Saturday, November 24, 2018

Digest for comp.lang.c++@googlegroups.com - 14 updates in 6 topics

David Brown <david.brown@hesbynett.no>: Nov 24 12:50AM +0100

On 24/11/2018 00:12, Chris M. Thomasson wrote:
 
> [...]
 
> No. acquires_count does not have to be atomic at all. A C11 mtx_t
> provides a standard mutex. In C++11, its equivalent is std::mutex.
 
I know that.
 
> C11
> https://en.cppreference.com/w/c/thread/mtx_lock
> https://en.cppreference.com/w/c/thread
 
I've read these. I've read the standards too, which are authoritative
(though I don't see anything wrong in those linked pages, and my
experience is that cppreference.com is very accurate).
 
I don't see anything in these that suggests that there is any kind of
synchronisation or ordering enforced on a non-atomic non-volatile
variable just because you have a mutex function. The mtx_lock()
function synchronises with other mtx_ functions on the same mutex. That
is /all/.
 
And in particular, even if mtx_trylock were defined to be a fence, and
even if the fence were defined to apply to all memory operations, not
just atomic ones (which is how C11 and C++11 fences are defined), it
still would not affect the validity of transforming:
 
if (res) {
++acquires_count;
}
 
into
 
auto tmp = acquires_count;
tmp += res ? 1 : 0;
acquires_count = tmp;
 
 
If I am wrong, show me /exactly/ which paragraphs of the C11 or C++11
standards make this wrong.
 
> language now. Variables protected by the mutex do NOT need to be atomic
> at all, no volatile or any crap like that. No extra memory barriers are
> needed at all.
 
There are no variables "protected" by a mutex. Mutex's do not "protect"
variables. They are locks, which are synchronisation primitives. You
can /use/ a mutex to protect access to a variable, but it does not
happen by itself. Think about it - how is "mutex" supposed to "know"
that it is protecting "acquires_count" ? How is "acquires_count"
supposed to "know" that is it protected by "mutex" ? These are
independent variables - they are unrelated.
 
And in particular, the code
 
if (res) {
++acquires_count;
}
 
is run whether the mutex is acquired or not. Is the mutex so magical
that it protects and controls this code even when you fail to get the lock?
 
 
> int a = 0;
> short b = 1;
> double c = 0.5;
 
Do you mean these variables to be at file-scope, while the rest of the
code is within a function? If the variables are local to a function
they will be eliminated entirely by the optimiser.
 
> ____________________
 
> a, b and c do not need to be atomic types at all. No extra memory
> barriers are needed. Period.
 
Wrong. And there is no "period" here - clearly the discussion is not over.
 
In practice, this will work, since there is little to be gained by
moving things around in a different way. But the standards and the
description of the mtx functions do not guarantee it.
Melzzzzz <Melzzzzz@zzzzz.com>: Nov 24 12:17AM


> In practice, this will work, since there is little to be gained by
> moving things around in a different way. But the standards and the
> description of the mtx functions do not guarantee it.
 
In practice, this always works... you have to provide *one* example
where this does not works and then you got point. That is any claim
that assumes for all mathematics say can be debuffed with one counter
example...
 
--
press any key to continue or any other to quit...
David Brown <david.brown@hesbynett.no>: Nov 24 02:13AM +0100

On 24/11/2018 01:17, Melzzzzz wrote:
> On 2018-11-23, David Brown <david.brown@hesbynett.no> wrote:
>> On 24/11/2018 00:12, Chris M. Thomasson wrote:
<snip>
> where this does not works and then you got point. That is any claim
> that assumes for all mathematics say can be debuffed with one counter
> example...
 
No, the onus is on Chris to provide a justification for claiming the
code here works as he thinks it does. That should be by pointing to the
relevant paragraphs in the standards (preferably C11, as it is smaller
and easier to navigate, but C++11 or later would be fine). Failing
that, clear documentation in a compiler reference would be useful, as
would some sort of official "C++11 memory model" paper.
 
Otherwise "it worked when I tried it" is not worth the pixels it is
written on.
"Chris M. Thomasson" <invalid_chris_thomasson@invalid.invalid>: Nov 23 05:17PM -0800

On 11/23/2018 5:13 PM, David Brown wrote:
> would some sort of official "C++11 memory model" paper.
 
> Otherwise "it worked when I tried it" is not worth the pixels it is
> written on.
 
C11 and C++11 have guarantees on mutex operations. The variables they
protect do not need any special decorations. Fwiw, a lock on a
std::mutex basically implies acquire semantics. An unlock implies
release. A conforming C11 or C++11 compiler shall honor the semantics of
a mutex. Period. Anything else, is non-conforming. End of story.
James Kuyper <jameskuyper@alumni.caltech.edu>: Nov 24 12:25AM -0500

On 11/23/18 20:17, Chris M. Thomasson wrote:
...
> C11 and C++11 have guarantees on mutex operations. The variables they
> protect do not need any special decorations.
 
What specifies which variables they protect? What is the nature of the
protection that they provide to those variables? I've reviewed every
line of the standard containing the word "mutex" without seeing any hint
of an answer to either of those questions - what did I miss?
 
"Alf P. Steinbach" <alf.p.steinbach+usenet@gmail.com>: Nov 24 06:42AM +0100

On 24.11.2018 06:25, James Kuyper wrote:
> protection that they provide to those variables? I've reviewed every
> line of the standard containing the word "mutex" without seeing any hint
> of an answer to either of those questions - what did I miss?
 
Mutexes are used to provide exclusive access to variables.
 
It's up to the programmer to establish the guard relationship.
 
 
>> std::mutex basically implies acquire semantics. An unlock implies
>> release. A conforming C11 or C++11 compiler shall honor the semantics of
>> a mutex. Period. Anything else, is non-conforming. End of story.
 
Cheers & hth.,
 
- Alf
Pavel <pauldontspamtolk@removeyourself.dontspam.yahoo>: Nov 23 02:51AM -0500

Chris M. Thomasson wrote:
 
> Well, there was a problem with a GCC optimization that broke a POSIX mutex. It
> was an old post on comp.programming.threads:
 
> https://groups.google.com/d/topic/comp.programming.threads/Y_Y2DZOWErM/discussion
 
I have never read that discussion before and got interested by your mentioning
of it. I found the claims of people who believed that acquiring POSIX mutex is
somehow supposed to protect some "associated" variable highly dubious. Just in
case, I double-checked and could not find any such guarantee in my copy of POSIX
standard (The Open Group Base Specifications Issue 6 IEEE Std 1003.1, 2004
Edition).
 
AFAIK (and I has been using POSIX synchronization primitives for decades), the
only thing that successful locking of a posix mutex by a thread guarantees is
that no other thread has or will have that mutex locked until the first thread
releases it. How to use this guarantee is entirely up to the programmer. In
particular, I believe that the OP's code was buggy if the intention was to
maintain a total count of successful locks by all threads in `acquires_count'
variable, exactly because the programmer did not take responsibility for memory
access sequencing. Respectively, IMHO the code optimization by GCC was
absolutely legal.
Tim Rentsch <txr@alumni.caltech.edu>: Nov 24 12:51PM -0800

> type to the type of the actual argument: it's already that type.
 
> So "the argument of the constructor" must be referring to the
> constructor's formal argument.
 
I don't think so. The quoted paragraph uses the term "argument"
and that does mean "argument" in its usual sense, not "parameter"
(or "formal argument" if someone prefers that term). I think
you may have missed a subtle but important point being made
here. Here is an example to illustrate (disclaimer: just typed
in, not compiled, though I have compiled similar examples):
 
class Foo {
public:
Foo( const long & ){}
};
 
int
takes_foo( Foo foo ){
(void) &foo; /* explicitly ignore the value of 'foo' */
return 47;
}
 
...
 
int
main(){
return takes_foo( 0 ) != 47;
}
 
The return statement in main() makes use of a user-defined
conversion sequence that is specified by a constructor. The
source type (for 0) is 'int'. The type of the (unnamed and
ignored) constructor parameter is 'const long &'. There must be
a conversion (or several) to get from the source type to the type
of the constructor parameter. The quoted paragraph says "the
initial standard conversion sequence converts the source type to
the type required by the argument of the constructor". The type
required by the argument of the constructor is (in this case)
'long'. The reason it is 'long' and not 'const long &' is that a
standard conversion sequence cannot get all the way from 'int' to
'const long &', only to 'long'. The definition of "standard
conversion sequence" is given in section 7, paragraph 1 (with
sub-paragraphs 1.1, 1.2, 1.3, and 1.4). Notice that it does not
include any conversions to get from a prvalue to the parameter
type 'const long &'. That step is done by a "temporary
materialization conversion", defined in 7.4. So the quoted
sentence, in talking about "the initial standard conversion
sequence", is referring to an intermediate _argument_ to the
constructor, said argument only subsequently being converted to
the type of the constructor parameter. There are three types:
 
the source type - int
the standard conversion sequence result type - long
the parameter type - const long &
 
The quoted paragraph is referring to the second of these, which
is the type of an intermediate argument to the constructor, not
the type of the parameter in the constructor's definition.
Evan Jack <evanjackspace5@gmail.com>: Nov 24 12:42PM -0800

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Tim Rentsch <txr@alumni.caltech.edu>: Nov 24 11:43AM -0800

> others. (Although I prefer to introduce temporaries for complex
> expressions.)
 
> 3. Enough people seem to feel that way, or gcc wouldn't warn.
 
I have a different reaction. Certainly the attitude you describe
is common. That is regrettable, especially for these operators,
which are placed where many or most programmers would expect them
to be, following a well-established mathematical custom going back
at least to the 1800's. Moreover these particular operators have
the same relative precedence in essentially all programming
languages (the only exceptions I'm aware of are APL and Smalltalk,
which have just one precedence level for all binary operators).
[Note: strictly speaking I believe neither APL nor Smalltalk has
short-circuiting binary logical operators, but I'm glossing over
that aspect.]
 
More generally, this sentiment appears to be a holdover from a
time 60 years ago, when compilers, languages, and documentation
could not be trusted to be accurate or correct. To me it seems
like the same attitude that produces sloppy code, habits like
making arrays "a little bigger" than they need to be. The C
language has had exactly the same operator ordering for 40 years
(and maybe more, but K&R came out in 1978). Anyone who is capable
enough to be a competent programmer can easily learn those rules
in about 15 minutes. I believe the programming community would do
well to adopt the attitude that any serious C programmer should be
expected to have learned C's precedence rules (and similarly for
C++, at least for those parts where C and C++ are the same).
 
It could be helpful to have a "meeting-of-the-minds" discussion to
decide where the dividing line should be for which aspects should
be expected to be known and which are more esoteric. To me though
the relative precedences of relational/equality operators, the &&
operator, and the || operator, are clearly and emphatically well
inside the line of what should be expected of any serious C or C++
programmer.
Elephant Man <conanospamic@gmail.com>: Nov 24 04:45PM

Article d'annulation posté via Nemo.
Horizon68 <horizon@horizon.com>: Nov 24 06:41AM -0800

Hello,
 
My Parallel C++ Conjugate Gradient Linear System Solver Library that
scales very well was updated to version 1.73
 
You can read about it and download it from my website here:
 
https://sites.google.com/site/scalable68/scalable-parallel-c-conjugate-gradient-linear-system-solver-library
 
 
Thank you,
Amine Moulay Ramdane.
robert badea <badea.robert92@gmail.com>: Nov 24 04:17AM -0800

Hello!
I was messing around with C++ and wanted to do a type_trait which would return true if a type is one of the given types.
basically, is_same_v<T,T> with variadic.
 
This would remove the following boiler plate
 
struct is_something {
static const bool value = false;
};

template <>
struct is_something<int> {
static const bool value = true;
};

template <>
struct is_something<short> {
static const bool value = true;
};

template <>
struct is_something<long> {
static const bool value = true;
};
 
 
with
 
template<class T>
struct is_something {
static const bool value = is_one_of<T, int, short, long>;
};
 
I've made the following combinations:
 
1)
 
template<class Expected, class... Received>
inline constexpr bool is_one_of = (std::is_same_v<Expected, Received> || ...);
 
2)
 
template<class Expected>
struct is_one_of_2
{
template <class ...Received>
static constexpr inline bool exec()
{
return (std::is_same_v<Expected, Received> || ...);
}
};
 
 
 
Usage:
 
template<class T>
constexpr void exec(T data)
{
if constexpr(is_one_of<int,int,float> || is_one_of_2<T>::exec<int, float>())
{
std::cout << "same";
}
else
{
std::cout << "not same";
}
}
 
int main()
{
exec(32);
std::cin.get();
return 0;
}
 
 
I was thinking of templated operator ().
First one is the cleanest, but you need to know that the first param is compared to all the others.
The second one is just messy.
 
Do you have any idea how can i improve the design ?
Sam <sam@email-scan.com>: Nov 24 09:21AM -0500

robert badea writes:
 
> I was messing around with C++ and wanted to do a type_trait which would
> return true if a type is one of the given types.
 
> Do you have any idea how can i improve the design ?
 
Something like the following. The odd "<= 0" comparison, instead of "> 0"
was needed because ">" gets parsed …differently in the context of a template
parameter.
 
To check if the given type occurs exactly once, just change that to "!= 1".
 
#include <utility>
#include <type_traits>
#include <iostream>
 
template<typename T, typename ...Types>
struct count_occurences {
 
static constexpr int how_many=((std::is_same_v<T, Types> ? 1:0) + ...);
};
 
template<typename T, typename first_type, typename ...Types>
using occurs_in=std::conditional_t<
count_occurences<T, first_type, Types...>::how_many <= 0,
std::false_type, std::type_type>;
 
int main()
{
std::cout << occurs_in<int, float, int, double, int>::value
<< std::endl;
std::cout << occurs_in<int, double>::value << std::endl;
return 0;
}
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