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Jul 6th, 2009

Checking if variable is a number or a letter: Error

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Hi, I'm looking for some help with this short program I am writing...

I'm trying to make the sure that the user enters a number when it asks for the temperature in Celsius, and when the user enters a number it works. However if a letter is entered it all goes to hell with the message scrolling down the screen very quickly! Please advise!

If you can think of an alternative way of going around this, and stopping the error from occurring in the first place that would be great!

C++ Syntax (Toggle Plain Text)
#include <stdio.h>
#include <iostream>
 
using namespace std;
int main(int nNumberofArgs, char * pszArgs[])
{
	for(;;)
	{
	// enter the temperature in celsius
	int nCelsius;
	int nIsANumber;
 
	nIsANumber = 0;
	cout << "Enter the temperature in Celsius: ";
	while(nIsANumber == 0)
	{
		cin >> nCelsius;
		if(!isdigit(nCelsius))
		{
			nIsANumber = 1;
		}
		else {
			cout << "That was NOT a number!!! \nPlease enter the temperature in Celsius: ";
		}
	}
 
 
	//Conversion factor for celsius to fahrenheit
	int nFactor;
	nFactor = 212 - 32;
	int nFahrenheit;
	nFahrenheit = nFactor * nCelsius/100 + 32;
	//output the results
	cout << "Fahrenheit temperature is:";
	cout << nFahrenheit;
	//loop?
	cout << "\n\nWould you like to preform another conversion? \nEnter Y for yes or N for no: ";
 
	char cDoContinue;
	cin >> cDoContinue;
	cout << "\n\n\n";
	if(tolower(cDoContinue) == 'n'){
		break;
	}
	else if(!(tolower(cDoContinue) == 'y')){
		cout << "You are an idiot... That was not Y or N!!!\n\n";
	}
 
	}
	return 0;
}
#include <stdio.h>
#include <iostream>

using namespace std;
int main(int nNumberofArgs, char * pszArgs[])
{
	for(;;)
	{
	// enter the temperature in celsius
	int nCelsius;
	int nIsANumber;
	
	nIsANumber = 0;
	cout << "Enter the temperature in Celsius: ";
	while(nIsANumber == 0)
	{
		cin >> nCelsius;
		if(!isdigit(nCelsius))
		{
			nIsANumber = 1;
		}
		else {
			cout << "That was NOT a number!!! \nPlease enter the temperature in Celsius: ";
		}
	}


	//Conversion factor for celsius to fahrenheit
	int nFactor;
	nFactor = 212 - 32;
	int nFahrenheit;
	nFahrenheit = nFactor * nCelsius/100 + 32;
	//output the results
	cout << "Fahrenheit temperature is:";
	cout << nFahrenheit;
	//loop?
	cout << "\n\nWould you like to preform another conversion? \nEnter Y for yes or N for no: ";
	
	char cDoContinue;
	cin >> cDoContinue;
	cout << "\n\n\n";
	if(tolower(cDoContinue) == 'n'){
		break;
	}
	else if(!(tolower(cDoContinue) == 'y')){
		cout << "You are an idiot... That was not Y or N!!!\n\n";
	}

	}
	return 0;
}

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Pokenerd

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since Jul 2009

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Jul 6th, 2009

Re: Checking if variable is a number or a letter: Error

The atoi function returns an integer from a string. So:

C++ Syntax (Toggle Plain Text)
char buf[256];
// Ask for input here
cin.getline(buf,256);
int num = atoi(buf);
if(num == 0)
    cout << "Error";
char buf[256];
// Ask for input here
cin.getline(buf,256);
int num = atoi(buf);
if(num == 0)
    cout << "Error";

Last edited by u8sand; Jul 6th, 2009 at 8:40 pm.

u8sand

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Jul 6th, 2009

Re: Checking if variable is a number or a letter: Error

Lines 10 and 17. You define nCelsius as an integer, then you read it in as integer. If it's NOT an integer, the >> operation fails. Since you can't ASSUME it's an integer (that's the whole reason you're checking), you can't read it in as an integer. Read it in as a string, THEN check to see whether it's an integer.

VernonDozier

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Jul 6th, 2009

Re: Checking if variable is a number or a letter: Error

Click to Expand / Collapse Quote originally posted by u8sand ...

The atoi function returns an integer from a string. So:
C++ Syntax (Toggle Plain Text)
char buf[256];
// Ask for input here
cin.getline(buf,256);
int num = atoi(buf);
if(num == 0) // atoi returns 0 if it is invalid
    cout << "Error";
char buf[256]; // Ask for input here cin.getline(buf,256); int num = atoi(buf); if(num == 0) // atoi returns 0 if it is invalid cout << "Error";

Which is why you retrieve the input with a string (because a string can handle any text input) and turn it into an integer if its valid.

Last edited by u8sand; Jul 6th, 2009 at 8:42 pm.

u8sand

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Jul 6th, 2009

Re: Checking if variable is a number or a letter: Error

Click to Expand / Collapse Quote originally posted by VernonDozier ...

Lines 10 and 17. You define nCelsius as an integer, then you read it in as integer. If it's NOT an integer, the >> operation fails. Since you can't ASSUME it's an integer (that's the whole reason you're checking), you can't read it in as an integer. Read it in as a string, THEN check to see whether it's an integer.

How?

Pokenerd

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Jul 6th, 2009

Re: Checking if variable is a number or a letter: Error

You're trying to test a numerical value for an ASCII character!

if(!isdigit(nCelsius))

isdigit( char ) thus "9834" first character '9'
But its encoded as 9834 an integer.

What they said! Simultaneous posting!

Last edited by wildgoose; Jul 6th, 2009 at 8:44 pm. Reason: note

wildgoose

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Jul 6th, 2009

Re: Checking if variable is a number or a letter: Error

Thanks for your help - I got it working!

Pokenerd

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