Nebie in need

I'm pretty new and teaching myself threw tutorials. One issue in the use of the programs i make is the user can easily crash the program by inputting a variable of the wrong data type. Currently the variable is an int, and if anyone inputs a char, it sends the program into a continuous loop.
If there is a way to limit the user or a data type that will support both integers and chars, i would love to know.

Thank you immensly in advance.

could you post your problematic code for me to see?? It is possible you aren't doing something right in the program itself.

The best way to avoid this problem is to read all input as string data and then convert it as necessary to integer or floating point values.

-Prelude

Quote:

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Originally posted by Unregistered
If there is a way to limit the user or a data type that will support both integers and chars, i would love to know.

Thank you immensly in advance.

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just fyi, in variable length functions, all variables are promoted upwards so that it can't tell the difference between int and char. (for instance, scanf and printf are variable length functions). there's probably a way to check for this kind of error using cin, though.


(in hindsight the above shouldn't apply anyway in your situation, but it's there for reference. follow prelude's advice)

Do you mean like cin.fail()?

i think hes in my problem in witch he has a c++ calculator the user is supposed to input a number but the user puts ina letter and its hard to add f+3. (unless f is a variable)

cin.fail() and/or cin.good() in combination with cin.clear() can be used to do at least some input error screening, although not all input error screening.

cin.good() works well for the first character. Absolutely agreed. After that, it's anyone's guess. Testing with cin.good() for a float value, as an example, "1abc.def" passes. Not at all a good thing.

Shadow makes a good point, as well. If the expected input is restricted to 0-9, cin.good() will do the job. If your expected input is of an 'int' type, potentially greater than 9, cin.good() will not do the job properly for you. Upshot: This will depend on your code and what you are trying to accomplish.

As usual, Prelude is correct. (There's a news flash :).) Simply put, you can isalpha()/isdigit()/iswhatIwant() yourself into Never-Never Land and you won't come up with pre-defined code that allows you to properly check for the input you want without "brute-forcing" it.

(Skipper's note: "brute-forcing" is programmer-dependent and a subjective term. What one person finds to be a daunting task is a 'no-brainer' for someone else. An uncomfortable truth, but a truth nonetheless, particularly at the stage that you're at.)

Hang in there. You'll get it. :)

-Skipper

Damn, you people know alot :)
The expected input is 1-9, so what do i do with the cin.good() ?

Thanks again.

>The expected input is 1-9, so what do i do with the cin.good() ?
This method of reading input is imprecise and easily messed up if you don't cover for every possible occurance. My recommendation of reading everything into a string with getline and then validating that string still holds. But if you must do it the other way, here is one possible implementation:

Code:

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#include <iostream>

using std::cin;
using std::cout;
using std::cerr;

int main()
{
  int input;
  bool done = false;

  while ( !done ) {
    cout<<"Enter a number 1 - 9: ";
    cin>>input;
    if ( !cin.good() ) {
      cerr<<"Invalid input\n";
      /*
      ** C I/O was always easier to work with. :p
      */
      while ( getchar() != '\n' );
      cin.clear();
    }
    else {
      if ( input < 1 || 9 < input )
        cerr<<"Invalid number\n";
      else
        done = true;
    }
  }
  cout<<"Your number is "<< input <<"\n";
  return 0;
}

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-Prelude