dev-cpp-users Mailing List for Dev-C++

It took me a whoole day  to find these,hope they are of some info for you 
all on how to make a OS.Some are for linux ,others are not,check em out.

http://www.codesites.com
http://www.kernel.org/pub/
http://www.cs.arizona.edu/people/bridges/os/free.html
http://www.acm.uiuc.edu/sigops/roll_your_own/
http://www.acm.uiuc.edu/sigops/roll_your_own/1.html
http://www.geocities.com/SiliconValley/Campus/1671/english/iindice.htm
http://www.linuxdoc.org/


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I saw that i made a mistake, so i resend.



> -----Original Message-----
> From: dev...@li...
> [mailto:dev...@li...] On Behalf 
> Of Green, Aaron
> Sent: Saturday, December 29, 2001 3:19 AM
> To: dev-c++ user's mailing list
> Subject: [Dev-C++] C++ pointers
> 
> 
> This may be OT, but I know I've seen programming questions
> here before, so I'll give it a try.
> 
> I wrote up a small program that tries to explain how pointers
> work to me.
> 
> #include <iostream>
> 
> double* f(double *pD2)
> {
> 	*pD2 = 15;
> 	return pD2;
> }
> 
> int main()
> {
> 	double d;
> 	double *pD = &d;
> 	d = 20;
> 	*pD = 10;
> 	std::cout << *pD << '\n';
> 	std::cout << d << '\n';

> 	double *pD3 = f(&d);


This points pD3 toy d which after the execution of this line is 15.
        

> 	std::cout << d << '\n';
> 	std::cout << *pD3 << '\n';
> 	d = 100;


Now you change d toy 100 so *pD3 is now 100 (pD3==&d).

> 	std::cout << d << '\n';
> 	std::cout << *pD3 << '\n';
> 	std::cout << *pD << '\n';
> 
> 
> 	return 0;
> }
> 
> 
> The last parts of the output are 100   100   100.  My 
> question is:  Why does *pD3 change when it points to a
> function that changes whatever is at &d ?  I thought calling 
> cout would run that function again, or at least give me the 
> same value as before.  


Calling cout<<*pD3; doesn't execute any function it simply the value of
d which is 100. If you wanted the function to be called again you should
do:

	std::cout << *(pD3=f(pD3)) << '\n';

    (or 	std::cout << *(pD3=f(&d)) << '\n';   )


or std::cout << *f(pD3) << '\n'

(    or  std::cout << *f(&d) << '\n'    )



Ioannis

* Ioannis Vranos
* Programming pages: http://www.noicys.f2s.com
* Alternative URL: http://run.to/noicys

> -----Original Message-----
> From: dev...@li... 
> [mailto:dev...@li...] On Behalf 
> Of Green, Aaron
> Sent: Saturday, December 29, 2001 3:19 AM
> To: dev-c++ user's mailing list
> Subject: [Dev-C++] C++ pointers
> 
> 
> This may be OT, but I know I've seen programming questions 
> here before, so I'll give it a try.
> 
> I wrote up a small program that tries to explain how pointers 
> work to me.
> 
> #include <iostream>
> 
> double* f(double *pD2)
> {
> 	*pD2 = 15;
> 	return pD2;
> }
> 
> int main()
> {
> 	double d;
> 	double *pD = &d;
> 	d = 20;
> 	*pD = 10;
> 	std::cout << *pD << '\n';
> 	std::cout << d << '\n';

> 	double *pD3 = f(&d);


This points pD3 toy d which after the execution of this line is 15.
        

> 	std::cout << d << '\n';
> 	std::cout << *pD3 << '\n';
> 	d = 100;


Now you change d toy 100 so *pD3 is now 100 (pD3==&d).

> 	std::cout << d << '\n';
> 	std::cout << *pD3 << '\n';
> 	std::cout << *pD << '\n';
> 
> 
> 	return 0;
> }
> 
> 
> The last parts of the output are 100   100   100.  My 
> question is:  Why does *pD3 change when it points to a 
> function that changes whatever is at &d ?  I thought calling 
> cout would run that function again, or at least give me the 
> same value as before.  


Calling cout<<*pD3; doesn't execute any function it simply the value of
d which is 100. If you wanted the function to be called again you should
do:

	std::cout << *(pD3=f(pD3)) << '\n';

    (or 	std::cout << *(pD3=f(d)) << '\n';   )


or std::cout << *f(pD3) << '\n'

(    or  std::cout << *f(d) << '\n'    )



Ioannis

* Ioannis Vranos
* Programming pages: http://www.noicys.f2s.com
* Alternative URL: http://run.to/noicys

This may be OT, but I know I've seen programming questions here before, so I'll give it a try.

I wrote up a small program that tries to explain how pointers work to me.

#include <iostream>

double* f(double *pD2)
{
	*pD2 = 15;
	return pD2;
}

int main()
{
	double d;
	double *pD = &d;
	d = 20;
	*pD = 10;
	std::cout << *pD << '\n';
	std::cout << d << '\n';
	double *pD3 = f(&d);
	std::cout << d << '\n';
	std::cout << *pD3 << '\n';
	d = 100;
	std::cout << d << '\n';
	std::cout << *pD3 << '\n';
	std::cout << *pD << '\n';


	return 0;
}


The last parts of the output are 100   100   100.  My question is:  Why does *pD3 change when it points to a function that changes whatever is at &d ?  I thought calling cout would run that function again, or at least give me the same value as before.  

My friend...

Jason is right. It is possible for you to create one for a particular system. But if it should support all systems, it will be hard.

Although you should pursue the venture. All I can advice for you is utilize whats already being used. Linux IS a good place to start. Building it under windows would be challenging. But Windows uses a default "Shell" which is known as explorer. You might consider using this to advantage (i.e. making a shell enhancement, or possibly... a shell replacement w/c is what I'm doing now). Windows already supports a variety of hardware so building on top of its shell or changing its shell might be good. Your windows "shell" would, in your description, do whatever you need it to do by providing a simple interface(complex if you want). Plus, it will still retain the abilities of windows.

But I encourage you to create one. Its quite interesting how those folks at Microsoft do such an OS. May I take a look at the code? or any plans that you made? Do you have any URL for that? I'm just curious and hoping also to help you...

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