public static int getNumber(int MAXNUM)throws IOException
{
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
String userNum;
int intUserNum;
while (true) {
System.out.println("Please enter a number: ");
userNum = input.readLine();
if (isNumber(userNum) != true) {
System.out.println("Invalid Entry. Please enter a whole number: ");
continue;
}
intUserNum = Integer.parseInt(userNum);
if (intUserNum > MAXNUM) {
System.out.println("Invalid Entry. Number too large.");
continue;
}
else {
return intUserNum;
}
}
}
getNumber methodIs there a more efficient way to do this?
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1 Replies - 964 Views - Last Post: 26 March 2010 - 11:49 PM
#1
getNumber method
Posted 26 March 2010 - 11:20 PM
Below is my getNumber method. It's used to get a number from the user, it checks if the value is a number using the isNumber method, which returns a true if it's a number and false if it is not. And if the value is less than the Maximum number passed to the method. What I have works, I just think there's probably a more efficient way to do it. Any suggestions on a way to make this more efficient?
Replies To: getNumber method
#2
Re: getNumber method
Posted 26 March 2010 - 11:49 PM
this should do it
public static int getNumber(int MAXNUM)
{
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
String userNum;
int intUserNum=0;
boolean keepGoing = true;
while (keepGoing) {
try{
System.out.println("Please enter a number: ");
userNum = input.readLine();
intUserNum = Integer.parseInt(userNum);
if(intUserNum <= MAXNUM)
keepGoing = false;
}
catch(Exception e){
System.out.println("invalid number");
}
}
return intUserNum;
}
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