Programming Forums > Application Development > Java

tumbleTetris · May 20th, 2006, 10:49 PM

I'm making a menu, it has 3 choices, the last is x which is cancel.

I don't understand how I can get the computer to understand x as an x since I have

(Toggle Plain Text)

int choice = keyboard.nextInt();

and I know that it won't except anything of type 'char'.

The menu looks something like this:

1. Choice A
2. Choice B
x. Cancel

Wizard1988 · May 20th, 2006, 10:57 PM

First get the ASCII value of the key pressed check to see if it is the same as x and then check if it is 1 or 2

tumbleTetris · May 20th, 2006, 11:00 PM

sorry, how do I go about that?

Wizard1988 · May 20th, 2006, 11:12 PM

I haven't toyed around with Java for quite some time but i think that you should read the input and then cast it to an integer.

(Toggle Plain Text)

char input = kbd.read();

if((int)input == 'x')
{
  //do x option
}
if(input == '1')
{
  //do option 1
}

titaniumdecoy · May 20th, 2006, 11:15 PM

Why not use the char returned by keyboard.nextInt()?

(Toggle Plain Text)

char choice = keyboard.nextInt();

switch (choice) {
    case '1':
      // do something
      break;
    case '2':
      // do something else
      break;
    default:
      // otherwise do this
}

tumbleTetris · May 20th, 2006, 11:27 PM

Ok, I got it to work.

But I don't understand, if I

(Toggle Plain Text)

int input = kbd.read();

if((char)input == 'x')
{
  //do x option
}

works, however

(Toggle Plain Text)

if(input = '1')
{
  //do option 1
}

doesn't, I made it 1 equal because it was of type int (I thinkt hats how its supposed to be).

I think I'm just making an obvious mistake here, but I'm kind of new to programming so I make a lot of mistakes like this

lectricpharaoh · May 21st, 2006, 7:37 AM

As Wizard1988 says, you need to use == for comparisons, not =. The former is a test for equality, and the latter is an assignment. In some languages, like C and C++, any nonzero result from an expression is treated as true. In Java, this is not the case, but if you do an assignment into a boolean variable, then the compiler will not flag this as an error.

One trick when comparing to a constant is to put the constant on the left hand side of the expression. For example, instead of

(Toggle Plain Text)

if(flag == true)

use

(Toggle Plain Text)

if(true == flag)

This way, if you mis-type the == as =, the compiler will catch it, and give you an error (since you cannot assign to a constant). You can use this trick with literal constants, variables declared as final in Java or const in C/C++, or with enumerated types. The trick works in Java, C, C++, C#, and I'd expect it works in any other languages that use a C/C++ style syntax.

Wizard1988 · May 20th, 2006, 11:43 PM

Don't worry about not knowing this, it is just part of programming. You learn things as you go along.

If you put anything in single quotes as in 'x' or '1' you take its ASCII value. You can see how this works by casting chars to integers and the other way around.

You should also fix your if statement it should be

(Toggle Plain Text)

if(input == '1')
{
  //do option 1
}

instead of

(Toggle Plain Text)

if(input = '1')
{
  //do option 1
}

tumbleTetris · May 21st, 2006, 8:42 AM

thank you everyone

I was wondering if I was doing what I was doing in the most efficient manner possible.

The program is supposed to calculate ticket prices, there are two different sessions and 3 different seating plans. What I'm doing is asking for the session time in and then in two different statements I am basically asking for the seats, I don't think this is a very efficient way as I basically repeat myself twice, and was wondering if anyone thought of a better way of doing it

tumbleTetris · May 21st, 2006, 9:33 AM

ooh, another little technicality.

if a user inputs something that is not 1 2 or x, how can I tell them they made a mistake and should reinput something 1 2 or x.

If I use else if's I can't ask again, can I?

May 20th, 2006, 10:49 PM	#1
tumbleTetris Programmer Join Date: May 2006 Posts: 39 Rep Power: 0	how can x be a valid choice in type int? I'm making a menu, it has 3 choices, the last is x which is cancel. I don't understand how I can get the computer to understand x as an x since I have (Toggle Plain Text) int choice = keyboard.nextInt(); int choice = keyboard.nextInt(); and I know that it won't except anything of type 'char'. The menu looks something like this: 1. Choice A 2. Choice B x. Cancel

May 20th, 2006, 11:12 PM	#4
Wizard1988 Professional Programmer Join Date: Oct 2005 Location: Chitown Posts: 422 Rep Power: 4	I haven't toyed around with Java for quite some time but i think that you should read the input and then cast it to an integer. (Toggle Plain Text) char input = kbd.read(); if((int)input == 'x') { //do x option } if(input == '1') { //do option 1 } char input = kbd.read(); if((int)input == 'x') { //do x option } if(input == '1') { //do option 1 }

May 20th, 2006, 11:15 PM	#5
titaniumdecoy Expert Programmer Join Date: Nov 2005 Posts: 931 Rep Power: 4	Why not use the char returned by keyboard.nextInt()? (Toggle Plain Text) char choice = keyboard.nextInt(); switch (choice) { case '1': // do something break; case '2': // do something else break; default: // otherwise do this } char choice = keyboard.nextInt(); switch (choice) { case '1': // do something break; case '2': // do something else break; default: // otherwise do this }

May 20th, 2006, 11:27 PM	#6
tumbleTetris Programmer Join Date: May 2006 Posts: 39 Rep Power: 0	Ok, I got it to work. But I don't understand, if I (Toggle Plain Text) int input = kbd.read(); if((char)input == 'x') { //do x option } int input = kbd.read(); if((char)input == 'x') { //do x option } works, however (Toggle Plain Text) if(input = '1') { //do option 1 } if(input = '1') { //do option 1 } doesn't, I made it 1 equal because it was of type int (I thinkt hats how its supposed to be). I think I'm just making an obvious mistake here, but I'm kind of new to programming so I make a lot of mistakes like this

May 21st, 2006, 7:37 AM	#7
lectricpharaoh SEXY SHOELESS GOD OF WAR! Join Date: Jun 2005 Location: Wet west coast of Canada Posts: 1,194 Rep Power: 5	As Wizard1988 says, you need to use == for comparisons, not =. The former is a test for equality, and the latter is an assignment. In some languages, like C and C++, any nonzero result from an expression is treated as true. In Java, this is not the case, but if you do an assignment into a boolean variable, then the compiler will not flag this as an error. One trick when comparing to a constant is to put the constant on the left hand side of the expression. For example, instead of (Toggle Plain Text) if(flag == true) if(flag == true) use (Toggle Plain Text) if(true == flag) if(true == flag) This way, if you mis-type the == as =, the compiler will catch it, and give you an error (since you cannot assign to a constant). You can use this trick with literal constants, variables declared as final in Java or const in C/C++, or with enumerated types. The trick works in Java, C, C++, C#, and I'd expect it works in any other languages that use a C/C++ style syntax. __________________ And once again, Probability proves itself willing to sneak into a back alley and service Drama as would a copper-piece harlot. - Vaarsuvius, Order of the Stick

May 20th, 2006, 10:57 PM	#2
Wizard1988 Professional Programmer Join Date: Oct 2005 Location: Chitown Posts: 422 Rep Power: 4	First get the ASCII value of the key pressed check to see if it is the same as x and then check if it is 1 or 2

May 20th, 2006, 11:00 PM	#3
tumbleTetris Programmer Join Date: May 2006 Posts: 39 Rep Power: 0	sorry, how do I go about that?

May 20th, 2006, 11:43 PM	#8
Wizard1988 Professional Programmer Join Date: Oct 2005 Location: Chitown Posts: 422 Rep Power: 4	Don't worry about not knowing this, it is just part of programming. You learn things as you go along. If you put anything in single quotes as in 'x' or '1' you take its ASCII value. You can see how this works by casting chars to integers and the other way around. You should also fix your if statement it should be (Toggle Plain Text) if(input == '1') { //do option 1 } if(input == '1') { //do option 1 } instead of (Toggle Plain Text) if(input = '1') { //do option 1 } if(input = '1') { //do option 1 }

May 21st, 2006, 8:42 AM	#9
tumbleTetris Programmer Join Date: May 2006 Posts: 39 Rep Power: 0	thank you everyone I was wondering if I was doing what I was doing in the most efficient manner possible. The program is supposed to calculate ticket prices, there are two different sessions and 3 different seating plans. What I'm doing is asking for the session time in and then in two different statements I am basically asking for the seats, I don't think this is a very efficient way as I basically repeat myself twice, and was wondering if anyone thought of a better way of doing it

May 21st, 2006, 9:33 AM	#10
tumbleTetris Programmer Join Date: May 2006 Posts: 39 Rep Power: 0	ooh, another little technicality. if a user inputs something that is not 1 2 or x, how can I tell them they made a mistake and should reinput something 1 2 or x. If I use else if's I can't ask again, can I?

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Programming Forums > Application Development > Java

how can x be a valid choice in type int?