I am converting a roman numeral input to it's integer value. In my convertChar method, I keep getting an error that it is an unreachable statement whenever I add a break statement in to the code. I don't know why this is. I'm a new student and I must have done something wrong and I was hoping maybe someone can show me what I did wrong and how to fix it. It must have something to do with the way I set the methods up right? I still get confused on what variables to input so maybe I messed up there but I'm not experienced enough to know exactly what I did wrong. Here is everything I have:
public class RomanNumeralConverter {
public int romanInput() {
return convert(getUserInput());
}
public String getUserInput() {
Scanner numberInput = new Scanner (System.in);
System.out.print("Enter a roman numeral in uppercase: ");
String userInput = numberInput.next();
numberInput.close();
return userInput;
}
public int convert (String userInput) {
int result = 0;
int subtractamount = 0;
int x = userInput.length();
while(x != 0) {
char romanConvert = userInput.charAt(x);
if(x >= 1) {
if(convertChar(romanConvert) >= convertChar(userInput.charAt(x - 1))) {
subtractamount += convertChar(userInput.charAt(x - 1));
}
}
result += convertChar(romanConvert);
x--;
}
result -= subtractamount;
return result;
}
public static char convertChar(char value) {
switch (value) {
case 'I':
return 1;
break;
case 'V':
return 5;
break;
case 'X':
return 10;
break;
case 'L':
return 50;
break;
case 'C':
return 100;
break;
case 'D':
return 500;
break;
case 'M':
return 1000;
break;
default:
System.out.println("Invalid character!");
return 0;
break;
}
return value;
}
public void printValue() {
System.out.println(romanInput());
}
public static void main(String[] args) {
new RomanNumeralConverter().printValue();
}
}
Your problem lies in your switch statement. You can minimize this occurring elsewhere by attempting to have methods return only once (which i think is best practice)
public static char convertChar(char value) {
char result;
switch (value) {
case 'I':
result = 1;
break;
case 'V':
result = 5;
break;
case 'X':
result = = 10;
break;
case 'L':
result = 50;
break;
case 'C':
result = 100;
break;
case 'D':
result = 500;
break;
case 'M':
result = 1000;
break;
default:
System.out.println("Invalid character!");
result = 0;
break;
}
return result
}
In Java, it is a compile error to have statements that will never be reached while execution. In your case, the break statement will never be reached as there is a return statement above it. Also that last return statement will never be reached as you already would have returned in any case by the end of the switch block.
The problem is in your switch statement.
A default case can be thought of like the else in an if-else statement; it will always execute if no other condition in the switch is satisfied. If you are performing a return (or throw) inside of a default case, any code that follows after that will not be reachable.
You have two options:
Change the return statements to only assign a value to result instead, meaning that there's only one point of return from your code, or
Remove the return result from after your switch.
Related
Im very new to java, newbee in short. I want to make a program that checks if the number is hexidecimal, binary, octal, decimal. And if its Hexdecimal/binary/octal convert it to decimal but if it is a decimal not convert it. in my case i made a jbutton that if its clicked it will first check the number then convert it to decimal if it is not a decimal. Now the problem is i don't know what to add to my code, at present i can only check if it is a hexidecimal and then convert it to decimal if it meets the requirements and if not it will just stay. here is my code.
private void del1ActionPerformed(java.awt.event.ActionEvent
evt) {
String hex=prime1.getText();
long dec=0;
int r=0,c=0,b,con,er;
con=(int)Math.round(inb);
String co2, tra=prime1.getText();
boolean valid;
if(valid=true){for (int i = 0; i < hex.length(); i++ ) {
int digit = hexValue( hex.charAt(i) );
if (digit == -1) {
return;
}dec = 16*dec + digit;
prime1.setText(String.valueOf(dec));
}
static int hexValue(char ch) {
switch (ch) {
case '0':
return 0;
case '1':
return 1;
case '2':
return 2;
case '3':
return 3;
case '4':
return 4;
case '5':
return 5;
case '6':
return 6;
case '7':
return 7;
case '8':
return 8;
case '9':
return 9;
case 'a': // Note: Handle both upper and lower case letters.
case 'A':
return 10;
case 'b':
case 'B':
return 11;
case 'c':
case 'C':
return 12;
case 'd':
case 'D':
return 13;
case 'e':
case 'E':
return 14;
case 'f':
case 'F':
return 15;
default:
return -1;
}
} // end hexValue
here prime1 is a jtextfield.
If I got you correctly, this should be what you are looking for:
int getBaseAndConvertToDecimal(String input) {
int output = -1;
for(int i=0; i<input.length(); i++) {
if(Character.isAlphabetic(input.charAt(i)) && Character.toUpperCase(input.charAt(i))<='F') {
output = 16;
}
else {
if(Character.isDigit(input.charAt(i)) && Integer.parseInt(String.valueOf(input.charAt(i)))>=8) {
output = 10;
}
else {
if(Character.isDigit(input.charAt(i)) && Integer.parseInt(String.valueOf(input.charAt(i)))>=2) {
output = 8;
}
else {
if(input.charAt(i)=='0' || input.charAt(i)=='1') {
output = 2;
}
else {
output = -1;
}
}
}
}
}
if(output != -1) {
//int this case, the output object represents the base of the input string
//Integer.parseInt(inputToConvert, destinationBase);
return Integer.parseInt(input, output);
}
return output;
}
Below I have a method named 'hextoBinary' that returns a hexadecimal to binary conversion through type void.
In order for me to continue with my program I need a conversion from hex to binary method that returns and int so I can convert that binary int into a decimal with my 'hextoDecimal' method.
Can anybody help me or guide me on what approach to take, i've been stuck on this for a while now. i am limited to doing this manually instead of using parse or java automatic conversions.
import java.io.*;
import java.util.Scanner;
import java.util.ArrayList;
public class Main
{
static void hexToBinary(char hexdec[])
{
for (char c: hexdec)
{
switch (c)
{
case '0':
System.out.print("0000");
break;
case '1':
System.out.print("0001");
break;
case '2':
System.out.print("0010");
break;
case '3':
System.out.print("0011");
break;
case '4':
System.out.print("0100");
break;
case '5':
System.out.print("0101");
break;
case '6':
System.out.print("0110");
break;
case '7':
System.out.print("0111");
break;
case '8':
System.out.print("1000");
break;
case '9':
System.out.print("1001");
break;
case 'A':
System.out.print("1010");
break;
case 'B':
System.out.print("1011");
break;
case 'C':
System.out.print("1100");
break;
case 'D':
System.out.print("1101");
break;
case 'E':
System.out.print("1110");
break;
case 'F':
System.out.print("1111");
break;
default:
System.out.print("\nInvalid hexadecimal digit " + hexdec[c]);
}
}
}
public static int hextoDecimal(int n)
{
int decimal = 0, p = 0;
while(n != 0)
{
decimal += ((n % 10) * Math.pow(2,p));
n = n / 10;
p++;
}
return decimal;
}
public static void main(String[] args) throws IOException
{
Scanner sc = new Scanner(new File("RAMerrors8x4c"));
ArrayList<String> hexValues = new ArrayList<>();
while(sc.hasNext())
{
hexValues.add(sc.nextLine());
}
hexToBinary(hexValues.get(0).toCharArray());
}
}
This code is based on some that came from here but that link no longer seems to be active. Anyway, from a hex string you can get an int like this:
int hexToDecimal(String s){
int result = 0;
int digit = 0;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c >= '0' && c <= '9')
digit = c - '0';
else
if (c >= 'A' && c <= 'F')
digit = 10 + c - 'A';
else
inputError(s);
result = 16 * result + digit;
}
return result
}
I modified your code a little.
a. In your code only the first hex was printed.
Change:
call hexToBinary for every hex String.
b. the binary value was discarded after printing, so it couldn't be reused.
Change:
Changed returntype of hexToBinary from void to String and returned the binary value calculated.
To be able to return a String I add the peaces(nibbles) of the hex/binary to a String in every switch(case) clause.(a Stringbuilder might be better than a String - you can additionally improve that)
in the main: additionally collect all the returned binary values in a arraylist called "binaryValues" in order to have them for the next step.
With the above (little) changes I now have all the binary values that had already been calculated.
So I am able to simply use them in a binaryToDecimal method which just sums up the binary values weighted by their position.
Why not do it again? Because youd need to convert the A-F to numbers what your hexToBinary already did. So storing the values saves you doing that step again. I have a feeling that is what your teacher had in mind when he/she combined the tasks like this.
The resulting code is:
import java.io.*;
import java.util.Scanner;
import java.util.ArrayList;
public class Main
{
static String hexToBinary(char hexdec[]) {
String hex = "";
for (char c : hexdec) {
switch (c) {
case '0':
System.out.print("0000");
hex += "0000";
break;
case '1':
System.out.print("0001");
hex += "0001";
break;
case '2':
System.out.print("0010");
hex += "0010";
break;
case '3':
System.out.print("0011");
hex += "0011";
break;
case '4':
System.out.print("0100");
hex += "0100";
break;
case '5':
System.out.print("0101");
hex += "0101";
break;
case '6':
System.out.print("0110");
hex += "0110";
break;
case '7':
System.out.print("0111");
hex += "0111";
break;
case '8':
System.out.print("1000");
hex += "1000";
break;
case '9':
System.out.print("1001");
hex += "1001";
break;
case 'A':
System.out.print("1010");
hex += "1110";
break;
case 'B':
System.out.print("1011");
hex += "1111";
break;
case 'C':
System.out.print("1100");
hex += "1100";
break;
case 'D':
System.out.print("1101");
hex += "1110";
break;
case 'E':
System.out.print("1110");
hex += "1110";
break;
case 'F':
hex += "1111";
System.out.print("1111");
break;
default:
System.out.print("\nInvalid hexadecimal digit " + hexdec[c]);
}
}
System.out.println();
return hex;
}
public static int binaryToDecimal(String binary) {
int decimal = 0;
for (int i = 1; i < binary.length()-1; i++) {
decimal += Math.pow(2, i-1) * (binary.charAt(binary.length()-i) - '0');
}
return decimal;
}
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(new File("RAMerrors8x4c"));
ArrayList<String> hexValues = new ArrayList<>();
ArrayList<String> binaryValues = new ArrayList<>();
while (sc.hasNext()) {
hexValues.add(sc.nextLine());
}
for (String hex : hexValues) {
String binary = hexToBinary(hex.toCharArray());
binaryValues.add(binary);
System.out.println(binary);
}
for (String binary : binaryValues) {
int decimal = binaryToDecimal(binary);
System.out.println(decimal);
}
}
}
}
Besides using a Stringbuilder another idea could be to do all the printing of the binary values in the main. The hexToBinary returns the String - so you can print it in the loop - if you want.
I have been working on a utility to parse text files in the format used by Paradox Interactive in their grand strategy games to be used with a visual-based modding tool I am also developing. I have a mostly-implemented, crude, early version of the parser written out and it is mostly working as intended. This is my second attempt at writing a text parser (the first, which ended up working just fine, parsed a subset of XML).
I speed-wrote my parser on the 9th and have spent all weekend trying to debug it, but all my efforts have failed. I have tracked the issue down to the 3rd line of nextChar(). It was throwing an ArrayIndexOutOfBounds error with a crazy small number (in the -2 millions). After I added a bounds check the program just... continues. It reads all the information as needed, it just doesn't ever exit the parse loop.
The format is basically this:
car = {
model_year = 1966
model_name = "Chevy"
components = {
"engine", "frame", "muffler"
}
}
though I have yet to add support for nested lists like I plan, so my test string is:
car = {
model_year = 1966
model_name = "Chevy"
}
For both my understanding and anybody who would see my code, I tried to generously comment my code where I thought it might be necessary, though if any clarification is needed I would be happy to provide it.
My code:
/**
* Parses text files in the format used by Paradox Interactive in their computer games EUIV, CK2, and Stellaris.
*
* #author DJMethaneMan
* #date 12/9/2016
*/
public class Parser
{
private int pos, line, len, depth;
public String text;
private char[] script; //TODO: Initialize in the parse method
public Parser()
{
pos = 0;
line = 1;
len = 0;
depth = 0;
text = "car = {\n" +
" model_year = 1966 \n" +
" model_name = \"Chevy\"\n" +
"}\u0003";
//text = "Hello World";
//Car c = new Car();
//parse(text, c);
}
public static void main()
{
Car c = new Car();
Parser p = new Parser();
p.parse(p.text, c);
System.out.println("The model name is " + c.model_name);
System.out.println("The model year is " + c.model_year);
}
//TODO: Work
public void parse(String text, Parseable parsed)
{
char[] script = text.toCharArray();
this.script = script;
boolean next_char = false;
PARSE_LOOP:while(true)
{
char c;
if(next_char)
{
c = nextChar();
}
else
{
c = script[0];
next_char = true;
}
switch(c)
{
case 'A':
case 'a':
case 'B':
case 'b':
case 'C':
case 'c':
case 'D':
case 'd':
case 'E':
case 'e':
case 'F':
case 'f':
case 'G':
case 'g':
case 'H':
case 'h':
case 'I':
case 'i':
case 'J':
case 'j':
case 'K':
case 'k':
case 'L':
case 'l':
case 'M':
case 'm':
case 'N':
case 'n':
case 'O':
case 'o':
case 'P':
case 'p':
case 'Q':
case 'q':
case 'R':
case 'r':
case 'S':
case 's':
case 'T':
case 't':
case 'U':
case 'u':
case 'V':
case 'v':
case 'W':
case 'w':
case 'X':
case 'x':
case 'Y':
case 'y':
case 'Z':
case 'z':
case '_'://TODO: HERE
if(depth > 0) //
{
parsed.parseRead(buildWordToken(true), this);//Let the class decide how to handle this information. Best solution since I do not know how to implement automatic deserialization.
}
continueUntilChar('=', false); //A value must be assigned because it is basically a key value pair with {} or a string or number as the value
skipWhitespace();//Skip any trailing whitespace straight to the next token.
break;
case '{':
depth++;
break;
case '}':
depth--;
break;
case '\n':
line++;
break;
case ' ':
case '\t':
skipWhitespace();
break;
case '\u0003': //End of Text Character... Not sure if it will work in a file...
break PARSE_LOOP;
}
}
}
//Returns a string from the next valid token
public String parseString()
{
String retval = "";
continueUntilChar('=', false);
continueUntilChar('"', false);
retval = buildWordToken(false);
continueUntilChar('"', false); //Don't rewind because we want to skip over the quotation and not append it.
return retval;
}
//Returns a double from the next valid token
public double parseNumber()
{
double retval = 0;
continueUntilChar('=', false); //False because we don't want to include the = in any parsing...
skipWhitespace(); //In case we encounter whitespace.
try
{
retval = Double.parseDouble(buildNumberToken(false));
}
catch(Exception e)
{
System.out.println("A token at line " + line + " is not a valid number but is being passed as such.");
}
return retval;
}
/**********************************Utility Methods for Parsing****************************************/
protected void continueUntilChar(char target, boolean rewind)
{
while(true)
{
char c = nextChar();
if(c == target)
{
break;
}
}
if(rewind)
{
pos--;
}
}
protected void skipWhitespace()
{
while(true)
{
char c = nextChar();
if(!Character.isWhitespace(c))
{
break;
}
}
pos--;//Rewind because by default parse increments pos by 1 one when fetching nextChar each iteration.
}
protected String buildNumberToken(boolean rewind)
{
StringBuilder token = new StringBuilder();
String retval = "INVALID_NUMBER";
char token_start = script[pos];
System.out.println(token_start + " is a valid char for a word token."); //Print it.
token.append(token_start);
while(true)
{
char c = nextChar();
if(Character.isDigit(c) || (c == '.' && (Character.isDigit(peek(1)) || Character.isDigit(rewind(1))))) //Makes sure things like 1... and ...1234 don't get parsed as numbers.
{
token.append(c);
System.out.println(c + " is a valid char for a word token."); //Print it for debugging
}
else
{
break;
}
}
return retval;
}
protected String buildWordToken(boolean rewind)
{
StringBuilder token = new StringBuilder(); //Used to build the token
char token_start = script[pos]; //The char the parser first found would make this a valid token
token.append(token_start); //Add said char since it is part of the token
System.out.println(token_start + " is a valid char for a word token."); //Print it.
while(true)
{
char c = nextChar();
if(Character.isAlphabetic(c) || Character.isDigit(c) || c == '_')//Make sure it is a valid token for a word
{
System.out.println(c + " is a valid char for a word token."); //Print it for debugging
token.append(c); //Add it to the token since its valid
}
else
{
if(rewind)//If leaving the method will make this skip over a valid token set this to true.
{
//Rewind by 1 because the main loop in parse() will still check pos++ and we want to check the pos of the next char after the end of the token.
pos--;
break; //Leave the loop and return the token.
}
else //Otherwise
{
break; //Just leave the loop and return the token.
}
}
}
return token.toString(); //Get the string value of the token and return it.
}
//Returns the next char in the script by amount but does not increment pos.
protected char peek(int amount)
{
int lookahead = pos + amount; //pos + 1;
char retval = '\u0003'; //End of text character
if(lookahead < script.length)//Make sure lookahead is in bounds.
{
retval = script[lookahead]; //Return the char at the lookahead.
}
return retval; //Return it.
}
//Returns the previous char in the script by amount but does not decrement pos.
//Basically see peek only this is the exact opposite.
protected char rewind(int amount)
{
int lookbehind = pos - amount; //pos + 1;
char retval = '\u0003';
if(lookbehind > 0)
{
retval = script[lookbehind];
}
return retval;
}
//Returns the next character in the script.
protected char nextChar()
{
char retval = '\u0003';
pos++;
if(pos < script.length && !(pos < 0))
{
retval = script[pos]; //It says this is causing an ArrayIndexOutOfBoundsException with the following message. Shows a very large (small?) negative number.
}
return retval;
}
}
//TODO: Extend
interface Parseable
{
public void parseRead(String token, Parser p);
public void parseWrite(ParseWriter writer);
}
//TODO: Work on
class ParseWriter
{
}
class Car implements Parseable
{
public String model_name;
public int model_year;
#Override
public void parseRead(String token, Parser p)
{
if(token.equals("model_year"))
{
model_year = (int)p.parseNumber();
}
else if(token.equals("model_name"))
{
model_name = p.parseString();
}
}
#Override
public void parseWrite(ParseWriter writer)
{
//TODO: Implement along with the ParseWriter
}
}
Use of the labeled break statement break PARSE_LOOP; is generally considered bad practice. You are essentially writing a "goto" statement: whenever the break PARSE_LOOP; condition is hit, it jumps back to the beginning of the while loop (because that's where you wrote PARSE_LOOP:). This is probably the reason for your infinite loop. I also don't understand why you would restart a while loop that is already infinite (while true).
Change your code to:
public void parse(String text, Parseable parsed)
{
char[] script = text.toCharArray();
this.script = script;
boolean next_char = false;
boolean parsing = true;
while(parsing)
{
char c;
if(next_char)
{
c = nextChar();
}
else
{
c = script[0];
next_char = true;
}
switch(c)
{
case 'A':
case 'a':
case 'B':
case 'b':
case 'C':
case 'c':
case 'D':
case 'd':
case 'E':
case 'e':
case 'F':
case 'f':
case 'G':
case 'g':
case 'H':
case 'h':
case 'I':
case 'i':
case 'J':
case 'j':
case 'K':
case 'k':
case 'L':
case 'l':
case 'M':
case 'm':
case 'N':
case 'n':
case 'O':
case 'o':
case 'P':
case 'p':
case 'Q':
case 'q':
case 'R':
case 'r':
case 'S':
case 's':
case 'T':
case 't':
case 'U':
case 'u':
case 'V':
case 'v':
case 'W':
case 'w':
case 'X':
case 'x':
case 'Y':
case 'y':
case 'Z':
case 'z':
case '_'://TODO: HERE
if(depth > 0) //
{
parsed.parseRead(buildWordToken(true), this);//Let the class decide how to handle this information. Best solution since I do not know how to implement automatic deserialization.
}
continueUntilChar('=', false); //A value must be assigned because it is basically a key value pair with {} or a string or number as the value
skipWhitespace();//Skip any trailing whitespace straight to the next token.
break;
case '{':
depth++;
break;
case '}':
depth--;
break;
case '\n':
line++;
break;
case ' ':
case '\t':
skipWhitespace();
break;
case '\u0003': //End of Text Character... Not sure if it will work in a file...
parsing = false;
break;
}
}
}
Put a debug statement in to prove that it's hitting your break, I'm guessing it's not (Although it could be the break label--I haven't had reason to look into that construct since I first learned java a couple decades ago). I have a couple suggestions though...
I'd use isAlpha instead of that part of the switch. Cleaner, shorter, probably about as efficient and language-agnostic.
Instead of using the break label (Which is very uncommon), You might want to use boolean parsing=true;while(parsing)... instead. It's not really wrong to use the break label, but... Anything that causes the next guy to spend a minute or two scratching his head is a few minutes wasted.
I have to make a program which tells if a String that I type in my keyboard is a number, by using a switch. I know how to do it with try and catch, but I don't know how to do it with switch.
Any tips?
You would need to check each characer in the String. Something like this would probably work.
static boolean isNumber(String s) {
if (s == null) {
// Debatable.
return false;
}
int decimalCount = 0;
for (int i = 0; i < s.length(); i++) {
switch (s.charAt(i)) {
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':
// These are all allowed.
break;
case '.':
if (i == 0 || decimalCount > 0) {
// Only allow one decimal in the number and not at the start.
return false;
}
decimalCount += 1;
break;
default:
// Everything else not allowed.
return false;
}
}
return true;
}
Up to Java7 you can use switch(String) statement.
But here you have enough with switch(int) and a little workaround:
public static void main(String[] args) throws Exception {
String a = "2";
switch (Integer.parseInt(a)) {
default:
System.out.print("is a number");
break;
}
}
This is the solution I got asking to some classmates and thinking it quietly.
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner entry = new Scanner(System.in);
String myNumber;
int tf;
myNumber = entry.next();
try {
Double.parseDouble(myNumber);
tf = 1;
}
catch (Exception e) {
tf = 0;
}
switch(tf) {
case 1:
System.out.println("Is a number");
break;
default:
System.out.println("No es un número");
break;
}
}
Thanks to the community for being so nice!
I came up with a shorter code BUT it uses regular expressions, which if Halo is just starting with Java, he may have not seen that topic yet. But then it answers the question too so here it is:
Scanner scanner = new Scanner(System.in);
String expression = scanner.nextLine();
String matches = new Boolean(expression.matches("\\d+")).toString();
switch (matches) {
case "true":
System.out.println("IT'S a number");
break;
case "false":
System.out.println("NOT a number");
}
scanner.close();
Here is the code:
public static void main(String args[])
{
int i=0;
int m=0;
double scale;
boolean exit;
Shape[] s = new Shape[10];
while(exit !=true)
{
System.out.print("\nChoose an option:\n"+
"1-Add a new circle\n"+
"2-Add a new rectangle\n"+
"3-Delete all shapes\n"+
"4-Scale all shapes\n"+
"5-Display perimeter of all shapes\n"+
"6-Display the area of all shapes\n"+
"7-Enter scale factor\n"+
"8-Exit program\n");
Scanner input = new Scanner(System.in);
m=input.nextInt();
if(i<=9)
{
switch (m)
{
case 1: Circle c = new Circle(0);
s[i]=c;
i++;
break;
case 2: Rectangle r = new Rectangle(1,1);
s[i]=r;
i++;
break;
case 3: s=null;
i=0;
break;
case 4: for(i=0; i<s.length; i++)
{
s[i].scaleShape();
}
break;
case 5: for(i=0; i<s.length; i++)
{
if(s[i] != null)
{
System.out.println(s[i].getPerimeter());
}
}
break;
case 6: for(i=0; i<s.length; i++)
{
if(s[i] != null)
{
System.out.println(s[i].getArea());
}
}
break;
case 7: do
{
System.out.println("\nEnter scale factor");
scale=input.nextDouble();
}
while(scale<0);
Shape.setScaleFactor(scale);
}
break;
case 8: System.out.println("Do you want to quit?");
break; //Nothing here since loop should terminate it.
//default: System.out.println("Number must be 1-8");
// break;
}
}
}
}
Oddly the compiler is giving me an error on case 8 saying:
Type mismatch can't convert from int to boolean.
But Im not converting anything to boolean
-syntax error on token "case" assert expected
-syntax error on token :, ; expected
But all the commands there have semi-colons
expression must return a value
Why is the compiler acting so funny? Normally errors like that are easy to find. What is going on?
Your problem is in the case for 7:
case 7: do
{
System.out.println("\nEnter scale factor");
scale=input.nextDouble();
}
while(scale<0);
Shape.setScaleFactor(scale);
}
Notice the extra close brace: that's closing your switch statement, orphaning your case 8.
} // <-- Why is this here?
break;
case 8: System.out.println("Do you want to quit?");
You're ending the switch statement with an extra }. Remove it and things may work.
Every one has pointed out that you have an extra parenthesis in your code, what we've failed to point out is where it's coming from...
while(scale<0); // <-- This isn't going to work....
Shape.setScaleFactor(scale);
}
It should be...
while(scale<0) {
Shape.setScaleFactor(scale);
}
The next question is, how is scale decremented? Cause this could cause an infinite loop if you're not careful.
while(scale<0);
Shape.setScaleFactor(scale);
} // Remove this parenthesis.
break;