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Cannot come up with why does my Java code loop infinitely

I tried to do this LeetCode daily challenge but I've found out that my code loops infinitely.
I looked through it multiple times, but I cannot find where the problem is. If anyone could spot it, please answer.
public int longestValidParentheses(String s) {
int count, highestOne = 0, index = 0;
boolean isSevered = false;
boolean theEnd = false;
while(!theEnd) {
count = 0;
while(!isSevered) {
if(index<s.length()-2) {
if(s.charAt(index) == '(' & s.charAt(index++) == ')') {count = count + 2;index = index+2;}
else {isSevered = true;}}
else theEnd=true;isSevered=true;
}
highestOne = count;
}
return highestOne;
}

I have 2 suggestions for you:
Use indentation and do not write if/else on the same line as the code associated with them
Always, ALWAYS use bracelets, even if you have only a single command. I think one of the wrongs java did is letting the programmers the free not to use bracelets if there is just a single command after it. It confusing.
So you have 2 mistakes here that make your code run for infinity:
isSevered will always be true after one loop exactly, as you change it to true no matter what happens as it is outside the if else statements, hence the reason I wrote the 2 advices above.
You never changing isSeveres or theEnd at the outside loop. Meaning that if isSevers is true and theEnd is false, you will never enter the internal while and will never exit the outside while.
The two of those combined means that if the condition that make theEnd be initialized with true won't happen at the first run, you will be stuck with infinity loop.

Core Java What is Wrong with this Simple Syntax

I have a code like this.
public void test()
{
final boolean passDay = true;
final int status;
//if the right side takes place means i need status value below and is set
//if only the left side is takes place i dont care about status because i will not use it
if(!passDay && ((status=loadStatusValue()))== Constants.YORK_CORK_EMPTY_SET)
System.out.println("inside 1");//I DONT CARE ABOUT STATUS VALUE
else
{
if(passDay)System.out.println("BAMBA");//I DONT CARE ABOUT STATUS VALUE
else
{
//HERE STATUS ALWAYS WILL HOLD A VALUE SIMPLYBECAUSE RIGHT SIDE
//IS ALREADY EVALUATED BECAUSE passDay=false and !passDay=true
System.out.println(status);
//I MEAN I USE STATUS ONLY AFTER BEING INITIALIZED
//WHY variable status might not have been initialized IS SHOW IF I AM HERE IS BECAUSE STATUS HAS A VALUE.
}
}
}
private boolean compute(){return true;}
private int loadStatusValue(){return Constants.BOTH_YORK_CORK_SET;}
What i think in this method everything works i use the status int variable when was already set even is not defined in the initialization.
As you can see the passDay is a boolean means only could hold true or false i have try hardcoded with true and false and not compilation error is show when shows a compilation error when instead being hardcode i do something like it.
final boolean passDay = compute();
in fact compute is also returning a boolean could be true or false but in this case a compilation error is show variable status might not have been initialized but can't java realize that even when true status is just not used and when false the status variable is initialized or set it and used later or i am wrong?

Please read about short circuit evaluation in Java -
(condition1 && condition2).
The condition preceding the && operator if evaluates to false; the execution does not evaluate the other condition as it is not required. Regardless of the output of condition2 - false && condition2 shall evaluate to false. Try rearranging the order of conditions or initialize the variable first.
hope this helps.

&& is a short-circuiting operator. This means that in the following expression:
!passDay && ((status=loadStatusValue()))== Constants.YORK_CORK_EMPTY_SET
The right-hand operand (((status=loadStatusValue()))== Constants.YORK_CORK_EMPTY_SET) is only evaluated if !passDay is true. This means that the assignment of status will only take place if passDay is false.
If you use final boolean passDay = compute();, there is no way of knowing at compile time if compute() will return true or false.
I suspect (without having tried it) that your code will compile with compute() if you use the non-short-circuiting AND operator, &, which will evaluate the right-hand operand even if the left-hand operand is false.
I think that a better approach is just to restructure your code to make this conditional more readable. Conditionals with side effects are error-prone because it is very easy to overlook the side effect when reading the code:
if (passDay) {
System.out.println("BAMBA");
} else {
status = loadStatusValue();
if (status == Constants.YORK_CORK_EMPTY_SET) {
System.out.println("inside 1");
} else {
System.out.println(status);
}
}

Within your code, you set passDay to final meaning it's value can never be changed throughout program execution. Bear this in mind whenever you're doing conditional checks, such as if(!passDay ... ) etc.
If you expect, or intend, to have the value passDay changed, then you can initialise to true at the start of execution, but don't delcare as final
Hopefully this clears some things up.

Can I return Boolean value in a loop statement? [closed]

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I just came up with a problem returning a Boolean value in accordance with a given condition. I thought in order to check the given condition for full possibilities I need to use for loop. But when I tried to compile it, it gives me error, possibly because there is uncertainty returning a Boolean value using for loop. Here is an original problem:
Return true if the given string contains a "bob" string, but where the middle 'o' char can be any char.
bobThere("abcbob") → true
bobThere("b9b") → true
bobThere("bac") → false
And here is my code:
public boolean bobThere(String str)
{
for(int i=0; i<str.length()-3; i++)
{
if (str.length()>=4): && str.charAt(i)=='b' && str.charAt(i+2)=='b')
{
return true;
}
else
return false;
else if (str.length()==4 && str.charAt(0)=='b' && str.charAt(2)=='b')
{
return true;
}
else
{
return false;
}
}
}
I just wanted to ask :
1. Can I fix the this code for returning a value. I mean, can I use for loop and return specific value for a given condition? If yes, please could you give me a sample.
2. Or are there any ways other than for loop to solve this problem.
Thanks in advance.

The compiler error is almost certainly because you have an elseif after an else. That's invalid.
Looking at your code, what you seem to want to do is loop through the string, and then return true if you're at the start of a b?b string. I'm not sure why you have your second if condition in there - at the moment your code would check the first and third characters of the string on every iteration of the loop, if the string happens to be exactly four characters long. Pointless, it doesn't need to be there. The check for length isn't necessary at all.
Additionally, your end condition for the loop is currently i < string.length()-3. This means that the final three characters of the string will not be checked. You would need to change this to either i <= string.length()-3 or i < string.length()-2 to solve this.
Your else return false stuff is going to give you a serious problem. Your code will enter the loop once, and then either return true or false, without ever going to the next phase of the loop. What you should do is loop through the string, and if you find what you're looking for, return true. Otherwise, don't return at all, and keep going with the loop. If you get to the end of the loop it means you never found what you were looking for, so you can at that point return false.
Taking those comments into account, your revised code would look like this (please note I haven't compiled or run this):
public boolean bobThere(String str)
{
for(int i = 0; i <= str.length() - 3; i++)
{
if (str.charAt(i) == 'b' && str.charAt(i + 2) == 'b')
{
return true;
}
}
return false;
}

Does Java waste time to check conditionA in "if(isOK && conditionA)" if isOK=false?

Ok, i am building program to check many fields. If at least 1 field is not ok then i don't want my program to spend time to check other fields. So let look at this code:
// Util.isReadyToUse method return true if the string is ready for using, & return false if it is not.
boolean isOK=true;
if(!Util.isReadyToUse(firstName)){
isOK=false;
}
else if(isOK && !Util.isReadyToUse(lastName)){
isOK=false;
}
else if(isOK && !Util.isReadyToUse(email)){
isOK=false;
}
.....more checking
if(isOK) {
//do sthing
}
Ok, when running, the program will first check !Util.isReadyToUse(firstName). Suppose it returns (isOK=false). Next the program will check isOK && !Util.isReadyToUse(lastName).
So my question here is that Since the isOK currently false, then will the program spend time to check the condition !Util.isReadyToUse(lastName) after &&?
Ok, As a human being, if you see isOK=false and now you see isOK && !Util.isReadyToUse(email), then you don't want to waste time to look at !Util.isReadyToUse(email) since isOK=false and u saw && after isOK.
Will machine also work like that?
I am thinking to use break but why people say break doesn't work in if statement:
if(!Util.isReadyToUse(firstName)){
isOK=false;
break;
}
else if(isOK && !Util.isReadyToUse(lastName)){
isOK=false;
break;
}......
What is the best solution in this situation?

So my question here is that Since the isOK currently false, then will
the program spend time to check the condition
!Util.isReadyToUse(lastName) after &&?
Java is smart, if you have a condition if(somethingFlase && something), then something won't be reached due to Short-circuit evaluation. Since the whole expression will be false regardless of the second condition, there is no need for Java to evaluate that.
From 15.23. Conditional-And Operator &&:
If the resulting value is false, the value of the conditional-and
expression is false and the right-hand operand expression is not
evaluated. If the value of the left-hand operand is true, then the right-hand expression is evaluated.
if(a && b) - if a is false, b won't be checked.
if(a && b) - if a is true, b will be checked, because if it's false, the expression will be false.
if(a || b) - if a is true, b won't be checked, because this is true anyway.
if(a || b) - if a is false, b will be checked, because if b is true then it'll be true.

No, it shortcuts the rest of the predicate.
That's you'll see things like
if(A != null && A.SomeVal == someOtherVal)

Java supports what is referred to as Short-Circuit Evaluation. See this page:
http://en.wikipedia.org/wiki/Short-circuit_evaluation
What this means is that if the first boolean in your statement is enough to satisfy the statement, then the rest of the values are skipped. If we have the following:
boolean a = false;
boolean b = true;
if(a && b) /*Do something*/;
'b' will never be checked, because the false value for 'a' was enough to break out of the if statement.
That being said, your program will never take advantage of this because the only time isOK is set to false is within one of your else if statements.

As the other responders mentioned Java will do the smart thing.
But it could be the case that you want Java to continue to check, in that case you can use & vs && or | vs ||.
if (someMethod() | anotherMethod() {
If the first method reutrns true, Java will still execute the second method.
if (someMethod() & anotherMethod() {
If the first method is false, Java will still execute the second method.

No, Java won't "waste time" for it. It's called short circuit evaluation.
This mechanism is commonly used e.g. for null checking :
if (foo != null && foo.neverFailsWithNPE()) {
// ...
}

You don't need to use break on an if..else if.. else statement because once it finds a condition which is true the rest aren't even looked at.

increment a number in java until it gets to 100 than decrement down to 0 continously

I'm making a game where there is a goalie. i want him to move back and forth forever. i have an int called goalieposx (goalie position on the x axis) and i want this is go up by 1 until it hits 200, then go down by one till its back a 0 and repeat. I've tried the folllowing
//this bit isnt in the method, its outside as global varibale
boolean forward=true
//this bit is in a method which is continiouly called nonstop
if (goalieposx<200){
forward=true;
}
else if (goalieposx>200){
forward=false;
}
System.out.println(forward);
if(forward=true){
goalieposx++;
System.out.println("forward");
}
else if (forward=false){
goalieposx--;
System.out.println("backwards");
}
}
this method is called continously. It prints true until it gets to 200, then it prints false. However, it always prints forward, never backward. So conclusion is: the boolean changes as expected but the first if is always called, it seems to ignore the condition
ive also tried this
if(forward = true){
if(goalieposx==200){
forward=false;
}
else{
goalieposx++;}
}
else{
if(goalieposx==0){
forward=true;
}
else{
goalieposx--;}
System.out.println(goalieposx);
}
but this doesnt work either, it prints 1 then 2 etc upto 200 then prints 200 forever. Anyone know how i can solve this? is an if statement the wrong idea altogether?

This is why you should never do comparison for boolean types in if, while, for, whatever. You have just done the assignment in your if statement:
if(forward=true)
the above if statement will always evaluate to true. The problem with this is, this compiles successfully in Java, as syntax wise it's alright. Compiler just checks the type of expression in if evaluates to boolean or not. And it does, so it's ok with it.
You need to do the comparison:
if(forward==true)
.. but as I said, you should not do comparison for boolean types. So, simply doing this:
if(forward)
would be enough.
You also don't need those else if in both the conditions. Just an else will work fine. Well, I don't understand the use of boolean variable at all. It seems like you don't need it. You can change your code to:
if (goalieposx<200){
// forward=true;
goalieposx++;
System.out.println("forward");
}
else {
// forward=false;
goalieposx--;
System.out.println("backwards");
}
What you were previously doing is, setting a boolean variable, based on a condition, and using that boolean variable as condition to execute another if-else block. Well, whatever you are executing in the 2nd if-else block, can simply be moved in the original if-else block, without taking the help of the middle-actor boolean variable.

if(forward=true) does not do what you thing it does.
In java = is the assignment operator and == is the comparison operator. What you are doing with that statement is saying "if assign forward to true" which will set forward to true and always return true.
What you mean to say is if(forward) and if(!forward).
In fact you don't need the else if just an else as if the boolean is not true it must be false.

A better way to do it is to get it to move to the left by adding a minus number, and to the right by adding a positive number. Here's an example of doing this with a loop:
for(int i = -10; i < 100; i++) {
xPosition += i;
}
This would add -10 then -9 etc. to the position.

In your if statements, you need to put two equal signs to check for equality.
if (forward == true){
// execute code
}
EDIT 1:
if (forward)
would be much simpler.

First let's examine what you have already written:
if (goalieposx<200){
forward=true;
}
else if (goalieposx>200){
forward=false;
}
The problem with this code being first is that it while it might set the direction to false once 'goalieposx' has reached 201, in the next call, it will set the direction back to true.
Instead, try using this clever alternative:
//This goes before the infinite loop method
counter = 0;
//Then in the infinite loop method
counter++;
if(counter > 100) {
counter = -100;
}
goalieposx = 100 + counter; //(this shifts counter from
// between -100 and 100 to 0 and 200)

The problem is you are setting the direction based on the value of the integer, instead of whether a condition has previously been met. Try this:
//this bit is in a method which is continiouly called nonstop
if (forward && (goalieposx>200)){
forward=false;
}
System.out.println(forward);
if(forward=true){
goalieposx++;
System.out.println("forward");
}
else if (forward=false){
goalieposx--;
System.out.println("backwards");
}
}