Simple try catch block problem.

[foco]

My goal is to have a simple piece of code that will raise an exception if input of non-int is entered.

My problem so far, is that I haven't been able to reach the catch block. I have read a little about bad_typeid, but do not know how to implement it.

#include <iostream>
#include <typeinfo>
using namespace std;

int isnum() //reads an int from user. Raises error if not of type int
{
  int temp;
  cout << "Please enter an int\n";
  cin >> temp;
  if (typeid(temp) != typeid(2)) //if input is not of same type as int
  {
    throw 4;
  }
  return temp;
}

int main()
{

  try
  {
    int temp = isnum();
  }
  catch (int e)
  {
    cout << "An exception occurred. Exception Nr. " << e << endl;
  }

  return 0;
}

Thanks.

[camelCase]

Hello,

Here is a source to do this task. Though it does not rely on a try,catch,throw block - it still gets the job done. I apologize if it does not meet your goal.

#include <iostream>
// this application by camelCase will resolve user input. If input must be translated into an ASCII table then the input was bad. Else, input is good.
using namespace std;

int main ()
{
    cout << "Please enter a number followed by a <return> or <enter>: ";
    double number;
    while (!(cin >> number))
    {
        // Enter this loop if input fails because of invalid data. Data will be evaluated on an ACSII standard.
        cout << "Sorry, I said enter a number!  Try again: ";
        cin.clear ();   // reset the "failure" flag

        // The input "cursor" is still positioned at the beginning of
        // the invalid input, so we need to skip past it.

        cin.ignore (1000, '\n');  // Skip to next newline or 1000 chars,
                                  // whichever comes first.  This is why
                                  // you have to follow the input with
                                  // <return> or <enter>.
    }
    cout << "OK, you entered " << number << "." << endl;
	getchar();
	getchar();
    return 0;
}

[vijayan121]

The result of a typeid expression is a std::typeinfo object representing the *type* of the expression.
Type of temp is an int, so typeid(temp) and typeid(4) both give typeid(int).

To check if the user entered an int or not, check the state of the stream after the attempted input. If an int could not be read, the stream would be put into a failed state.

if( !( cin >> temp ) ) throw 4 ;