This thread was really interesting until it became confrontational. Is
this the programmers' version of "my car's bigger than yours"? Please
try to bite your tongues and stop insulting one another. "What the hell
are you talking about" should never be a part of any message on here. Is
this list being moderated?
Cheers, Lynn
-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of Christopher
Coale
Sent: 14 April 2011 04:34
To: programmingblind@xxxxxxxxxxxxx
Subject: Re: again a newbee question
Now what are you talking about? My game engine has a scene graph of
inherited nodes (sometimes 3 or 4 deep) that all get allocated and
destroyed, with the destructors called in the proper order? So again,
what the hell are you talking about? By the way, I never call "exit" in
it nor do I register callbacks to destroy nodes using the atexit
function. And 32 callbacks is slightly limiting, wouldn't you say?
As for what Tyler said, and it totally emphasized my point, was how
exactly do you want these callbacks to destroy resources? Shall we store
global pointers to the resources so that the callback can access them?
Here is what I mean:
class SomeGraphicsClass {
IDirect3DVertexBuffer9 *pVB = .....; // assume this is a created
resource };
void myOnExit()
{
// ok.. now how do I destroy pVB?
}
On 4/13/2011 8:29 PM, Sina Bahram wrote:
I don't see how that's the same thing though.
That's platform specific, no? i could be totally off on that point,
but I thought it was.
This allows you to register 32 callbacks, and all your objects have
destructors which get called by parents, etc, etc, so you should be
good to go.
Of course destructors of large object trees is a joke in C++, as
anyone who has had to deal with multiple inherited classes and
destructor order resolution issues will tell you. throw in some
virtual functions on top of that, and the only thing you'll be
destroying reliably is a bottle of something.
Take care,
Sina
-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of Ken Perry
Sent: Wednesday, April 13, 2011 11:25 PM
To: programmingblind@xxxxxxxxxxxxx
Subject: RE: again a newbee question
I think Chris wants people to use the onDestroy stuff but in truth
OnDestroy is not going to free up nothing if the person doesn't' add
the code to it so I figure it doesn't matter either way.
Ken
-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of Sina
Bahram
Sent: Wednesday, April 13, 2011 11:20 PM
To: programmingblind@xxxxxxxxxxxxx
Subject: RE: again a newbee question
Here:
void exit ( int status );
Terminates the process normally, performing the regular cleanup for
terminating processes.
First, all functions registered by calls to atexit are executed in the
reverse order of their registration. Then, all streams are closed and
the temporary files deleted, and finally the control is returned to
the host environment.
The status argument is returned to the host environment.
http://www.cplusplus.com/reference/clibrary/cstdlib/exit/
and here:
int atexit ( void ( * function ) (void) ); Set function to be executed
on exit
The function pointed by the function pointer argument is called when
the program terminates normally.
If more than one atexit function has been specified by different calls
to this function, they are all executed in reverse order as a stack,
i.e. the last function specified is the first to be executed at exit.
One single function can be registered to be executed at exit more than
once.
C++ implementations are required to support the registration of at
C++ least 32
atexit functions.
http://www.cplusplus.com/reference/clibrary/cstdlib/atexit/
take care,
Sina
-----Original Message-----
From: Sina Bahram [mailto:sbahram@xxxxxxxxx]
Sent: Wednesday, April 13, 2011 11:17 PM
To: 'programmingblind@xxxxxxxxxxxxx'
Subject: RE: again a newbee question
I'd recommend looking up how exit works.
Take care,
Sina
-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of
Christopher Coale
Sent: Wednesday, April 13, 2011 11:06 PM
To: programmingblind@xxxxxxxxxxxxx
Subject: Re: again a newbee question
Also, just because exit() cleans up some standard system resources,
does not mean it cleans up ALL resources. I do most of my programming
with XNA, Direct3D, and the like. When you create objects that exist
in graphics memory instead of RAM, you are responsible for managing
that object. If I create a vertex buffer on the GPU, I am responsible
for freeing that resource, not the "exit" function.
Teaching beginner programmers to call exit to leave the program is
just simply a horrible decision. But hey, what do I know..
On 4/13/2011 7:41 PM, Sina Bahram wrote:
I don't understand what that means.
Totally destroys what structure?
It's an exit ... of course it destroys everything.
Furthermore, exit is an extremely clean way of exiting a program.
It's a
million times better than return 0.
It actually calls, in reverse order of course, all functions which
registered via atexit. It also closes streams, not only IO ones
for that matter, and it gets rid of/cleans temporary files.
Quite puzzled ...
Take care,
Sina
-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of
Christopher Coale
Sent: Wednesday, April 13, 2011 9:38 PM
To: programmingblind@xxxxxxxxxxxxx
Subject: Re: again a newbee question
Alright, I just wanted to force you to clarify, so you don't give
Ashish the wrongi dea. ;) I grade C++ programming assignments, and I
see new
C++ programmers using the exit function simply because it's
C++ convenient,
but it totally destroys the structure.
On 4/13/2011 6:35 PM, Ken Perry wrote:
Well I was using exit as a concept more than a method since
depending on
the
gui you code in there are things like finish, exit, delete and all
kinds
of
way to make a program die.
Ken
-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of
Christopher Coale
Sent: Wednesday, April 13, 2011 9:28 PM
To: programmingblind@xxxxxxxxxxxxx
Subject: Re: again a newbee question
Careful with saying "until you call exit." You don't want to give
the impression that the exit() function is okay in a structured
program, do we? ;)
On 4/13/2011 6:25 PM, Ken Perry wrote:
Nod you will not need things like scanf and getch and things of the
like when you get into serious coding because you will be probably
building windows apps of some kind and the window will remain open
till you call exit. This is just while you're learning so feel
free to use scanf,
getch,
getchar, or any of the things like read to just hold the program
till
your
ready to close.
Ken
-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of ashish
rohtagi
Sent: Wednesday, April 13, 2011 9:05 PM
To: programmingblind@xxxxxxxxxxxxx
Subject: Re: again a newbee question
sorry, I corrected the mistake about printf and scanf. thanks for
the advice. take care, regards. ashish
On 4/14/11, Littlefield, Tyler<tyler@xxxxxxxxxxxxx> wrote:
I'm sorry, getch should be getchar. Also: it's fairly bad coding
style to put a printf on the same line as a scanf. Unless you have
good reason, try to keep your code separated. It makes it easier
to read and people don't miss things quite as easily.
On 4/13/2011 6:50 PM, ashish rohtagi wrote:
like this?? #include<stdio.h>
int main()
{
char me[20];
printf("What is your name?"); scanf("%s",&me); printf("Darn glad
to meet you, %s!\n",me);
scanf("s%",&me);
return(0);
} but result is still the same. can I use getch fungtion? if yes
please tell me how to use it . take care, regards.
On 4/14/11, Jared Wright<wright.jaredm@xxxxxxxxx> wrote:
It goes right between the last printf statement and the return
statement. It's job is to hold the program open at the end, so
it
should
be the last statement before return, which ends the program.
On 4/13/2011 12:09 PM, ashish rohtagi wrote:
hi Tyler, thanks for your suggestion, but I am unable to
understand where to put second scanf. will you just once show
me? take care, regards. ashish
On 4/13/11, Littlefield, Tyler<tyler@xxxxxxxxxxxxx>
wrote:
Like ken said, use scanf to catch it before the window closes,
or
run
it
in a command line. Here's what's happening:
You open the process, and the first scanf call blocks; that is
to
say,
it will not process anything, it is simply waiting for the
user to input something. But after it's done printing, there
is no reason why it should stay open because you don't tell it
to. So run from the
command
prompt, use getch() to let you hit enter before it closes or
use another scanf.
On 4/13/2011 9:32 AM, ashish rohtagi wrote:
friends, thanks for all the encouragement. here is my
problem, I
want
to use scanf fungtion. in input and output but as soon as I
give
my
input and press enter the window disappears. it does not give
output
based on my input. here is the code. #include<stdio.h>
int main()
{
char me[20];
printf("What is your name?"); scanf("%s",&me); printf("Glad
to meet you, %s!\n",me); return(0); }
On 4/13/11, Littlefield, Tyler<tyler@xxxxxxxxxxxxx>
wrote:
O. oops. In my defense, I haven't found coffee. :) On
4/13/2011 7:46 AM, Jared Wright wrote:
It's there, just on the same line as the printf statement.
On 4/13/2011 9:35 AM, Littlefield, Tyler wrote:
good job on using int main, just don't forget a return 0;
when
you
are
done. between the printf and the closing brace.
On 4/13/2011 6:36 AM, ashish rohtagi wrote:
here is my code it compiles correctly but text just shows
on
the
screen for 1 second, I want it to stay. #include<stdio.h>
int main()
{
printf("I will learn programming!\n"); return(0); }
On 4/13/11, DaShiell, Jude T. CIV NAVAIR 1490, 1, 26
<jude.dashiell@xxxxxxxx> wrote:
You didn't put your printf statement in your message,
but let
me
make a
guess. You didn't put a (\n) in before the closing quote
mark
so
you
just put text on the screen without a line ending. That
could
be
one
contributing factor. Another might be you didn't put an
#include
<stdio.h> statement in at the top of your program
and
maybe
used
#include<conio.h>. If that's the case, don't use printf
with
conio.h
unless you also include stdio.h. For conio.h, cprintf is
the
function
you want. Hope this helps.
-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf
Of
ashish
rohtagi
Sent: Wednesday, April 13, 2011 8:04
To: programmingblind@xxxxxxxxxxxxx
Cc: punitdiwan@xxxxxxxxx
Subject: again a newbee question
friends, sorry here I am with another foolish question,
sorry
but
here
is no one to teach and I am learning myself. when I
print any
text
on
screen using printf it quickly disappears. what should I
do
that
it
should stay? take care, regards. ashish
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