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cin and scanf which is best for simple input

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  cin and scanf which is best for simple input papoupapa <papoupapoug@gmail.com> - 2011-06-11 03:53 -0700
    Re: cin and scanf which is best for simple input "osmium" <r124c4u102@comcast.net> - 2011-06-11 12:06 -0500
      Re: cin and scanf which is best for simple input "Sprechen Sie C++" <spamabuse@hotmail.com> - 2011-06-11 10:50 -0700
    Re: cin and scanf which is best for simple input Kai-Uwe Bux <jkherciueh@gmx.net> - 2011-06-11 20:29 +0200
      Re: cin and scanf which is best for simple input papoupapa <papoupapoug@gmail.com> - 2011-06-12 16:25 -0700

#6615 — cin and scanf which is best for simple input

Hi,

Can someone tell from the code below why scanf solves the issue of a
number out of

the range and cin seems to read twice a great number such as
100000000000 or go into a wild loop if a character or some text is
entered.

 #include "std_lib_facilities.h"

int main()
{
	int val1 = 0 ;
	int val2 = 0 ;
	int nb_try ;
	bool ok = 0 ;

	cout << "\nThis code reads 2 numbers and takes these 2 numbers into
simple

processes : \n" ;

	nb_try = 3 ;
	do
	{
		cout << "\nPlease enter 1st number (between 0 and 65,535) : \n" ;
		//scanf("%d", &val1) ;
		cin >> val1 ;

		if ((val1 >= 0) && (val1 < 65536))
			ok = 1 ;
		else
		{
			ok = 0 ;
			nb_try-- ;
		}

		if (!ok)
		{
			cout << "\nEntry outside 0-65535 range\n" ;
			cout << nb_try << " left\n\n" ;
		}

	} while ((!ok) && (nb_try >= 0)) ;

	nb_try = 3 ;

	do
	{
		cout << "\nPlease enter 2nd number between 0 and 65,535) : \n" ;
		//scanf("%d", &val2) ;
		cin >> val2 ;

		if (((val1 >= 0) && (val1 < 65536)) && ((val2 >= 0) && (val2 <
65536)))
			ok = 1 ;
		else
		{
			ok = 0 ;
			nb_try-- ;
		}

		if (!ok)
		{
			cout << "\nEntry outside 0-65535 range\n" ;
			cout << nb_try << " left\n\n" ;
		}

	} while ((!ok) && (nb_try >= 0)) ;

	cout << "\n\n" ;
	system("pause") ;
	return 0 ;
}

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#6623

"papoupapa" wrote::

<snippage>
> Can someone tell from the code below why scanf solves the issue of a
> number out of
>
> the range and cin seems to read twice a great number such as
> 100000000000 or go into a wild loop if a character or some text is
> entered.
>
> #include "std_lib_facilities.h"
>
> int main()
> {
> int val1 = 0 ;
> int val2 = 0 ;
> int nb_try ;
> bool ok = 0 ;
>
> cout << "\nThis code reads 2 numbers and takes these 2 numbers into
> simple

I suggest you simplify things.  You seem to think cin does something wrong. 
Write a *current* conforming C++ program with cin and only cin which 
exhibits the problem.  Don't complicate things with a second program using 
scanf.  In general, the use of scanf is discouraged.  WRT conforming:
o your include file is not standard.
o your program will not compile.  A simple cout will no longer be accepted.

My wild guess is that your use of "solve" is misguided. 

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#6625

Might be a better idea to use

#include <iostream>;

std::cout << mydata << std::endl;


easier to work with and 100x more maintainable

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#6626

papoupapa wrote:

> Hi,
> 
> Can someone tell from the code below why scanf solves the issue of a
> number out of
> 
> the range and cin seems to read twice a great number such as
> 100000000000 or go into a wild loop if a character or some text is
> entered.
> 
>  #include "std_lib_facilities.h"
> 
> int main()
> {
> int val1 = 0 ;
> int val2 = 0 ;
> int nb_try ;
> bool ok = 0 ;
> 
> cout << "\nThis code reads 2 numbers and takes these 2 numbers into
> simple
> 
> processes : \n" ;
> 
> nb_try = 3 ;
> do
> {
> cout << "\nPlease enter 1st number (between 0 and 65,535) : \n" ;
> //scanf("%d", &val1) ;
> cin >> val1 ;
> 
> if ((val1 >= 0) && (val1 < 65536))

You test whether val1 is in the required range. However, you have no test 
for whether the reading from cin succeeded. You could try:

  if ( ( cin >> val1 )  &&  ( val1 >= 0 )  &&  ( val1 < 65536 ) ) {
    ...
  }

[snip]


Best,

Kai-Uwe Bux

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#6665

thanks

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