how i can control that ?

cout<<"\nEnter your <student number>:";
cin>>studentnumber;

if user enter alphabetic values program will want user and will ask again user ?
how i can do that

Lynx876 (742)

Have a read through the following link. It will help you with what you need and stop an infinite loop when a char is entered in to an int.

http://www.cplusplus.com/forum/articles/6046/

dangrr888 (80)

Hi malikrgc,

this can be tricky, here's something I came up with.

#include <iostream>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstddef>

using std::string; using std::isdigit;
using std::endl; using std::cin;
using std::cout; using std::size_t;

bool all_numbers(const string& s)
{
  const char* str = s.c_str();
  bool ret = true;

  for(size_t i = 0; i != s.size(); ++i)
    ret = (ret && isdigit(str[i])) == 0 ? false : true;
    
  return ret;
}

int main()
{
  string studentNumber;

  cout << "\nEnter your <student number>: ";
  cin >> studentNumber;
  
  while(!all_numbers(studentNumber))
    {
      cout << "Sorry, you entered >=1  non-digit, please try again!: ";
      cin >> studentNumber;
    }
  
  cout << "Thank You!" << endl;

 return 0;
}

With an example output as follows:


Enter your <student number>: asdasda213123
Sorry, you entered >=1  non-digit, please try again!: 231231#
Sorry, you entered >=1  non-digit, please try again!: 123123+
Sorry, you entered >=1  non-digit, please try again!: 123123'
Sorry, you entered >=1  non-digit, please try again!: 23123122?
Sorry, you entered >=1  non-digit, please try again!: 123123~
Sorry, you entered >=1  non-digit, please try again!: 31231
Thank You!

Hope this helps, good luck!

malikrgc (34)

thank you very much dngr88 that help but i have delivered my project it was a part of my project i

packetpirate (388)

Here's a more advanced solution to your problem:

template<typename T>
T grab(const char * prompt)
{
    T thing;         
    std::istream::iostate old_state = cin.exceptions();
    cin.exceptions(std::istream::failbit);
            
    while(true)
    {
        try
        {
            cout << prompt;
            cout.flush();
            cin >> thing;
            cin.exceptions(old_state);
            
            return thing;    
        } catch(std::istream::failure &e) {
            cout << "Invalid input. Try again." << std::endl;
            cin.clear();
            cin.ignore(1024, '\n'); 
        }
    }
    
    return thing;
}

Then just call the function like this:

int i = grab<int>("Enter a number: ");

dangrr888 (80)

@packetpirate
Yes, I like your solution. Although the last return statement is redundant I think.

packetpirate (388)

No it isn't. You can't put a return statement in just the conditional block, otherwise it will have a condition in which it does not return anything, which is a syntax error.

dangrr888 (80)

@packetpirate

No it isn't. You can't put a return statement in just the conditional block, otherwise it will have a condition in which it does not return anything, which is a syntax error.

I am not sure I agree. You always enter the loop, and either don't throw an exception, in which case you exit the function via the first return, or throw an exception, in which case you recheck the loop condition, which is always true, and enter the loop again. How can you ever get to the last return statement?

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how i can control that ?