Check for an integer, fail if anything else

I'm working on an integration program, that works out the integation by approximating with the rectangle rule.

I am trying to ask how many strips do you want? This MUST be an integer, yet if i enter a float, it will work. Is there a way of checking? i have looked but it seems that this integer check business is not as simple a check as it could be.

just need some code that will say

Code:

---


********* code*********

cout << "Please enter the number of strips:  ";
cin >> n;

is n an integer?

      no - please retry

yes, carry on........

*********code*********

---

What might n be?

sorry, n is an integer.

Code:

---

#include<iostream>
using namespace std;
main()
{
float  c;
int d;

retry:
cout<<"enter a number";
cin>>c;
cout<<"\n";
d=c;
if((d-c)!=0)goto retry;



printf("%f",c);
}

---

this goes on and on until the user inputs and integer.

What the hell? I already answered this question for you yesterday and you even thanked me.

Code:

---

#include <limits>

/* *** CODE HERE *** */

std::cout << "Please enter the value for a, the lower bound  ";
while(!(std::cin >> a))    // If the stream fails from bad input
{
  std::cin.clear();            // Clear the fail state
  std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');  // Clear the buffer
  std::cout << "Invalid statement, please re-enter: ";    // Reprompt
}

---

...and please don't listen to qqqqxxxx. goto is not the best way to do anything.

Read the user input to a string. Check that each character in the string is valid, say by using std::isdigit() from <cctype> on each of them. If this checking passes, then use a stringstream to convert the string to an integer.

slymaelstrom: yes, but for some reason it wouldn't work when i entered a float.

no idea why. prob me!!

but it worked for when i entered a character, just not a float, and it needs to be a float.

There is no reason that code wouldn't work for a float. Post what you had.

Quote:

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There is no reason that code wouldn't work for a float. Post what you had.

---

From what I understand, the OP needs an integer, but your example will still accept floating point input, although it will only read the integer portion.

QUOTED BY SLYMAELSTORM
>...and please don't listen to qqqqxxxx. goto is not the best way to do anything.

hmmmm

ok, this is what i have. apologies for incorect formatting, i'm new to this. i'm having to go out in a few minutes so i won't reply for a while either.

Code:

---


#include <iostream.h>
#include <iomanip.h>
#include <cmath.h>
#include <fstream.h>
#include <limits.h>
#include <stdio.h>
#include <conio.h>

float funcj(float);

int main()
{
    int n, i, m;
    float a, b, h, x, s, integral;
                           
   

    clrscr();
    cout << "************************************************************************\n\n";
    textcolor (12);
    cprintf ("The function you are integrating is x^2 - 5x - 3");

    cout << "\n\nWould you like to input your values?\n";
    cout << "OR would you like to input the values from the file rectangle.dat?";

    cout << "\n\nOption 1: Input the values\n";
    cout << "\nOption 2: Read the values\n";

    cout << "\n\nPlease enter your choice, 1 or 2:  ";
    cin >> m;

        if (m == 1)

        {

    cout << "\n\nPlease input a, the lower bound:  ";

   
    while(!(cin >> a))

        {

        textcolor (2);
        cprintf ("You have inputted the data incorrectly.");
        cout << "\nPlease try again.  ";
        return 0;
        }
     

    cout << "\nPlease input b, the upper bound:  ";

    //cin >> b;

    while(!(cin >> b ))

        {

        textcolor (2);
        cprintf ("You have inputted the data incorrectly.");
        cout << "\nPlease try again. ";
        return 0;
       
        }
   
    cout << "Please enter the number of strips:  ";
       
    while(!(cin >> n))    // If the stream fails from bad input
       
        {
         
        cin.clear();            // Clear the fail state
          cin.ignore(std::numeric_limits<streamsize>::max(), '\n');  // Clear the buffer
          cout << "Invalid statement, please re-enter: ";    // Reprompt
       
        }


        }

      else if (m == 2)

        {

        ifstream read ("rectangle.dat");

        read >> a >> b >> n;
       
        }

        else

        {
                textcolor (2);
                cprintf ("You have inputted data incorrectly.");
                cout << "\nPlease try again.";
                return 0;
        }

    if (a >= b)

    {
      textcolor (2);
      cprintf ("\nYou have entered the data incorrectly.");
      cout << "\na must be lower than b.\n";
      cout << "Please Try Again.\n";

    }

  // else if ( n < 1)

  //    {
  //    textcolor (2);
  //    cprintf ("\nYou have entered the data incorrectly.");
  //    cout << "\nYou must have a positive (integer) number of strips.\n";
  //    cout << "Please try again.";


  //    }

    else

    {
        integral = 0;

                                                                     
            h = fabs(a - b) / n;
       
            for  (i = 0; i < n; i++ )

            {
                                                       
                x = a + i * h;
       
                s = h * funcj(x);
                                                       
                integral = integral + s;
       
                     

            }

          textcolor (4);

          cprintf ("\nThe answer is: ");
          cout << integral << "\n";
    }
   
    return 0;
                                       
}         

float funcj(float x)
{
    float f;

    f = (x*x) - (5*x) - 3 ;

    return f;
}

---

i'm aware i haven't used the std that you have, so i will alter it if you need me too when i get back.

its not the finished article by any means. if you do run the program, if u use option 2, you'll need a file called rectangle.dat

which just contains 3 numbers.

Code:

---


2 5 100

---

// where a =2, b = 5, n = 100]

this just shows how i can read data from a file. its not needed but i'm just showing my teacher i can read data from a file as well as from inputting the data in the prog.

its by no means finished, but if i can just check for n being an integer, then i can then tweak it, so its more user friendly.

thanks for the help. :) its much appreciated.

sorry for the rushed messgae.

Quote:

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Originally Posted by derek tims

Code:

---

#include <iostream.h>
#include <iomanip.h>
#include <cmath.h>
#include <fstream.h>
#include <limits.h>
#include <stdio.h>
#include <conio.h>

---

i'm aware i haven't used the std that you have, so i will alter it if you need me too when i get back.

---

You should understand a few things here, First of all, "std" is the namespace which contains the C++ standard library.
(Note - Only in newer compilers. compilers older than about 1999 don't use namespaces)

However, you aren't actually using any of the C++ standard library, most of the headers you've got there are outdated - iostream, iomanip, cmath, fstream, limits and cstdio must be used without an .h
(Note conio.h is non-standard anyway, so that can be left alone, and in C++, all C headers are prefixed with "c", eg cstdio)

There are dozens of FAQ links and tutorials which explain this in more detail, so I won't go into that, but I suggest you do go and read up on the std namespace.

When using the C++ standard library, you can optionally add a line after your includes of

Code:

---

using namespace std;

---

which removes the need for you to put the std:: prefix in front of names such as std::cout.

Quote:

---

Originally Posted by qqqqxxxx

Code:

---

#include<iostream>
using namespace std;
main()
{
float  c;
int d;

retry:
cout<<"enter a number";
cin>>c;
cout<<"\n";
d=c;
if((d-c)!=0)goto retry;



printf("%f",c);
}

---

this goes on and on until the user inputs and integer.

---

Please stop trying to screw with people.

double post

HOMEYG
cut the crap,come up with some code first.
and the goto can be easily replaced by a while loop,but i suppose you r just lacking in talent.

ok, well what i have done is when i enter n, i store that value as a string.

now then, i read from that file, an integer q, and read the same value, a float p.

then i set up a while loop saying that when q-p is not equal to zero, then it brings about an error.

thanks for all the advice.

In your posted code, you use while(!(cin >> a)), but you didn't put the cin.clear() and cin.ignore() inside that while loop. It won't work without them.

If you want to make sure n is an int, you can change your loop to the following. It checks to see if there is anything typed after the integer is read in, so if the user types 3.6, the 3 will be read in and the . will cause it to give the error:

Code:

---

    while(!(cin >> n) || cin.get() != '\n')    // If the stream fails from bad input
        {
        cin.clear();            // Clear the fail state
          cin.ignore(1000, '\n');  // Clear the buffer
          cout << "Invalid statement, please re-enter: ";    // Reprompt
        }

---

You want to make a cast...

Code:

---

float x = p - (float)q;

---

or put your intenger on a float

Code:

---

float aux = q;
result = p - aux;

---

that is...

so bb..

------------------------------------------

Sorry for my poor english...

Or for another simple way of doing it, you could read in a string, and check for the presence of a '.' indicating a floating point number or any other non-numeric character, and reject it as you see it.

Really though, I've not seen a valid use of goto's that come to mind, I am sure in some rare case they could be useful, but in every instance I've been tempted to use them, I find a loop much easier to read in any event.

aside to qqqqxxxx: What you really wanted to do is use a do-while loop.

Code:

---

do {
cout<<"enter a number";
cin>>c;
cout<<"\n";
d=c;
} while ((d-c)!=0);

---