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Check number of character from command line

I am trying to solve this question: Write a program that passes a string to the command line and displays the number of uppercase letters in the string.
The test is ProgrammingIsFun, so the count would be three.
Here is what I have so far
public static void main(String[] args) {
int count = 0;
for(int i = 0; i < args.length;i++) {
for(int j = 0; j >= i; j++)
if(Character.isUpperCase(args[i].charAt(j)))
count++;
}
System.out.print("The number of upper cases are: " + count);
}
when this runs it does count the correct number of uppercases but it runtimes when it reaches the end of the characters(16). I can't figure out how to stop the loop right after the last character. I know the count is correct because I can see it in the debug. I can see something called arg0 in the variables. That has the correct number of characters but I can't use it.
I am still learning and I can't use objects or anything like that for now.
Any help would be appreciated. or if you could point me somewhere I could read. I tried the docs but it was a lot and I got overwhelmed fairly quickly.

You have an error in the inner loop:
for(int j = 0; j >= i; j++)
Note that the outer loop is run for each of the command line argument and the inner loop is run for each character of an argument given by i.
So change it to
for(int j = 0; j < args[i].length(); j++)

for(int j = 0; j < args[i].length(); j++)?
This will sum all uppercases for all strings passed as arguments.
Your code could be simpler if you only expect one string.

To make your code easier to read (and write) you should use foreach loops, one for each string in the args array, and one for each character in the array that we get by doing foo.toCharArray().
The syntax of a foreach loop is as follows:
for (type var : iterable) {
// Code here
}
Where iterable is something with the Iterable interface (usually an array or ArrayList).
We can switch out these in your code and make it a lot easier to read:
String[] args = {"ProgrammingIsFun!", "When It Works..."};
int count = 0;
for (String e : args) {
for (char f : e.toCharArray()) {
if (Character.isUpperCase(f)) {
count++;
}
}
}
System.out.print("The number of upper cases are: " + count);

Palindromes using Stacks and Queues

My NEW sample text i was testing: My mom is a good cook. Although sometimes at around noon she will leave and forget to make me lunch and some pop. #Old homework become relevant again
my problem is just that i am not getting the correct output, as my method only prints *Mom a noon i
This is all GUI based.I am reading in a file and checking for palindromes and printing them out in my JTextArea afterwards using Stacks and Queue's.
Issue is, all of this is working and when i start the program and attach the text file, i only get the first palindrome. SO it will print "mom" which is my first testcase, but it won't go to any of the other palindromes following it?
I thought i might have got bogged down in my code blocking at some point but after tinkering with it for a day now i'm sort of stuck.
EDIT 1: I am now getting more results
my method is,
public void fileDecode() throws FileNotFoundException
{
if (fileChoice.showOpenDialog(null) == JFileChooser.APPROVE_OPTION)
{
file = fileChoice.getSelectedFile();
scanInput = new Scanner(file);
while(scanInput.hasNext())
{
int nextPalindrome = 0;
int counter = 0;
String token = scanInput.next();
Stack<Character> stk = new Stack<Character>();
Queue<Character> que = new LinkedList<Character>();
for (int i = 0; i < token.length(); ++i)
{
stk.push(token.charAt(i)); //pushing all char's onto the stack/queue
que.add(token.charAt(i));
}
for (int j = 0; j < token.length(); ++j)
{
char tempStk = stk.pop(); //removing them one at a time and checking if they are equal and incrementing counter
char tempQue = que.remove();
if (tempStk == tempQue)
{
counter++;
}
}
if (counter == token.length())
{
build.append(token + " "); //if the counter # was the same as the token's length, than it is indeed a palindrome and we append it into our StringBuilder that we pass to the JTextArea
nextPalindrome = token.length();
}
}
}
}

You set counter to 0 outside your while loop, so the count of the second word is going to start at the count of the first word. Move counter = 0 inside the while loop and it should work.
Also, nextPalindrome doesn't appear to be used, and even if it is, if you set it at the bottom of the loop, it's immediately set to 0 at the top, so it will only remain non-zero if the last word is a palindrome.
Also, think about what's happening in the second for loop. You're looping over all the characters and comparing the one from the stack and the one from the queue. If ever those are different, you know you don't have a palindrome, so once you find a difference, it's pointless to continue with the loop. You also already have a loop counter, j, so you don't need another one. So I'd rewrite the second loop and following condition as follows:
for (int j = 0; j < token.length(); ++j)
{
char tempStk = stk.pop(); //removing them one at a time and checking if they are equal and incrementing counter
char tempQue = que.remove();
if (tempStk != tempQue)
break;
}
if (j == token.length())
That works because the only way j can equal token.length() after the loop is done is if the loop completed, which can only happen if no pairs of characters aren't equal (in other words, all pairs of characters are equal, which is what you want).

Java infinite loop issue

So our teacher gave us this homework, we had to write a program that went something along the lines of
Write an application that reads a line of text from the keyboard and prints a table indicating the number of occurrences of each letter
of the alphabet in the text, For example, the phrase
To be, or not to be: that is the question:
Contains one “a,” two “b’s,” no “c’s,” and so on.
Well I've written the code, but I've ran into one small problem when I enter the to be or not to be part the code continually loops forever. I've looked at this program forever, I even tried asking some folks at Yahoo (but I think I confused them). So I am hoping someone here will spot something I missed or have some advice to give me.
public class occurances {
public static void main(String[] args) {
Scanner inp = new Scanner(System.in);
String str;
char ch;
int count = 0;
System.out.println("Enter the string:");
str = inp.nextLine();
while (str.length() > 0) {
ch = str.charAt(0);
int i = 0;
while (i < str.length() && str.charAt(i) == ch) {
count = count++;
i++;
}
str = str.substring(count);
System.out.println(ch);
System.out.println(count);
}
}
}

There are many problems with your code, but start with count = count++. It will always have its initial values (0 in your case). That causes the infinite loop. If you manage this one you'll be good to go with debugging your code further. Learn how to use debugger and/or print for checking your code. Good luck.

It seems your approach is not necessarily the one the teacher wants. Your approach (if it worked at all) would display the character counts in the order the characters appear in the string. So, for example, for "To be, or not to be: that is the question" you would show the character count for "T" first, whereas the teacher probably wants you to show the character count for "a" first. Your approach also doesn't show the character counts for the characters that are missing in the answer.
It has been suggested in the other two answers to use a Map. I recommend that approach, although you could use a simple int[] array where the index is (ch - 'a') assuming that it is between 0 and 25. See Character.toLowerCase() for how to convert a character into a lowercase one, because the correct answer would probably treat "T" and "t" as the same.
You need only one loop through the array, incrementing the counts for the characters that appear. Obviously all of the counts should be initialized to 0 prior to the loop.

I've not done JAVA in a while i came here to help someone with Javascript... but i'll give it go
while(str.length()>0)
This will obviously enter an endless loop because you have no breakout terms... i'd probably use a for loop instead... also you need something to store your counts inside of for each unique letter you find.
-- EDIT based on the comment below i find myself agreeing that answering your homework is not a good thing but here is what i'd suggest you look at...
Look at Hashtables or Maps and thing about "storing the dictionary counts as you find them"
Bare in mind upper and lower case letter, i'd just cast to lowercase and count
use a for loop instead of the while loop, use the strings length as the limit e.g. count from 0 to string.length()
This should get you on the right path :http://www.java2s.com/Tutorial/Java/0140__Collections/GettingelementskeyvaluepairsfromaHashtabletheentrySetmethod.htm

Some things to resolve in your code:
while(str.length()>0) : Infinite loop here. You could go through the string, with a for (int index = 0; index < str.length(); index++).
ch = str.charAt(0); : Here you are always taking the first letter of the String, you should take each letters with ch = str.charAt(index);.
count = count++; : count++; would be enough or count = count + 1;.
count = count++; : count should be re-initialized for each letter count = 0; before your while inner loop.
str = str.substring(count); : No need to substring here. You would loose letters for next computation.
Your code should work with one disavantages. The letters and their count will be written as many times as their is their occurence. If a is 3 times in the String, a : 3 will be written 3 times.
So you should whether you had already print the letter (maybe an array), but performance would be bad. For a String of n letters, you would have n * n computations (Complexity O(n^2)). If you take an HashMap between letter and their count, you will just of to go through the String one time, incrementing your counter while iterating on the loop. Then you would have only n computations. (Complexity O(n)).
HashMap<Char, Integer> letters = new HashMap<Char, Integer>();
for (int i = 0; i < str.length(); i++) {
char letter = str.charAt(i);
if (!letters.contains(letter)) {
letters.put(letter, 1);
} else {
letters.put(letter, letters.get(letter) + 1);
}
}
Then you go through the map to display the count.

Here is your corrected code for the problem you asked..Try this..
import java.util.*;
public class Occurence {
public static void main(String[] args) {
Scanner inp = new Scanner(System.in);
String str;
char ch;
int count = 0;
System.out.println("Enter the string:");
str = inp.nextLine();
int length = str.length();
int i = 0, j = 0;
for (i = 0; i < length; i++) {
j = 0;
count=0;
for (j = 0; j < length; j++) {
if (str.charAt(j) == str.charAt(i)) {
count++;
}
System.out.println(str.charAt(i)+" "+count);
}
}
}
}

String Switch Statement Not Working [closed]

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This program's purpose is to translate Morse code into English. I've divided the program's execution logic into three cases: the translation of one Morse code character, two characters and three+ characters. This program tries to translate characters that fall under case three. If you can help, and decide to test the program, please input at least four morse code characters (between a and d)! Otherwise, expect some outofbounds exceptions!
What's wrong with the switch I used? I've read that, up until a few years ago, Java did not support switch statements that used Strings. Java now supports String switch statements and, if I'm not mistaken, my syntax agrees with the not-very-complex conventions of these statements. To determine whether the compiler first recognizes the switch's subject (i.e. switch (aReferenceThatPointsToaString), I printed the reference in the line preceding the switch. No problem there. I also looked at whether cases could be executed if the switch's subject were explicitly stated. Instead of saying switch (array3[i]), I wrote in switch ("._") This test did result in a working switch, leaving me with the impression that switch(array3[i]) isn't quite the same as switch(aStringexpression)
import java.util.Scanner;
public class MorseToEnglish1 {
public static void main (String[]args){
System.out.println("enter morse here");
Scanner input = new Scanner (System.in);
String container = input.nextLine();
char [] array = container.toCharArray();
int count = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] == ' ')
count++ ;
}
// count counts the number of times a space appears
System.out.println(count);
int [] array2 = new int [count];
int counter = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] == ' '){
array2[counter] = i;
counter ++ ;
}
}
// array2 assigns the indexes of spaces to its members
System.out.println(counter);
String [] array3 = new String [array2.length - 1];
for (int i = 0; i < array3.length ; i ++)
array3[i] = container.substring(array2[i], array2[i+1]);
// array3 creates substrings based on these spaces
System.out.print (array3[1]);
System.out.print(array3[0]);
// tests above
for (int i = 0; i < array3.length; i ++){
switch (array3[i]){
case ("._"):
{ System.out.println("a");
break; }
case "_...":
{ System.out.println("b");
break; }
case "_._.":
{ System.out.println("c");
break;}
case "_..":
{ System.out.println("d ");
break;}
}
}
input.close();
}
}

There's nothing wrong with your switch statement, there is a lot wrong with the rest of the program...
The problem is, your custom "split" process is leaving spaces within the text, so, if were to enter ._ _... _._. _.., the array would be split into...
" _...", " _._."
Which is wrong on two accounts, one, it's missing stuff I entered and two, it's leaving the leading space character in the String, so when you try and compare the Strings in the switch statement, you are actually trying to do something like...
"_...".equals(" _...")
Which will fail...
To start with, this "manual" splitting code...
char[] array = container.toCharArray();
int count = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] == ' ') {
count++;
}
}
// count counts the number of times a space appears
System.out.println(count);
int[] array2 = new int[count];
int counter = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] == ' ') {
array2[counter] = i;
counter++;
}
}
// array2 assigns the indexes of spaces to its members
System.out.println(counter);
String[] array3 = new String[array2.length - 1];
for (int i = 0; i < array3.length; i++) {
array3[i] = container.substring(array2[i], array2[i + 1]);
}
// array3 creates substrings based on these spaces
System.out.println(array3[1]);
System.out.println(array3[0]);
Should be replaced with...
String[] array3 = container.split(" ");
As I, frankly, have no idea what it's trying to achieve and it breaks if there is less than 4 "codes" on the input String and doesn't correctly parse the input anyway (throwing away values)

Your splitting code is leaving leading space characters which cause them to not correctly match. You should instead be using a version of String.split in order to produce your array3. This will create the correct substrings with a lot less work.
Edit:
Your splitting code also appears to leave off the first and last morse codes which is why you get ArrayIndexOutOfBoundsExceptions when you input less than 4.

From looking at the code you've got too many } curly braces. You only need one of those at the end of the switch statement.
case ("._"):
{System.out.println("a");
break;
case "_...":
{System.out.println("b");
break;
case "_._.":
{System.out.println("c");
break;
case "_..":
{System.out.println("d ");
break;
}
It should work. My assumption is of course that your problem is that your code doesn't compile. Which I'm pretty sure that the code you posted won't do.

How to go about making this type of combination?

I have a four letter string of letters, such as ASDF and I want to find all 3 (three) letter combinations using these letters and the three letter combination does not need to form a real word.Ex.
AAA AAS AAD AAF
ADA ADS ADD ADF
...............
SSA SSD SSF SSS
I am relatively new to Java, having just learned how to use the String class and using loops and conditional statements. The only way that I know how to do this is by a massive and very tedious set of for loops and if statements that account for every possibility that could arise. This would look like:
public static void main(String[] args)
{
String combo = "";
for(int counter = 1; counter <= 16; counter++){
combo = "A";
if(counter == 1){
combo = combo + "AA";
}
// This would continue on for all the possibilities starting with "A" and
// then move on to "S" as the lead character
}
}
I know that this is one of the worst ways to go about this problem, but I am really stuck as to how to do it another way. It would be easier if I had 3 letters and made the 3 letter combos, as then I could just get each letter from the array and just rearrange them, but as I'm only using 3 of the 4 letters it is more difficult. Any advice on how to get this done in an more efficient manner?

Use a recursive function.
Like this (not tested, don't have a Java compiler on my laptop).
Performance could probably be boosted by using StringBuilder.
static void printAllPossibilities(String charSet, int length) {
printAllPossibilities_(charSet, length, "");
}
static void printAllPossibilities_(String charSet, int length, String temp) {
if (length == 0) {
System.out.println(temp);
return;
}
for (int i = 0; i < charSet.length(); i++)
printAllPossibilities_(charSet, length - 1, temp + charSet.charAt(i));
}
Usage:
printAllPossibilities("ASDF", 4); // Print all 4-letter combinations out of "ASDF"
printAllPossibilities("bar", 2); // Print all 2-letter combinations out of "bar"

For the general case (all combinations of N chars out of M), the solution by #Qntm is exactly what you need... but, use a StringBuilder as he said, and just change the last character instead of constructing strings like 'temp + charSet.charAt(i)'.
If you need exactly 3 chars out of N, it's easier to just do 3 nested loops:
for (int char1 = 0; char1 < charSet.length(); char1++) {
for (int char2 = 0; char2 < charSet.length(); char2++) {
for (int char3 = 0; char3 < charSet.length(); char3++) {
System.out.println(""+charSet.charAt(char1)+charSet.charAt(char2)+charSet.charAt(char3));
}
}
}