stop user from entering a letter when there should enter a number

Question

kankaortiz 0 Newbie Poster

12 Years Ago

how would i stop a user from entering a letter when they should be entering a number?

#include <iostream>
      #include <cstdlib>
      #include <cstdio>
      #include <cstring>
     
      usin namespace std;

      int main(int nNumberofArgs, char* pszArgs[])

{
char nAns;

do{
  
     int nPly1;
     int nPly2;
     int nPly3;
     int nPly4;
     int nPly5;
     int nPly6;
     int nMandril;

cout<< "this program calculates an O.D. for you\n\n";

cout<< " enter seven numbers:\n\n";

cout<< " ply 1 if no ply enter 0-:";
cin>>nPly1;
cout<<\n;
cout<<" ply 2 if no ply enter 0-:";
cin>> nPly2;
cout<<\n;
cout<< "ply 3 if no ply enter 0-:";
cin>>nPly3;
cout<<\n;
cout<< " ply 4 if no ply enter 0-:";
cin>> nPly4;
cout<<\n;
cout<< "ply 5 if no ply enter 0-:";
cin>> nPly5;
cout<<\n;
cout<< "ply 6 if no ply enter 0-:";
cin>>nPly6;
cout<<\n;
cout<<"mandril -:";
cin>>nMandril;
cout<<\n;

cout<<"outside diameter = ";
cout<< (nPly1+nPly2+nPly3+nPly4+nPly5+nPly6)*2+nMandril;

cout<<" do you want to do this again? (y/n)";
cin>>nAns;

}while(nAns=='y'||nAns=='Y');

}

c++

7 Contributors
10 Replies
5K Views
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Latest Post 12 Years Ago Latest Post by adityatandon

Dman01 10 Light Poster

12 Years Ago

I would use getch() from conio.h.

For example :

#include <conio.h> // for getch()
#include <iostream>
using namespace std;

int main()
{

    int number=0;
    int help=0;
    char buffer=0;

    cout << "Type in any number : ";

    do
    {
        buffer = getch(); // reads the key from user
        help = buffer - '0';

        if (help >= 0 && help <= 9)
        {
            number *= 10;
            number += help;
            cout << help;
        }
        else
        {
            //cout << "\nERROR : No letters are allowed!";
        }

    }while(buffer != 13); // 13 = Return

    cout << "\nEntered number : " << number;

    getch();

    return 0;
}

Edited 12 Years Ago by Dman01 because: n/a

WaltP 2,905 Posting Sage w/ dash of thyme

12 Years Ago

Dman01's suggestion only works on a select few compilers, therefore is not a good solution.

You need to read the input as a string of characters, test the characters to make sure they all all of the correct type, then either
1) show an error
2) convert to integer

Narue 5,707 Bad Cop

12 Years Ago

Let me know, if you think its not correct.

It's not correct. Let's ignore that unistd.h isn't a standard header (you don't use anything from it anyway) and go right to the first real error. Boiled down, it looks like this:

char *no;

cin >> no;

int len = strlen(no);

First, you forgot to include <cstring> for strlen(). Second, you clearly don't understand how pointers work because an uninitialized pointer does not point to infinite memory. A pointer must be initialized to memory that your process owns before it can be used.

C++ is not just C with cin and cout, there are many more things that can make your life easier:

#include <algorithm>
#include <iostream>
#include <string>
 
using namespace std;
 
int main()
{
    string line;
    
    while (true) {
        cout << "Enter a number (EOF to quit): ";
        
        if (!getline(cin, line))
            break;
        
        bool not_num = any_of(line.begin(), line.end(), [](char c) { return !isdigit(c); });
        
        cout << "'" << line << "' is" << (not_num ? " not " : " ") << "a number\n";
    }
}

Notice how you don't have to worry about pointers with the std::string class. That's one of the key benefits for people who are low level programming challenged. ;)

All of this is assuming that the need for a "number" means a series of digits rather than a value suitable for conversion to one of the integer or floating-point types. If you need to convert to a numeric type, the naive algorithm of "only allow digits" isn't sufficient as it rejects valid numbers and doesn't protect against numbers beyond the supported range.

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vijayan121 1,152 Posting Virtuoso · Answer 1 · 2011-12-25T10:50:59+00:00

> how would i stop a user from entering a letter when they should be entering a number?

There is no really portable way of doing that. What can be done is: let the user enter any set of characters, and if those do not form a valid number, ask for input again.

This may be all that you need:

int number ;
while( !( std::cout << "number? "  && std::cin  >> number ) )
{
    std::cin.clear() ;
    std::cin.ignore( std::numeric_limits<std::streamsize>::max(), '\n' ) ;
}

With this, an input of 129XY would accepeted and treated as equivalent to 129. If 129XY is to be treated as invalid input, then validating the input as a string is required.

Use std::istringstream for this; it is easier than trying to validate it character by character yourself. 567, -567, +567 are all valid; 5-67, +-567, +5+67 are not.

int get_int()
{
    int number ;
    std::string str ;
    std::cout << "number? " && std::cin >> str ;
    std::istringstream stm(str) ;
    return stm >> number && stm.eof() ? number : get_int() ;
}

kankaortiz 0 Newbie Poster · Answer 2 · 2011-12-28T00:47:48+00:00

> how would i stop a user from entering a letter when they should be entering a number?
There is no really portable way of doing that. What can be done is: let the user enter any set of characters, and if those do not form a valid number, ask for input again.
This may be all that you need:
int number ;
while( !( std::cout << "number? "  && std::cin  >> number ) )
{
    std::cin.clear() ;
    std::cin.ignore( std::numeric_limits<std::streamsize>::max(), '\n' ) ;
}
With this, an input of 129XY would accepeted and treated as equivalent to 129. If 129XY is to be treated as invalid input, then validating the input as a string is required.
Use std::istringstream for this; it is easier than trying to validate it character by character yourself. 567, -567, +567 are all valid; 5-67, +-567, +5+67 are not.
int get_int()
{
    int number ;
    std::string str ;
    std::cout << "number? " && std::cin >> str ;
    std::istringstream stm(str) ;
    return stm >> number && stm.eof() ? number : get_int() ;
}

being that i am new to programing i dont really know what all this means or where i would place this code, if you could explain a bit i would really appreciate it.

vijayan121 1,152 Posting Virtuoso · Answer 3 · 2011-12-28T12:00:52+00:00

> if you could explain a bit i would really appreciate it.

Star by reading http://www.learncpp.com/cpp-tutorial/135-stream-states-and-input-validation/
and http://latedev.wordpress.com/2011/11/16/c-stringstreams/

Google would throw up many more links.

KMat 0 Newbie Poster · Answer 4 · 2011-12-28T14:17:44+00:00

Hi,

I hope the below code will help you.

#include <iostream>
#include <unistd.h>

using namespace std;

int main( int argc , char **argv)
{

   char want;
   char *no;

   do {

      cout << "Enter a number : ";
      cin >> no;

      int len = strlen(no);
 
      int i=0;
      bool stat = true;

      while( i < len ) {
         if( isdigit((int)no[i]) ) {
            stat = true;
         } else {
            cout << "You have entered " << no[i] << ", this is not a number. so please type a number\n";
            stat = false;
            break;
         }
         i++;
      }

      if( stat == true ) {
        cout << "Yes you entered the input as numbers...\n";
      }

      cout << "do you still want to continue.. press 'y'";

      cin.clear();
      cin >> want;

   } while( want == 'y' );
}

Let me know, if you think its not correct.

BR/ KMat

how would i stop a user from entering a letter when they should be entering a number?

#include <iostream>
      #include <cstdlib>
      #include <cstdio>
      #include <cstring>
     
      usin namespace std;

      int main(int nNumberofArgs, char* pszArgs[])

{
char nAns;

do{
  
     int nPly1;
     int nPly2;
     int nPly3;
     int nPly4;
     int nPly5;
     int nPly6;
     int nMandril;

cout<< "this program calculates an O.D. for you\n\n";

cout<< " enter seven numbers:\n\n";

cout<< " ply 1 if no ply enter 0-:";
cin>>nPly1;
cout<<\n;
cout<<" ply 2 if no ply enter 0-:";
cin>> nPly2;
cout<<\n;
cout<< "ply 3 if no ply enter 0-:";
cin>>nPly3;
cout<<\n;
cout<< " ply 4 if no ply enter 0-:";
cin>> nPly4;
cout<<\n;
cout<< "ply 5 if no ply enter 0-:";
cin>> nPly5;
cout<<\n;
cout<< "ply 6 if no ply enter 0-:";
cin>>nPly6;
cout<<\n;
cout<<"mandril -:";
cin>>nMandril;
cout<<\n;

cout<<"outside diameter = ";
cout<< (nPly1+nPly2+nPly3+nPly4+nPly5+nPly6)*2+nMandril;

cout<<" do you want to do this again? (y/n)";
cin>>nAns;

}while(nAns=='y'||nAns=='Y');

}

adityatandon 23 Junior Poster in Training · Answer 5 · 2011-12-31T01:11:47+00:00

Agree with narue's code !
Another modification that can be used, that is dman01's idea using getch()

#
#        \\include necessary headers
#

void main()
{clrscr();

 int i=0;
 char a[50];

 while(a[i]!='\r')  \\ reading carriage return(enter key)
   {a[i]=getch();
        
     if(isdigit(a[i])==0)
          {cout<<"You have entered a character, please enter only numbers";
            break;
           }
      ++i;
    }
}

The above program breaks from the loop if a character is entered.

Advantage : It tells the user at that instance whether he has given an incorrect input rather than after entering the whole string

Disadvantage : Not a very reusable code, may not work on all versions of c++

WaltP 2,905 Posting Sage w/ dash of thyme Team Colleague · Answer 6 · 2011-12-31T02:26:50+00:00

Agree with narue's code !
Another modification that can be used, that is dman01's idea using getch()
Advantage : It tells the user at that instance whether he has given an incorrect input rather than after entering the whole string
Disadvantage : Not a very reusable code, may not work on all versions of c++

Disadvantage : this is why we do NOT make this suggestion... It works on very few versions of C++.

adityatandon 23 Junior Poster in Training · Answer 7 · 2011-12-31T08:08:52+00:00

Dude, get real... portability of a code IS a problem faced by every programmer.. And like i said, using dman01's idea... and on the version he uses, it IS possible.. And i can say that coz he used conio.h which isnt a part of the VC ++ library or the newer versions but instead for older versions like turbo c++. All the other codes also arent reusable if u think they are...

Only difference - I pointed it out !