Develop Reference

Weird thing happens in a for loop, java

I'm fairly new to programming and I started to write some basic codes. I tried to create an algorithm that found every position of the character you wanted in a string. If I were to enter "helllo" as a string and wanted to search for the character l, it would output 2, 3 and 4. The code worked fine until I tried to trick the code a bit to change something. To do so, I first removed the termination condition in the "for loop" and instead added an if statement during the loop to break it. I spent about 2h on the error that occurred after and I still can't find out what's happening.
Here's the code and the output it gives me. (I know my code is a mess and not properly optimized, but right now I would just like to know what is happening. I'll rearrange it later if I get it to work. thank you^-^.)
When I run the code, instead of displaying as it should, "7, 8, 9, 10"(it searches for the character ^ in the string Oioriew^^^^) it outputs "7-1,8,9,10". To fix it, I can simply insert the termination condition in the loop again, which was, "pow != -1" but at this point, I really want to know why it happens.
public class Tests {
static void zeMethod(String mainString,char charToFind) {
int a = 0;
String b, c;
char chartToConvert;
c = "";
b = "";
for (int pow = mainString.indexOf(charToFind);
; // *the condition was here.*
pow = mainString.indexOf(charToFind, pow + 1)) {
a++;
if (a == 1){
System.out.println("String: "+mainString);
System.out.println("il y a un "+charToFind+" à la/aux position(s)");
}
if (a == 1){
System.out.print(pow);
}
if (a%2 == 0 && pow != -1) {
c = b+", "+pow;
}
if (a%2 != 0 && a != 1 && pow != -1) {
b = c+", "+pow;
}
if (pow == -1){
System.out.print(pow);
break;
}
//*end of loop*
}
if (a%2 != 0){
System.out.println(c);
}
else {
System.out.println(b);
}
}
public static void main(String[] args){
String string = "Oioriew^^^^";
char chara = '^';
zeMethod(string, chara);
}
}
I'm sorry if my question is a bit incoherent or not properly asked. This is my first time on the site and English isn't my mother language. Thank you for your time!
Edit:
I know the question wasn't clear at first, but what I meant is, why does pow become -1 after the second iteration of the loop. Also, why does the break after the System.out.print(pow); doesn't make it leave the loop. (I'm looking how to make a debugger work atm too.)

[...] instead of displaying as it should, "7, 8, 9, 10" [...] it outputs "7-1,8,9,10 [...] I really want to know why it happens
Sounds like you're asking why it prints -1, which happens because you explicitly asked it to:
if (pow == -1){
System.out.print(pow);
break;
}
If pow is -1 then print pow (aka -1) and exit the loop.
UPDATE
As for why the order of the output, use a debugger and step through the code one statement at a time, and you'll see.
But, lets see: We can agree that loop will iterate with pow having these numbers in this order: 7, 8, 9, 10, -1
Why does it print 7 first?
Because you have this code inside the loop:
a++;
if (a == 1){
System.out.print(pow);
}
Why does it print -1 next?
Because there are no other print statements inside the loop, which would be a lot more apparent if you indented the code correctly, i.e. indented the content of the loop.
Why does it print 8,9,10 last?
Because you print the content of b or c after the loop, and that is the content of whichever one of them is being printed.
Note that first value (-7) is not added to b, because you explicitly exclude it in the if statement.

I'm all good, thank you all for your time and Andreas, for the debugger option. It's a must-have for anyone programming. I don't know how I didn't see this earlier, but to break, it had to become -1 and since I had to break before I could print my answer, -1 was first.

Improving methods in Java

I have recently started a course where the main language we are learning at the moment is Java.
I have been tasked with creating a program that allows people to vote on two candidates - the program then counts the votes, and depending on how man votes have been made depends on what is displayed.
Here is the part I am concerned with at the moment:
public String printResults(){
if(candidate1Votes == 0 && candidate2Votes == 0)
{
System.out.println ("No votes cast, results cannot be displayed.");
return "No votes cast, results cannot be displayed.";
}
else if(this.completed == false)
{
System.out.println ("Voting has not finished");
return "Voting has not finished";
}
else if(this.completed == true)
{
System.out.println ("Voting has finished, no more votes will be allowed.");
return "Voting has finished, no more votes will be allowed";
}
{
double totalVotes = this.candidate1Votes + this.candidate2Votes;
double cand1Share = (double) this.candidate1Votes/totalVotes*100;
double cand2Share = (double) this.candidate2Votes/totalVotes*100;
System.out.format(candidate1 + " received %3.1f percent of the votes\n", cand1Share);
System.out.format(candidate2 + " received %3.1f percent of the votes\n", cand2Share);
return "v";
}
}
Originally I used void in this method, but part of our task was to then change it to a string value. This is where I am struggling - once I set completed to true, it is still allowing me to cast votes. I know that this code is incomplete but I can't finish it as I am unsure what to do! These were the next parts to the questions.
Modify your printResults method so that it applies the first two rules. Note that the value of the completed field indicates whether or not voting is complete. The method should be modified to return a String which indicates whether printing has been successful.
Modify your vote method to apply the third rule.
Test your methods by creating an instance and doing the following – before
doing each test note the result you expect to get, and compare this with what you actually get:
• Try to print results immediately
• Cast votes for both candidates and try to print results
• Set the completed field to true by calling setCompleted
• Try to cast a vote for a candidate
• Print the results
I am new to this (this is my first year) and have managed to do okay in my books to get this far, however any help on this next issue would be greatly appreciated!

First of your code is unnecessary complicated, which makes it hard to read/enhance. It can easily simplified, like
public String printResults(){
if(candidate1Votes == 0 && candidate2Votes == 0) {
System.out.println ("No votes cast, results cannot be displayed.");
return "No votes cast, results cannot be displayed.";
} // you returned ... NO need for ELSE!
if(this.completed == false) {
System.out.println ("Voting has not finished");
return "Voting has not finished";
}
// it is very clear here that completed must be true!
double totalVotes = this.candidate1Votes + this.candidate2Votes;
double cand1Share = (double) this.candidate1Votes/totalVotes*100;
double cand2Share = (double) this.candidate2Votes/totalVotes*100;
System.out.format(candidate1 + " received %3.1f percent of the votes\n", cand1Share);
System.out.format(candidate2 + " received %3.1f percent of the votes\n", cand2Share);
return "v";
}
Probably that easier-to-read code is all that you need to get you going!

Looking at the code the last block will never be reached because either you have no votes or you have votes and in that case completed will be either true or false and will thus reach always one of the else ifs and they all return a string. So I wonder why how you can cast any votes at all.
You could also post the code where you call printResults and setCompleted to see where the problem lies.
Some more hints for improving your code:
Sometimes you have the opening bracket on the same line and sometimes on the next. You should probably choose one style
It is not necessary to surround the last code block with brackets
if (this.completed == true) and else if (this.completed == false) is a bit redundant and can be written like: if (this.completed) and if (!this.completed). Also you can write
if (this.completed) {
...
} else {
....
}
because if completed is not true it can only be false.
Instead of writing every String two times and having to edit it two times in case you want to change something you could also do the following:
String msg = "Voting has not finished"
System.out.println(msg);
return msg;

Half String every second char

I an coding beginner.I have started practicing SPOJ basic problems.This was the one I was trying to solve , But the code is incorrect.
Please help me where I have coded this question wrong as I am unable to figure out:
public class Print2ndChar {
public static void main(String[] args) throws java.lang.Exception {
Print2ndChar mainObj = new Print2ndChar();
java.io.BufferedReader inputReader = new java.io.BufferedReader(new java.io.InputStreamReader(System.in));
String noOfTestCase;
if(((noOfTestCase = inputReader.readLine()) == null))
System.exit(0);
int noOfLines = 0;
try{
noOfLines = Integer.parseInt(noOfTestCase);
}catch(Exception e){
System.exit(0);
}
if(noOfLines<0 || noOfLines>100)
System.exit(0);
String [] randomWords = new String[noOfLines];
for(int i=0;i<noOfLines;i++){
randomWords[i] = inputReader.readLine();
if(randomWords[i] == null || randomWords[i].length()<2 || randomWords[i].length()%2!=0 || (randomWords[i].length()/2)>100)
System.exit(0);
}
for (String word : randomWords){
mainObj.letsBegin(word.substring(0, word.length() / 2));
System.out.println();
}
}
private void letsBegin(String data) {
if (data.length() <= 0) {
return;
} else {
System.out.print(data.charAt(0));
if (data.length() >= 3)
letsBegin(data.substring(2, data.length()));
}
}
}
EDIT :
I/P : 4
your
progress
is
noticeable
O/P
y
po
i
ntc

OK! So after a lot of hit and trials, I know what is wrong with your code. The code that you have written fails because of the condition randomWords[i].length()%2!=0 inside your if. There is nothing wrong with you putting this condition to check the input, but if you will select sample test case, inside the highlighted blue area you will notice an extra space after every string. Like this :
You can see that other than the last input all other input strings have a space character at the end. So, when you read the string from stdin the length of the string is 2*k + 1 (because of the space), and your program will exit without any output. Hence you get a wrong answer.
This problem exists with other test cases as well probably. And how do I know this? After spoj shows you wrong answer, if you click on the wrong answer, it will show you 2 failed test cases, something like this:
It shows your program's output is empty because your code exited because of the extra space at the end of strings.
So, I believe the person who wrote the test cases should be given a WT Error (Wrong Test Cases) :P :D
So, the possible correction is you remove the mentioned condition from the if and you will get AC. Because now you will be dividing 2*k + 1 by 2, which will not be an integer and which will get rounded to the nearest smallest integer, which will be same as dividing 2*k by 2 and the program will give the correct result.
A few things that you should take care while solving questions on spoj, you do not have to verify that every input lies within the range specified in the question, or if it is a valid data type. The range is given to tell you that Spoj will only test your program with cases which lie between those ranges and will not exceed them. So, even if you remove all the code where you check for exceptions and ranges of input data, you will get an AC. Moreover, writing such code only adds to the burden.
Hope this helps. :)

Split an excel content and display in console ignore Blank Cell

In my Excel cell I've data as below. And it is in a single cell.
01/2015 to present
02/2010 - 04/2010
01/2008 - 12/2009
I want to split this cell and display result as
Date 1 - 01/2015 to present
Date 2 - 02/2010 - 04/2010
Date 3 - 01/2008 - 12/2009
I'm not sure of how many different lines of data will be in a single cell. There may be up to 10 Lines, then it should print the same like in the below format
Date (counter) - (Date Value)
Thanks for the hint #Saurabh
I'm using the below code now. and able to print the dates in the above format.
if (!covDate.equals(null) || !(covDate == "")|| !(covDate == null) || !covDate.equals(""))
{
String[] date = covDate.split("\\n");
for(int i=0;i<parts.length; i++){
System.out.println("Date "+ i + " - "+date[i])
}
}
else{
System.out.println("Dates Missing");
}
Here Basically I've 2 columns in my Excel and in the second column I've these dates. And there are instances where in there is data in first column but data is missing in Dates column. And there it is throwing me a null pointer Exception Also in my eclipse my else block is showing error as a dead block
please let me know how can i fix these.
Thanks

I have simplified your test expression below, the date variable inside the for loop is replaced with parts :
if (covDate != null && !covDate.equals(""))
{
String[] parts = covDate.split("\\n");
for(int i=0;i<parts.length; i++){
System.out.println("Date "+ i + " - "+parts[i])
}
}
else{
System.out.println("Dates Missing");
}

Firstly, you should not such a long ternary operator, instead you can use, StringUtils.isEmpty(object) ..
This is available in org.apache.maven.shared.utils.StringUtils. Once you have this, you can read the object to this method, and perform accordingly.
Secondly, coming to second column, you can try, if(StringUtils.isEmpty(cellValue)) if this gives you true, then go ahead and split the string, else continue.

if statement with integers [duplicate]

This question already has answers here:
Why does my if condition not accept an integer in java?
(7 answers)
Closed 3 years ago.
I'm new at Java. I'm looking for some help with homework. I wont post the full code I was doing that originally but I dont think it will help me learn it.
I have a program working with classes. I have a class that will validate a selection and a class that has my setters and getters and a class that the professor coded with the IO for the program (it's an addres book)
I have a statement in my main like this that says
//create new scanner
Scanner ip = new Scanner(System.in);
System.out.println();
int menuNumber = Validator.getInt(ip, "Enter menu number: ", 1, 3);
if (menuNumber = 1)
{
//print address book
}
else if (menuNumber = 2)
{
// get input from user
}
else
{
Exit
}
If you look at my if statement if (menuNumber = 1) I get a red line that tells me I cannot convert an int to boolean. I thought the answer was if (menuNumber.equals(1)) but that also gave me a similar error.
I'm not 100% on what I can do to fix it so I wanted to ask for help. Do I need to convert my entry to a string? Right now my validator looks something like:
if (int < 1)
print "Error entry must be 1, 2 or 3)
else if (int > 3)
print "error entry must 1, 2, or 3)
else
print "invalid entry"
If I convert my main to a string instead of an int wont I have to change this all up as well?
Thanks again for helping me I haven't been diong that great and I want to get a good chunk of the assignment knocked out.

if (menuNumber = 1)
should be
if (menuNumber == 1)
The former assigns the value 1 to menuNumber, the latter tests if menuNumber is equal to 1.
The reason you get cannot convert an int to boolean is that Java expects a boolean in the if(...) construct - but menuNumber is an int. The expression menuNumber == 1 returns a boolean, which is what is needed.
It's a common mix-up in various languages. I think you can set the Java compiler to warn you of other likely cases of this error.
A trick used in some languages is to do the comparison the other way round: (1 == menuNumber) so that if you accidentally type = you will get a compiler error rather than a silent bug.
This is known as a Yoda Condition.
In Java, a similar trick can be used if you are comparing objects using the .equals() method (not ==), and one of them could be null:
if(myString.equals("abc"))
may produce a NullPointerException if myString is null. But:
if("abc".equals(myString))
will cope, and will just return false if myString is null.

I get a red line that tells me I cannot convert an int to boolean.
Thats because = is an assignment operator. What you need to use is == operator.

A single equal sign is assignment: you assign value to a variable this way. use two equal signs (==) for comparison:
if ($menuNumber = 1) {
Update: forgot dollar sign: $menuNumber