Solved

If statement accepting a through z only... need help!

Posted on 2007-12-06
8
292 Views
Last Modified: 2007-12-09
I need help with this code I need my if statement to accept a through z only and if a number or something else is entered it would fall into the else statement. When I entered a word if goes into the else I do not know what it is doing this.

Here is what I have for this file:
#include <iostream>
#include <string>
#include <fstream>
#include "game.h"
 
#define MaxWordSize 10
#define MaxWords 50
typedef char String[MaxWordSize];
String Words[MaxWords - 1];
int Count;
char Word[MaxWordSize];
int Size;
int Loop;
Game myGame;
 
using namespace std;
 
void Game::addWord()
{
    myGame.loadFile();
    ofstream Datfile("words.txt");
try
{
    cout << "\tEnter a new word, keep it below " << MaxWordSize << " characters: ";
    cin >> Word;
	if(Word == "ABCDEFGHIJKLMNOPQRSTUVWXYZ")
	{
    Size = strlen(Word);
 
    strcpy_s(Words[Count++],MaxWordSize,Word);
 
    for (int i = 0; i < Count; ++i) Datfile << Words[i] << endl;
	}
	else
	{
		throw 1;
	}
}
	catch(int e)
	{
		cout << "Error...\n";
	}
}
void Game::deleteWord()
{
	cout << "you have chosen to delete a word...\n";
}
 
void Game::loadFile()
{
	char C;
	ifstream Datfile;
 
	Count = 0;
 
	Datfile.open("words.txt");
 
	while((C=Datfile.peek()) !=EOF)
	{
		Datfile >> Words[Count++];
 
		if(Count > MaxWords - 1)
		{
			cout << "\nToo many words in the file, stopping with " << MaxWords << " words." << endl;
			Count = MaxWords;
			break;
		}
	}
 
	Count--;
 
	Datfile.close();
}

Open in new window

0
Comment
Question by:jschmuff
  • 4
  • 3
8 Comments
 
LVL 55

Accepted Solution

by:
Jaime Olivares earned 300 total points
ID: 20425122
>>      if(Word == "ABCDEFGHIJKLMNOPQRSTUVWXYZ")

this won't work at all, you can design your own function, so you can use  like:

      if(validateWord(Word))

Here is the validating funcion:


// this will require: #include <stdio.h>
 
bool validateWord(char *word)
{
       for (int i=0; word[i]; i++)
            if (!isletter(word[i])
                  return false;
       return true;
}

Open in new window

0
 

Author Comment

by:jschmuff
ID: 20425168
what does that have to do with taking the input only a through z?
0
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20425181
it validates if all characters in Word are A though Z and returns you a boolean value.
your code:
if(Word == "ABCDEFGHIJKLMNOPQRSTUVWXYZ")
doesn't work as you expected and always return false, so exception will never occur.
0
Announcing the Most Valuable Experts of 2016

MVEs are more concerned with the satisfaction of those they help than with the considerable points they can earn. They are the types of people you feel privileged to call colleagues. Join us in honoring this amazing group of Experts.

 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20425186
sorry, in this case, exception will always occur.
0
 

Author Comment

by:jschmuff
ID: 20425202
so that is something that would be done before the if statement since if(validateWord(Word)) is that correct.
so to actually do a - z I have to use a for loop? What is the isletter in this example you have given me? cause it is something that will need to be declared.
0
 

Author Comment

by:jschmuff
ID: 20425211
Is there another way this can be done like if(number is input) { throw 1; } else then the code to add and save word to file.. catch blah blah blah?
0
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20425236
>>so that is something that would be done before the if statement since if(validateWord(Word)) is that correct
validateWord is a separate function.

>>so to actually do a - z I have to use a for loop?
yes, you have to use a loop

>>What is the isletter in this example you have given me
isletter is a function declared in the stdio.h header, it test if a character is a letter or not. But you can use isalpha instead, it is more compatible.
If you don't want to use it, you can do this instead:


bool validateWord(char *word)
{
       char c; 
       for (int i=0; c = word[i]; i++)
            if (!(c>='A' && c<='Z') && !(c>='a' && c<='z'))
                  return false;
       return true;
}

Open in new window

0
 
LVL 40

Assisted Solution

by:evilrix
evilrix earned 200 total points
ID: 20434994
use std::string::find_first_not_of: -

http://www.cppreference.com/cppstring/find_first_not_of.html
char const * Word = "qwerty"; // Your word!
 
if(std::string::npos != std::string(Word).find_first_not_of("abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"))
{
	throw std::runtime_error("Invalid data");
}

Open in new window

0

Featured Post

Gigs: Get Your Project Delivered by an Expert

Select from freelancers specializing in everything from database administration to programming, who have proven themselves as experts in their field. Hire the best, collaborate easily, pay securely and get projects done right.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
Better understanding on C++ Class serialization and formats 9 65
base64 decode encode 12 133
c++ reading data from file into two dimensional array 3 107
Math Equation 23 82
Article by: SunnyDark
This article's goal is to present you with an easy to use XML wrapper for C++ and also present some interesting techniques that you might use with MS C++. The reason I built this class is to ease the pain of using XML files with C++, since there is…
If you haven’t already, I encourage you to read the first article (http://www.experts-exchange.com/articles/18680/An-Introduction-to-R-Programming-and-R-Studio.html) in my series to gain a basic foundation of R and R Studio.  You will also find the …
The viewer will learn how to implement Singleton Design Pattern in Java.
This video teaches viewers about errors in exception handling.

813 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

12 Experts available now in Live!

Get 1:1 Help Now