I want to get only y and n from user to enter
when user try to enter another it get the input again until user enter
"y" or "n"
i use this logic but its not working

do
{
   cout<<"\n\n Enter Number: ";
   number=getche();
   cout<<"\n Want to enter another:(y/n)";
   ans=getche();
      if(ans!='n'||ans!='y')
        ans=getch();
}
  while(ans!='n');

Put in a cin.ignore() after the ans = getche(); line. The newline ('\n') from the when you entered y or n the first time was staying in the input stream and getting picked up by the getch() which took it as input and moved on. I don't know offhand if the cin.ignore() plays nice with the conio functions.

That being said, I would avoid the conio.h functions unless it's absolutely necessary to do so. Use cin.get() or getchar() (getchar requires <cstdio>) for portability.

Put in a cin.ignore() after the ans = getche(); line. The newline ('\n') from the when you entered y or n the first time was staying in the input stream and getting picked up by the getch() which took it as input and moved on. I don't know offhand if the cin.ignore() plays nice with the conio functions.

That being said, I would avoid the conio.h functions unless it's absolutely necessary to do so. Use cin.get() or getchar() (getchar requires <cstdio>) for portability.

There was no \n left in the buffer unless it was before the do-while . getche() is the same as getch() with echo.

But, I agree with jonsca about not using conio.h functions. They are non-standard and not used in the industry.

Looking at your IF statement, put it in a truth table. You will find it will always be true.

Here was my interpretation (I reserve the right to be wrong, so I'm just seeking for my own clarification here, so I'll beg the OPs pardon)

ans=getche(); //user inputs 'g' <enter> (only room for g in the 'ans')
if(ans!='n'||ans!='y')   //this is true
        ans=getch(); //I was saying enter gets picked up by this
                //apparently that's not the case?

On a peripheral matter, I thought this was funny (from the conio.h of a recent mingw):

/*
 * conio.h
 * This file has no copyright assigned and is placed in the Public Domain.
 * This file is a part of the mingw-runtime package.
 * No warranty is given; refer to the file DISCLAIMER within the package.
 *
 * Low level console I/O functions. Pretty please try to use the ANSI
 * standard ones if you are writing new code.
 *
 */

(my color emphasis added)
You can't make this stuff up...

Edited 6 Years Ago by jonsca: n/a

my question is still there that how can i place check on the input
that user can enter only 'y' or 'n'

my question is still there that how can i place check on the input
that user can enter only 'y' or 'n'

Consider using logical AND (&&) ..

do
{
  prompt the user ..
  get the input ..
}
while(ans != 'y' && ans != 'n');

or logical OR (||) ..

do
{
  prompt the user ..
  get the input ..
}
while((ans == 'y' || ans == 'n') == false);

Try to think through your original if-statement to understand why it will not work.

I got the answer

do
{
cout<<"\n\n Enter Number: ";
number=getche();
cout<<"\n Want to enter another´┐╝y/n)";
ans=getche();
while(ans!='n'||ans!='y')
{
cout<<"enter only y or n";
ans=getche()
if(ans=='n'||ans=='y')
break;
}
}
while(ans!='n');

Basically, a thing that you need to understand is the fact that

while( ans != 'n' || ans != 'y' )

is effectively the same as writing

while( true )

And there is no way around that. :)

I got the answer

But your ending loop is too complex.

while(ans!='n'||ans!='y')
{
  cout<<"enter only y or n";
  ans=getche()
  if(ans=='n'||ans=='y') 
     break;
}

If the while statement is supposed to exit when Y or N is pressed, why the if statement that breaks out of the loop? Get rid of the if statement and let the while do it's job.

Basically, a thing that you need to understand is the fact that

while( ans != 'n' || ans != 'y' )

is effectively the same as writing

while( true )

And there is no way around that. :)

That was why I suggested a Truth Table... :icon_wink:

Here was my interpretation (I reserve the right to be wrong, so I'm just seeking for my own clarification here, so I'll beg the OPs pardon)

ans=getche(); //user inputs 'g' <enter> (only room for g in the 'ans')
if(ans!='n'||ans!='y')   //this is true
        ans=getch(); //I was saying enter gets picked up by this
                //apparently that's not the case?

No, getche() and getch() work the same way. Hit a key and the program continues. You won't get a chance to hit the ENTER.

Edited 6 Years Ago by WaltP: n/a

That was why I suggested a Truth Table... :icon_wink:

I know you did, but it just so much seems that the point still needs to be made, so to speak...

No, getche() and getch() work the same way. Hit a key and the program continues. You won't get a chance to hit the ENTER.

Sorry, when I wrote "enter" on line 3, I meant the stray newline in the stream from line 1. I may still missing something obvious, it's not intentional I promise.

do
{
cout<<"\n\n Enter Number: ";
number=getche();
cout<<"\n Want to enter another´┐╝y/n)";
ans=getche();
while(ans!='n'||ans!='y') // this while loop get the input again and again until the n or y are pressed
{
cout<<"enter only y or n";
ans=getche()     // if we dont again get the input again the loop continues
if(ans=='n'||ans=='y') // this is the check on the 2nd input that brings the control out from the loop
break;
}
}
while(ans!='n');

any other way u have??
plz write the code

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