rxjwg98@gmail.com>: May 25 05:10PM -0700

Hi,
 
I get a piece of code which is useful for my project. The code is incomplete.
Thus, I want to modify it to work for me. The problem is from a matrix usage.
It has the following snippet:
 
 
 
 
 
matrix state(6,1);
 
/*initializes the state*/
state[0][0]=0.1;
 
 
 
 
 
The matrix is known as two dimension, which looks just like a two dimension
array. Is it a pseudo code on the state variable? If it is the real code,
I don't know how to define the class to make it like a two dimensional array.
 
Could you help me on the problem?
 
Thanks,
Ian Collins <ian-news@hotmail.com>: May 26 12:18PM +1200

fl wrote:
 
> The matrix is known as two dimension, which looks just like a two dimension
> array. Is it a pseudo code on the state variable? If it is the real code,
> I don't know how to define the class to make it like a two dimensional array.
 
constexpr size_t X = some size;
constexpr size_t Y = some size;
 
using Matrix = std::array<std::array<double,Y>,X>;
 
Matrix matrix;
 
--
Ian Collins
fl <rxjwg98@gmail.com>: May 25 07:17PM -0700

On Monday, May 25, 2015 at 5:18:44 PM UTC-7, Ian Collins wrote:
 
> Matrix matrix;
 
> --
> Ian Collins
 
Thank Ian. Your reply is really helpful to me. I am still new to C++.
Could you make it a real specific example of your pseudo code? I have tried
some time, but it does not work yet. For example, what constexpr would be?
size_t is a type?
fl <rxjwg98@gmail.com>: May 25 07:24PM -0700

On Monday, May 25, 2015 at 5:18:44 PM UTC-7, Ian Collins wrote:
 
> Matrix matrix;
 
> --
> Ian Collins
 
I have tried this:
 
Matrix = std::array<std::array<double,6>,6>;
 
 
missing type specifier - int assumed.
'array' : is not a member of 'std'
'array' : undeclared identifier
type 'double' unexpected
 
The above message is from MSVC compiler.
 
Could you help me through the build?
 
Thanks
Ian Collins <ian-news@hotmail.com>: May 26 04:16PM +1200

fl wrote:
> Could you make it a real specific example of your pseudo code? I have tried
> some time, but it does not work yet. For example, what constexpr would be?
> size_t is a type?
 
It wasn't pseudo code!
 
Your C++ book should explain constexpr and size_t.
 
Let's try again with a complete example:
 
#include <array>
 
constexpr size_t X = 10;
constexpr size_t Y = 10;
 
using Matrix = std::array<std::array<double,Y>,X>;
 
int main()
{
Matrix state;
state[0][0] = 0.1;
}
 
--
Ian Collins
Ian Collins <ian-news@hotmail.com>: May 26 04:17PM +1200

fl wrote:
>> Ian Collins
 
> I have tried this:
 
> Matrix = std::array<std::array<double,6>,6>;
 
Why did you omit the "using"? It wasn't prose, it's a C++ keyword!
 
--
Ian Collins
fl <rxjwg98@gmail.com>: May 25 11:28PM -0700

On Monday, May 25, 2015 at 9:18:08 PM UTC-7, Ian Collins wrote:
 
> Why did you omit the "using"? It wasn't prose, it's a C++ keyword!
 
> --
> Ian Collins
 
Excuse me. I did use using at first, but it still can't pass build with
MSVC. Now, it is found that MSVC does not support c++11.
 
 
 
Whether below line is equivalent to typedef in MSVC 2010?
 
using Matrix = std::array<std::array<double,Y>,X>;
 
 
 
Now I find that a two-dimensional array is similar to a Matrix in the
project. My new question is whether a two-dimensional array can be
redefined as a Matrix in order to support matrix multiplication in the
following (A, P, Q are nXn Matrix, ' is transpose operation)?
 
 
p = A * P * A' + Q;
 
 
 
 
Thanks again.
Paavo Helde <myfirstname@osa.pri.ee>: May 26 02:20AM -0500

fl <rxjwg98@gmail.com> wrote in
 
> Excuse me. I did use using at first, but it still can't pass build
> with MSVC. Now, it is found that MSVC does not support c++11.
> Whether below line is equivalent to typedef in MSVC 2010?
 
 
So why don't you upgrade to a newer version of MSVC, there has been at
least 3 releases after MVSC 2010 and the C++11 standard is 4 years old.
Nobody in public forums is interested in providing outdated and inferior
solutions.
 
Cheers
Paavo
Chris Vine <chris@cvine--nospam--.freeserve.co.uk>: May 26 10:50AM +0100

On Mon, 25 May 2015 23:28:45 -0700 (PDT)
> redefined as a Matrix in order to support matrix multiplication in
> the following (A, P, Q are nXn Matrix, ' is transpose operation)?
 
> p = A * P * A' + Q;
 
With C++98/03, you can define an alias for a matrix type using the
built-in array syntax:
 
const size_t X = 6;
const size_t Y = 4;
 
typedef double Matrix[Y][X]; // Matrix is now a type alias for a
// statically sized 4×6 matrix of doubles
 
Using this approach you will need to write vector and matrix
multiplication functions yourself, which take and return such matrices.
They can accordingly be tedious to use. A more advanced approach is to
write a Matrix class, with multiplication and addition as overloaded
operators for the class.
 
In such a class, for a dynamically sized matrix you probably would not
use a multi-dimensional matrix internally, but instead use pointer
arithmetic on a single dimensional array to arrive at the correct cell.
This is because matrix[2][3] is equivalent to arr[ncols * 2 + 3],
where ncols is the number of columns (the X value above).
 
Read also:
https://isocpp.org/wiki/faq/freestore-mgmt#multidim-arrays
https://isocpp.org/wiki/faq/freestore-mgmt#multidim-arrays2
and the immediately following FAQs. Matrices are tricky in C++, with a
number of different approaches which involve different trade-offs for
static and dynamic sizing and exception safety, which is why they are
often part of a homework assignment :)
 
Chris
"Öö Tiib" <ootiib@hot.ee>: May 26 03:46AM -0700

On Tuesday, 26 May 2015 09:29:05 UTC+3, fl wrote:
 
> > Why did you omit the "using"? It wasn't prose, it's a C++ keyword!
 
> Excuse me. I did use using at first, but it still can't pass build
> with MSVC. Now, it is found that MSVC does not support c++11.
 
'std::array' is a simple template mostly copied from 'boost::array' and
it was certainly present in C++ library of MSVC 2010.
 
> Whether below line is equivalent to typedef in MSVC 2010?
 
> using Matrix = std::array<std::array<double,Y>,X>;
 
Yes. It is just easier to read than:
 
typedef std::array<std::array<double,Y>,X> Matrix;
 
Why you start a new project with 3 generations old compiler?
 
> Now I find that a two-dimensional array is similar to a Matrix in the
> project.
 
What you mean by a two-dimensional array? You mean raw array? It is
rather tricky to work with raw arrays in C++ since those decay to
pointer of first element on slightest provocation (that means almost
always when you try to use one for anything). It is therefore
advisable to encapsulate raw array to a class to make it bit more
manageable (and that is what std::array does).
 
> redefined as a Matrix in order to support matrix multiplication in the
> following (A, P, Q are nXn Matrix, ' is transpose operation)?
 
> p = A * P * A' + Q;
 
No. There are no such thing as post-fix operator ' in C++ and there
are no such thing as matrix operations in C++ and also you can't make
operators (or functions) that return raw arrays in C++. Raw arrays
can't be passed or returned by value (as legacy from C). You should
read a good beginner level book about C++; you won't get anywhere by
asking these novice level questions in forums.
 
When we need to do linear algebra with matrices in C++ then we usually
use something outside of standard C++ for example some linear algebra
library (Armadillo, Eigen or Boost.UBLAS). These already contain
classes for representing matrices so you should not invent your own
types of matrices but simply use theirs.
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