random_shuffle seed

[DrPizza]

Re: random_shuffle seed

"Pete Becker" <petebecker@acm .org> wrote in message
news:3FAA5FC8.8 2984BEB@acm.org ...[color=blue]
> DrPizza wrote:[color=green]
> >
> > "Pete Becker" <petebecker@acm .org> wrote in message
> > news:3FA9B097.7 46AD23B@acm.org ...[color=darkred]
> > > DrPizza wrote:
> > > >
> > > > You wouldn't want to permit your C++ library to call strtok()
> > > > internally, would you?
> > >
> > > Sure. Why not?[/color]
> > Because it means you can't use strtok() yourself?[/color]
> Non sequitur.[/color]
If the implementation can call strtok() then I can't be sure what state
strtok()'s static buffer is in -- it need not be in the state I left it in.
This renders strtok() useless. If I use it I risk fucking up the library; if
the library uses it it risks fucking up my program.
[color=blue]
> The C++ standard does not change that constraint. No C function calls
> rand.[/color]
Except the "not called by the library" statement refers only to the library of
the implementation, not the C library. Given that in C++ the implementation
is one of C++, and the library is that of C++, the restriction should constain
the C++ library too.


--
Now Playing: Daft Punk - Around The World


char a[99999],*p=a;main(c,V) char**V;{char*v =c>0?1[V]:V;if(c)for(;(c =*v)&&93^
c;p+=!(62^c)-!(60^c),*p+=!(4 3^c)-!(45^c),44^c||r ead(0,p,1),46^c ||putchar(*p)
,91^c||(v=*p?ma in(-1,v+1),v-1:main(0,v)),++ v);else for(;c+=!(91^*v )-!(93^*v)
;++v);return v;} /* drpizza@battlea xe.net brainf*** program as argv[1] */

[Pete Becker]

Re: random_shuffle seed

DrPizza wrote:[color=blue]
>
> "tom_usenet " <tom_usenet@hot mail.com> wrote in message
> news:nckkqvsrn5 lghoe2ire026mqv qlvqsr332@4ax.c om...[color=green]
> > The C standard only says that the implementation of the C library
> > won't call them.[/color]
> Well, no, it doesn't, it simply says that the implementation shall behave as
> if no library function calls rand. It doesn't specify which library.
>[color=green]
> > Taking the original quote:
> > "The implementation shall behave as if no library function calls the
> > rand function"
> >
> > The "library" referred to is the C library, not the C++ one. Just
> > because C++ includes the (slightly modified) C library doesn't mean
> > that it *is* the C library.[/color]
> Upon incorporation into C++, "the library" refers to the C++ library. The C++
> library is the only library there is, so it's the only one it could refer to.
>[/color]

Where in the C++ standard do you find this assertion?

--

Pete Becker
Dinkumware, Ltd. (http://www.dinkumware.com)

[Pete Becker]

Re: random_shuffle seed

DrPizza wrote:[color=blue]
>
> "Pete Becker" <petebecker@acm .org> wrote in message
> news:3FAA5FC8.8 2984BEB@acm.org ...[color=green]
> > DrPizza wrote:[color=darkred]
> > >
> > > "Pete Becker" <petebecker@acm .org> wrote in message
> > > news:3FA9B097.7 46AD23B@acm.org ...
> > > > DrPizza wrote:
> > > > >
> > > > > You wouldn't want to permit your C++ library to call strtok()
> > > > > internally, would you?
> > > >
> > > > Sure. Why not?
> > > Because it means you can't use strtok() yourself?[/color]
> > Non sequitur.[/color]
> If the implementation can call strtok() then I can't be sure what state
> strtok()'s static buffer is in -- it need not be in the state I left it in.[/color]

The implementation is not allowed to change the state you left it in.
That's obvious from the semantics of strtok. Whether it calls strtok
internally is irrelevant, so long as it satisfies that requirement.

That's the problem with redundant requirements: people look for meaning
in them.

--

Pete Becker
Dinkumware, Ltd. (http://www.dinkumware.com)

[DrPizza]

Re: random_shuffle seed

"Pete Becker" <petebecker@acm .org> wrote in message
news:3FAB9516.9 6D4EA1C@acm.org ...[color=blue]
> DrPizza wrote:[color=green]
> >
> > "tom_usenet " <tom_usenet@hot mail.com> wrote in message
> > news:nckkqvsrn5 lghoe2ire026mqv qlvqsr332@4ax.c om...[color=darkred]
> > > The C standard only says that the implementation of the C library
> > > won't call them.[/color]
> > Well, no, it doesn't, it simply says that the implementation shall behave[/color][/color]
as[color=blue][color=green]
> > if no library function calls rand. It doesn't specify which library.
> >[color=darkred]
> > > Taking the original quote:
> > > "The implementation shall behave as if no library function calls the
> > > rand function"
> > >
> > > The "library" referred to is the C library, not the C++ one. Just
> > > because C++ includes the (slightly modified) C library doesn't mean
> > > that it *is* the C library.[/color]
> > Upon incorporation into C++, "the library" refers to the C++ library. The[/color][/color]
C++[color=blue][color=green]
> > library is the only library there is, so it's the only one it could refer[/color][/color]
to.[color=blue][color=green]
> >[/color]
>
> Where in the C++ standard do you find this assertion?[/color]

It's the only thing that the restriction could mean.

--
char a[99999],*p=a;main(c,V) char**V;{char*v =c>0?1[V]:V;if(c)for(;(c =*v)&&93^
c;p+=!(62^c)-!(60^c),*p+=!(4 3^c)-!(45^c),44^c||r ead(0,p,1),46^c ||putchar(*p)
,91^c||(v=*p?ma in(-1,v+1),v-1:main(0,v)),++ v);else for(;c+=!(91^*v )-!(93^*v)
;++v);return v;} /* drpizza@battlea xe.net brainf*** program as argv[1] */

[Pete Becker]

Re: random_shuffle seed

DrPizza wrote:[color=blue]
>
> "Pete Becker" <petebecker@acm .org> wrote in message
> news:3FAB9516.9 6D4EA1C@acm.org ...[color=green]
> > DrPizza wrote:[color=darkred]
> > >
> > > "tom_usenet " <tom_usenet@hot mail.com> wrote in message
> > > news:nckkqvsrn5 lghoe2ire026mqv qlvqsr332@4ax.c om...
> > > > The C standard only says that the implementation of the C library
> > > > won't call them.
> > > Well, no, it doesn't, it simply says that the implementation shall behave[/color][/color]
> as[color=green][color=darkred]
> > > if no library function calls rand. It doesn't specify which library.
> > >
> > > > Taking the original quote:
> > > > "The implementation shall behave as if no library function calls the
> > > > rand function"
> > > >
> > > > The "library" referred to is the C library, not the C++ one. Just
> > > > because C++ includes the (slightly modified) C library doesn't mean
> > > > that it *is* the C library.
> > > Upon incorporation into C++, "the library" refers to the C++ library. The[/color][/color]
> C++[color=green][color=darkred]
> > > library is the only library there is, so it's the only one it could refer[/color][/color]
> to.[color=green][color=darkred]
> > >[/color]
> >
> > Where in the C++ standard do you find this assertion?[/color]
>
> It's the only thing that the restriction could mean.
>[/color]

Several people have already disagreed with you here, so you are merely
repeating your position. That's a waste of time.

--

Pete Becker
Dinkumware, Ltd. (http://www.dinkumware.com)