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Write a method that returns a number, given an uppercase letter, as follows.
int getNumber (char uppercaseLetter)
Write a test program that prompts the user to enter a phone number as a string. The input number may contain letters. The program translates a letter (uppercase or lowercase) to a digit and leaves all other characters intact.
Sample run from textbook
Enter a string: 1-800-Flowers
1-800-3569377
Enter a string: 1800flowers
18003569377
Here is what I have so far
import java.util.Scanner;
public class Assignment {
public static int correspondingNumber(char uppercaseLetter){
int correspondingNumber=0;
switch (uppercaseLetter)
{
case 'A':
case 'B':
case 'C': correspondingNumber=2; break;
case 'D':
case 'E':
case 'F': correspondingNumber=3; break;
case 'G':
case 'H':
case 'I': correspondingNumber=4; break;
case 'J':
case 'K':
case 'L': correspondingNumber=5; break;
case 'M':
case 'N':
case 'O': correspondingNumber=6; break;
case 'P':
case 'Q':
case 'R':
case 'S': correspondingNumber=7; break;
case 'T':
case 'U':
case 'V': correspondingNumber=8; break;
case 'W':
case 'X':
case 'Y':
case 'Z': correspondingNumber=9; break;
}
return correspondingNumber;
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String num;
char num1 = 0;
System.out.print("Enter a string: ");
num = input.next();
num.toUpperCase();
int i=0;
while(i!=num.length()){
num1=num.charAt(i);
}
System.out.print(correspondingNumber(num1));
}
}
steps need tobe done
scan input let's say as String
Convert string to character array (srcArray)
change method return of correspondingNumber to Char
default return to input and apply switch case
call method correspondingNumber, store return char in stringbuilder or Array of Char array
repeat step-5 until character array(srcArray) is completely processed
print the output
import java.util.Scanner;
public class Assignment {
// changed return type
public static char correspondingNumber(char uppercaseLetter) {
char correspondingNumber = uppercaseLetter;// default the return value
// to input
switch (uppercaseLetter) {
case 'A':
case 'B':
case 'C':
correspondingNumber = '2';
break;
case 'D':
case 'E':
case 'F':
correspondingNumber = '3';
break;
case 'G':
case 'H':
case 'I':
correspondingNumber = '4';
break;
case 'J':
case 'K':
case 'L':
correspondingNumber = '5';
break;
case 'M':
case 'N':
case 'O':
correspondingNumber = '6';
break;
case 'P':
case 'Q':
case 'R':
case 'S':
correspondingNumber = '7';
break;
case 'T':
case 'U':
case 'V':
correspondingNumber = '8';
break;
case 'W':
case 'X':
case 'Y':
case 'Z':
correspondingNumber = '9';
break;
}
return correspondingNumber;
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String num;
char num1 = 0;
System.out.print("Enter a string: ");
num = input.next();
num.toUpperCase();
int i = 0;
while (i != num.length()) {
num1 = num.charAt(i);
System.out.print(correspondingNumber(num1)); // moved print
// statement to
// appropriate place
i++; // iterate loop
}
input.close();
}
}
Related
I am writing a program that simulates the translation of an alphabetic phone number into just numbers. For example: 888-get-food == 555-438-3663.
Initially, the user should enter the phone number on the following format: 888-GET-FOOD (With the dashes). When I try to check if there are dashes on the user input, it prints the dashes, but with the number 1 in front of it. Very annoying.
This is what I have so far:
// Ask the user to enter the phone number
System.out.println("Please enter the phone number: ");
// Save the phone number into a string
String initialPhoneNumber = input.nextLine();
// Convert user input to UPPERCASE
initialPhoneNumber = initialPhoneNumber.toUpperCase();
// This will be the phone number converted
String finalPhoneNumber = fullPhoneNumber(initialPhoneNumber);
// Print number
System.out.println(initialPhoneNumber);
System.out.println(finalPhoneNumber);
for (int i = 0; i < strLength; i++) {
char letter = initialPhoneNumber.charAt(i);
if (Character.isLetter(letter)) {
switch (letter) {
case 'A': case 'B': case 'C': number = 2; break;
case 'D': case 'E': case 'F': number = 3; break;
case 'G': case 'H': case 'I': number = 4; break;
case 'J': case 'K': case 'L': number = 5; break;
case 'M': case 'N': case 'O': number = 6; break;
case 'P': case 'Q': case 'R': case 'S': number = 7; break;
case 'T': case 'U': case 'V': number = 8; break;
case 'W': case 'X': case 'Y': case 'Z': number = 9; break;
case '-':number='-';
}
}
else if (Character.isDigit(letter)) {
number = Character.getNumericValue(letter);
}
else if (initialPhoneNumber.charAt(i) == '-') {
number = Character.getNumericValue(letter);
}
and this is the output:
Please enter the phone number:
555-GET-FOOD
555-GET-FOOD
555-1438-13663
Why does the number 1 show up in front of the dashes? How can I make it so it doesn't show up? In other words, how can I print or separate the numbers separated by dashes?
Thanks
The problem is within the statement:
else if (initialPhoneNumber.charAt(i) == '-') {
number = Character.getNumericValue(letter);
}
The method Character.getNumericValue() returrns -1 if the character passed as parameter is not numeric.
Since you're passing -, the method returns -1.
That's why you get the 1 after the dash.
UPDATE
Supposing that the for-loop is part of your method fullPhoneNumber, you could resolve it as follows:
public String fullPhoneNumber(String initialPhoneNumber)
{
StringBuilder result;
result = new StringBuilder();
for (int i = 0; i < initialPhoneNumber.length(); i++)
{
char letter = Character.toUpperCase(initialPhoneNumber.charAt(i));
switch (letter)
{
case 'A':
case 'B':
case 'C':
letter = '2';
break;
case 'D':
case 'E':
case 'F':
letter = '3';
break;
case 'G':
case 'H':
case 'I':
letter = '4';
break;
case 'J':
case 'K':
case 'L':
letter = '5';
break;
case 'M':
case 'N':
case 'O':
letter = '6';
break;
case 'P':
case 'Q':
case 'R':
case 'S':
letter = '7';
break;
case 'T':
case 'U':
case 'V':
letter = '8';
break;
case 'W':
case 'X':
case 'Y':
case 'Z':
letter = '9';
break;
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':
case '-':
break;
default:
throw new IllegalArgumentException("");
}
result.append(letter);
} // for
return (result.toString());
} // fullPhoneNumber
Modify the program to accept a telephone number with any number of letters. The output should display a hyphen after the first 3 digits and subsequently a hyphen (-) after every four digits. Also, modify the program to process as many telephone numbers as the user wants.
Below is what i have to edit.
Scanner keyboard = new Scanner(System.in);
String telephone_letter;
int index;
System.out.print("Enter Telephone letters : ");
telephone_letter = keyboard.nextLine();
char aChar[] = telephone_letter.toCharArray();
int[] number = new int[100];
for(index=0;index<telephone_letter.length();index++){
switch (aChar[index])
{
case 'A':
case 'a':
case 'B':
case 'b':
case 'C':
case 'c':
number[index] = 2;
break;
case 'D':
case 'd':
case 'E':
case 'e':
case 'F':
case 'f':
number[index] = 3;
break;
case 'G':
case 'g':
case 'H':
case 'h':
case 'I':
case 'i':
number[index] = 4;
break;
case 'J':
case 'j':
case 'K':
case 'k':
case 'L':
case 'l':
number[index] = 5;
break;
case 'M':
case 'm':
case 'N':
case 'n':
case 'O':
case 'o':
number[index] = 6;
break;
case 'P':
case 'p':
case 'Q':
case 'q':
case 'R':
case 'r':
case 'S':
case 's':
number[index] = 7;
break;
case 'T':
case 't':
case 'U':
case 'u':
case 'V':
case 'v':
number[index] = 8;
break;
case 'W':
case 'w':
case 'X':
case 'x':
case 'Y':
case 'y':
case 'Z':
case 'z':
number[index] = 9;
break;
default:
break;
}
}
System.out.println("=======================================");
System.out.println("The Telephone letter is : " + telephone_letter);
System.out.println("The Phone number is : " + number[0]+number[1]+number[2]+"-"+number[3]+number[4]+number[5]+number[6]);
System.out.println("=======================================");
}
This part is what i tried.
Scanner keyboard = new Scanner(System.in);
String telephone_letter;
int index;
System.out.print("Enter Telephone letters : ");
telephone_letter = keyboard.nextLine();
char aChar[] = telephone_letter.toCharArray();
int[] number = new int[100];
while(telephone_letter !="2")
{
for(index=0;index<telephone_letter.length();index++){
switch (aChar[index])
{
case 'A':
case 'a':
case 'B':
case 'b':
case 'C':
case 'c':
number[index] = 2;
break;
case 'D':
case 'd':
case 'E':
case 'e':
case 'F':
case 'f':
number[index] = 3;
break;
case 'G':
case 'g':
case 'H':
case 'h':
case 'I':
case 'i':
number[index] = 4;
break;
case 'J':
case 'j':
case 'K':
case 'k':
case 'L':
case 'l':
number[index] = 5;
break;
case 'M':
case 'm':
case 'N':
case 'n':
case 'O':
case 'o':
number[index] = 6;
break;
case 'P':
case 'p':
case 'Q':
case 'q':
case 'R':
case 'r':
case 'S':
case 's':
number[index] = 7;
break;
case 'T':
case 't':
case 'U':
case 'u':
case 'V':
case 'v':
number[index] = 8;
break;
case 'W':
case 'w':
case 'X':
case 'x':
case 'Y':
case 'y':
case 'Z':
case 'z':
number[index] = 9;
break;
default:
break;
}
}
System.out.println("=======================================");
System.out.println("The Telephone letter is : " + telephone_letter);
System.out.println("The Phone number is : " + number[0]+number[1]+number[2]+"-"+number[3]+number[4]+number[5]+number[6]);
System.out.println("=======================================");
System.out.print("Enter Telephone letters : ");
telephone_letter = keyboard.nextLine();
}
My output is wrong.
Enter Telephone letters : fewfwef
======================================
The Telephone letter is : fewfwef
The Phone number is : 339-3933
=======================================
Enter Telephone letters : wsqsq
=======================================
The Telephone letter is : wsqsq
The Phone number is : 339-3933
=======================================
And another Question is how do i do this The output should display a hyphen after the first 3 digits and subsequently a hyphen (-) after every four digits.
This seems like a homework problem so here are some hints to help you solve it yourself.
(1) Using a fixed number like 100 is rarely a good idea:
int[] number = new int[aChar.length];
(2) The scanner can tell you if there is more to be scanned(read):
Scanner keyboard = new Scanner(System.in);
System.out.print("Enter Telephone letters : ");
while (keyboard.hasNextLine()) {
String telephone_letter = keyboard.nextLine();
...
System.out.println("");
System.out.print("Enter Telephone letters : ");
}
(3) A loop is good for counting. If the first count is different, then start different:
int todash = 3;
System.out.print("The Phone number is : ");
for (int i=0; i < number.length; ++i) {
if (todash == 0) {
System.out.print("-");
todash = 4;
}
System.out.print(number[i]);
--todash;
}
System.out.println("");
I've rewritten you're code. It may be easier to work with.
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Telephone {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print("Enter Telephone letters : ");
String input = keyboard.nextLine();
TelephoneProcessor telephoneProcessor = new TelephoneProcessor(input);
String result = telephoneProcessor.proccess();
System.out.println("=======================================");
System.out.println("The Telephone letter is : " + input);
System.out.println("The Phone number is : " + result);
System.out.println("=======================================");
}
}
class TelephoneProcessor {
Map<String, String> matches = new HashMap<>();
String toProcess;
public TelephoneProcessor(String toProcess) {
this.toProcess = toProcess;
matches.put("[a-c]", "2");
matches.put("[d-f]", "3");
matches.put("[g-i]", "4");
matches.put("[j-l]", "5");
matches.put("[m-o]", "6");
matches.put("[p-r]", "7");
matches.put("[s-w]", "8");
matches.put("[x-z]", "9");
}
public String proccess() {
if(toProcess != "2") {
matches.forEach((regex, replacement) -> {
toProcess = toProcess.replaceAll(regex, replacement);
});
}
return toProcess.substring(0,3) + "-" + toProcess.substring(3,7);
}
}
It only asks once but that's easy to solve by making a class that contains the main method in a method and loops it.
I hope this helps for further extensions of the program :)
Am trying to create a method to encrypt a word with Caesar cipher
here is the code:
public static void main (String [] args)
{
String s = "ahmed";
int k = 2;
char[] c = new char[5];
int[] a = new int[s.length()];
for (int i = 0; i<s.length();i++){
a[i] = Secret_Code_Library.getDigit(s.charAt(i));
a[i] = i + k ;
c[i] = Secret_Code_Library.getLetter(a[i]);
}
String ss = String.valueOf(c);
}
Secret_Code_Library is a class that contains the two methods
(getDigit(char): returns the number of the character a =0, b= 1...etc, and
getLetter(int): returns the letter corresponding to the number).
my problem is when ever i try to encrypt a word it encrypts only
the first letter correctly and save it in to the array a.
P.S: I have to use those two methods.
the output of the above code is: cdefg
while it should be: cjogf
this is the class :
public class Secret_Code_Library {
public static final char [] LETTERS={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z' };
public static char getLetter(int digit){
return LETTERS[digit]; }
public static int getDigit(char ch){
int digit=0;
switch (ch){
case 'A': digit=0; break; case 'B': digit=1; break; case 'C': digit=2; break;
case 'D': digit=3; break; case 'E': digit=4; break; case 'F': digit=5; break;
case 'G': digit=6; break; case 'H': digit=7; break; case 'I': digit=8; break;
case 'J': digit=9; break; case 'K': digit=10; break; case 'L': digit=11; break;
case 'M': digit=12; break; case 'N': digit=13; break; case 'O': digit=14; break;
case 'P': digit=15; break; case 'Q': digit=16; break; case 'R': digit=17; break;
case 'S': digit=18; break; case 'T': digit=19; break; case 'U': digit=20; break;
case 'V': digit=21; break; case 'W': digit=22; break; case 'X': digit=23; break;
case 'Y': digit=24; break; case 'Z': digit=25; break;
}// switch
return digit;
}}
a[i]= i + k ;
should be
a[i] = a[i] + k;
because you're not adding the key to the position of the current encrypted character, but the integer value of the character itself.
But that is not enough. What happens when you're encrypting 'z' with k = 2? You should wrap around in the alphabet, if the Secret_Code_Library.getLetter(a[i]); doesn't already do this for you.
If your alphabet is 26 characters long, then this should do it:
a[i] = (a[i] + k) % 26;
I just saw some of my work today, and i want to enter the user how many alphabet they want and continue converting. Can i have help? It would be fantastic. I tried anything. I'm just new in java programming that's why. Thanks :)
This is my whole programming
import java.util.Scanner;
public class Try2 {
public static void main(String[] args) {
int counter = 0;
int it = 0;
Scanner keyboard = new Scanner(System. in );
System.out.println("Enter words to digits: ");
String alpha = keyboard.nextLine();
alpha = alpha.toLowerCase();
String num = (" ");
while (counter < alpha.length()) {
switch (alpha.charAt(it)) {
case 'A':
case 'a':
case 'B':
case 'b':
case 'c':
case 'C':
num += "2";
counter++;
break;
case 'D':
case 'E':
case 'F':
case 'd':
case 'e':
case 'f':
num += "3";
counter++;
break;
case 'G':
case 'H':
case 'I':
case 'g':
case 'h':
case 'i':
num += "4";
counter++;
break;
case 'J':
case 'K':
case 'L':
case 'j':
case 'k':
case 'l':
num += "5";
counter++;
break;
case 'M':
case 'N':
case 'O':
case 'm':
case 'n':
case 'o':
num += "6";
counter++;
break;
case 'P':
case 'R':
case 'S':
case 'p':
case 'r':
case 's':
num += "7";
counter++;
break;
case 'T':
case 'U':
case 'V':
case 't':
case 'u':
case 'v':
num += "8";
counter++;
break;
case 'W':
case 'w':
case 'X':
case 'x':
case 'Y':
case 'y':
case 'Z':
case 'z':
case ' ':
num += "9";
counter++;
break;
}
if ((counter % 4) == 3) {
num += "-";
}
it++;
}
System.out.println(num);
}
}
You can wrap the read line code with another while loop:
Scanner keyboard = new Scanner(System.in);
while(true) {
System.out.println("Enter words to digits: ");
String alpha = keyboard.nextLine();
alpha = alpha.toLowerCase();
String num = " ";
int counter = 0;
int it = 0;
while (counter < alpha.length()) {
...
}
//new line (optional)
num += "\n";
}
You can check if the user entered a flag to end input:
//You can choose any flag you want
if(alpha.equals("finish")) {
break;
}
About your switch:
You can remove all the upper case letters because you wrote: alpha = alpha.toLowerCase();
You forgot 'q'
If alpha don't match: [a-zA-z ] (contains only letters and spaces) an IndexOutOfBoundsException will be thrown from alpha.charAt(it) because you don't increase counter and the loop don't exit in time.
To prevent the last problem you can remove counter and it and change the while loop to for loop:
for (int i = 0; i < alpha.length(); i++) {
if ((i % 4) == 3) {
num += "-";
}
switch (alpha.charAt(i)) {
...
}
}
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 8 years ago.
Improve this question
I have to keep it in the case 'A' Format i just don't understand what I should have in the (char c) because that seems to be where the error is coming from. If anyone has any suggestions it would be greatly appreciated.
public class PhoneNumber {
private int areacode;
private int number;
private int ext;
PhoneNumber() {
areacode = 0;
number = 0;
ext = 0;
}
PhoneNumber(int newnumber) {
areacode = 216;
number = newnumber;
ext = 0;
}
PhoneNumber(int newarea, int newnumber, int newext) {
areacode = newarea;
number = newnumber;
ext = newext;
}
PhoneNumber(String newnumber) {
String areacode = str[0];
String number = str[1];
String[] str = newnumber.split("-");
String[] number = newnumber;
boolean b1, b2;
int i = 0;
int place = 0;
for (int x: newnumber){
newnumber.charAt[i] = place;
b1 = Character.isDigit(place);
if (b1 == true){
number = place;
i++;
} else {
b2 = Character.isLetter(place);
} if (b2 == true) {
number = decode(place);
i++;
} else {
System.out.print("invalid phone number!");
}
}
System.out.print(areacode.concat(number));
return newnumber;
}
private String decode(place) {
switch (c) {
case 'A': case 'B': case 'C': return "2";
case 'D': case 'E': case 'F': return "3";
case 'G': case 'H': case 'I': return "4";
case 'J': case 'K': case 'L': return "5";
case 'M': case 'N': case 'O': return "6";
case 'P': case 'Q': case 'R': case 'S': return "7";
case 'T': case 'U': case 'V': return "8";
case 'W': case 'X': case 'Y': case 'z': return "9";
default: return "";
}
}
public boolean equals(PhoneNumber pn) {
//not complete
}
public String toString() {
//not complete
}
}
Here is the error:
G:\CIS260\Assignments>javac PhoneNumber.java
PhoneNumber.java:53: error: <identifier> expected
private String decode(place) {
^
1 error
You have one error that jump out at me. Note that if you have a syntax error, most compilers will freak out and mark everything past the actual error as errors. The best way to debug accidental syntax errors is to look at the first error :)
boolean = b1, b2;
Makes no sense. a variable declaration is as follows
[Type] [Variable Name] | ,[Additional Vars];
so it should be boolean b1, b2;
private String decode(char c) {
switch (c) {
case 'A': case 'B': case 'C': return "2";
case 'D': case 'E': case 'F': return "3";
case 'G': case 'H': case 'I': return "4";
case 'J': case 'K': case 'L': return "5";
case 'M': case 'N': case 'O': return "6";
case 'P': case 'Q': case 'R': case 'S': return "7";
case 'T': case 'U': case 'V': return "8";
case 'W': case 'X': case 'Y': case 'z': return "9";
default: return "";
}
}
I suggest you use a lookup table(hashmap) to do this set of code, it looks much less ugly and is easier on sore eyes.
An example of that would be
private static final Map<Char, String> myDecodeLookup = new Hashmap<Char,String>();
static{ //initializer block
myDecodeLookup.put('A', "2");
myDecodeLookup.put('B', "2");
//and so and and so forth
}
private String decode(char c) throws KeyNotFoundException{
return MyClassName.myDecodeLookup.get(c); //you should code this to support the keynotfoundexception that this might throw, or not.
}
If you ever find yourself debugging a syntax error for 5 hours, you should probably take a step back and look up the syntax of the language constructs themselves... that's good advice to take going forward.