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  #1  
Old January 19th, 2006, 09:34 PM
devshaikh11 devshaikh11 is offline
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Question Why was it able to pass through?

Code:
#include <iostream>
#include <cctype>
using namespace std;

int main()
{
	cout << "Enter a line of input:\n";
	char next;
	
	do
	{
		cin.get(next);
		cout << next;
	} while ((! isdigit(next)) && (next != '\n'));
	cout << "<END OF OUTPUT";
	return 0;
}


My input: I'll see you at 10:30 AM.
The Output: I'll see you at 1<End OF OUTPUT

My question is, why didn't the code prevent the digit 1 from being displayed? Why was it able to pass through? How would you track this code?

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Old January 19th, 2006, 10:35 PM
destin destin is offline
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This happens because you are doing cin.get(next);, and then checking after. So you've already stored and printed the character.

What you could do to fis this, is just switch cout << next; and cin.get(next);
Comments on this post
Itsacon agrees: Don't forget to put a cin.get(next); before the loop though.

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  #3  
Old January 20th, 2006, 10:16 AM
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Itsacon Itsacon is offline
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Even nicer, and based on my sick use of for() loops for everything (really, it's the greatest function in C).

cpp Code:
Original - cpp Code
  1. #include <iostream>
  2. #include <cctype>
  3. using namespace std;
  4.  
  5. int main()
  6. {
  7.     char next;
  8.  
  9.     cout << "Enter a line of input:\n";
  10.  
  11.     for(cin.get(next); ((! isdigit(next)) && (next != '\n')); cin.get(next))
  12.         cout << next;
  13.  
  14.     cout << "<END OF OUTPUT";
  15.     return 0;
  16. }
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This is my code. Is it not nifty?

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Old January 20th, 2006, 10:17 AM
devshaikh11 devshaikh11 is offline
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Quote:
Originally Posted by destin
This happens because you are doing cin.get(next);, and then checking after. So you've already stored and printed the character.

What you could do to fis this, is just switch cout << next; and cin.get(next);


thanks destin. I get it now.

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