Consider the following method:
void a ()
{
int x;
boolean b = false;
if (Math.random() < 0.5)
{
x = 0;
b = true;
}
if (b)
x++;
}
On x++ I get the "Local variable may not have been initialized" error. Clearly x will never be used uninitialized. Is there any way to suppress the warning except by initializing x? Thanks.
No, there is no way Java can examine all possible code paths for a program to determine if a variable has been initialized or not, so it takes the safe route and warns you.
So no, you will have to initialize your variable to get rid of this.
There is one :
void a () {
if (Math.random() < 0.5) {
int x = 1;
}
}
The compiler isn't responsible for devising and testing the algorithm. You are.
But maybe you should propose a more practical use case. Your example doesn't really show what's your goal.
Why don't you simply use
void a ()
{
int x;
boolean b = false;
if (Math.random() < 0.5)
{
x = 0;
b = true;
x++;
}
if (b) {
//do something else which does not use x
}
}
In the code why do you want to use x outside the first if block, all the logic involving x can be implemented in the first if block only, i don't see a case where you would need to use the other if block to use x.
EDIT: or You can also use:
void a ()
{
int x;
boolean b = (Math.random() < 0.5);
if (b) {
x=1
//do something
}
}
You can and should be defining the value of x unconditionally if it will be used later in your code.
There are a few ways to do this:
On initialization
int x = 0;
Because this is outside the conditional (if), Java won't complain.
Add else clause to conditional
if (Math.random() < 0.5)
{
x = 0;
b = true;
} else
{
x = 1;
}
Because there is an else to this if, and both code paths initialize x, Java will also be happy with this.
Move your usage of the variable into the conditional block
Clearly the question has a minimally-reproducible example, not a full one, but if you only ever want to use the variable conditionally, then it belongs in the conditional block.
if (Math.random() < 0.5)
{
x = 0;
x++;
}
If you don't aren't conditionally using the variable, then you need to provide an integer value to use in case Math.random() >= 0.5, using one of the solutions above.
Related
Why is x not initialized in the following ?
public class rough {
public static void main(String[] args) {
int x;
boolean found = false;
for (int i = 0; i < 10; i++) {
if (Math.random() < 0.5) {
found = true;
x = 10;
break;
}
}
if (!found)
x = -1;
System.out.println(x);//x isn't initialized here
}
}
On average, for half of the iterations, the if inside the for loop would be true, thus initializing x. For the other half, found stays false therefore the outer if would initialize. Therefore, I don't understand why the compiler is annoyed.
As the ultimate distillation (see successive simplifications below), consider
public static void main(String[] args) {
int x;
boolean found = false;
if (!found)
x = -1;
System.out.println(x);
}
which also gives the error that x isn't init.
previous simplifications
Even more surprisingly, changing
if (Math.random() < 0.5) to if(true) also has the same problem.
In fact, investigating further, replacing the original for loop by these
for (int i=0;i<1;i++)
x = 10;
for (; !found; ) {
x = 10;
break;
}
is equally worse. Only for(;;){... break;} & for(;true;){... break;} don't raise any init. errors.
The compiler can't easily detect all branches lead to x being initialized, but you can fix that (and the code) pretty easily by assigning -1 to x to begin with. Something like
public static void main(String[] args) {
int x = -1;
for (int i = 0; i < 10; i++) {
if (Math.random() < 0.5) {
x = 10;
break;
}
}
System.out.println(x);
}
And now you don't need found (so I removed it too).
(Writing this up as a separate answer since I think it'll benefit from being taken out of comments).
This is the language-lawyer's answer.
The language specification requires initialization of variables before they are used.
The rules include:
The variable must be given a value on all possible paths through the code. The specification refers to this as 'definite assignment'.
The compiler does not consider the values of expressions in this analysis. See Example 16.2 for this.
The second rule explains why even in cases that are 'obvious' to us, the compiler can't use that knowledge. Even if the compiler-writer cared to do a deeper analysis, adherence to the Java specification forbids it.
If the next question is 'but why?' then I'd have to guess, but the point of a standard is to get consistent behaviour. You don't want one compiler accepting as legal Java something that another compiler rejects.
since I desperately search for a solution for 3 hours, but didn't find anything, I'm gonna try it here.
public static void Assemble(boolean isAnnuityLoan, double K, double P, int N) {
double loan = K;
int year = 1;
Output.PrintHeader();
if (isAnnuityLoan) {
double *rpy* = Calculation.AnnRatePerYear(K, P, N);
} else {
double *T* = Calculation.AmortRate(N, K);
}
for (int x = 0; x < N; x++) {
double I = Calculation.Interest(loan, P);
if (isAnnuityLoan) {
double T = Calculation.AnnAmortRate(**rpy**, loan, P);
} else {
double rpy = Calculation.RatePerYear(loan, P, **T**);
}
Output.PrintTableLine(year, loan, I, **T**, **rpy**);
loan = loan - **T**;
year++;
System.out.println("\n");
}
Thats the code. Oh cool looks like formatting doesnt work in the code segment. Anyway, problem is, compiler marks the italic variables (K and rpy) as not used, and the bold ones (double stars) as not found. Now this doesnt make any sense to me, because lets trace both paths.
isAnnuityLoan is false. itll calculate T, then enter the for loop and since isAnnuityLoan is still false, the else statement will trigger and rpy will be calculated.
is vice versa, rpy gets calculated first, the T in the loop, and therefore both variables should be available.
Buuuut theyre not. And I have no idea why. Now where is the issue with this?
You help is highly apprechiated and will help me not to riddle about this in my sleep.
"rpy" variable is defined inside an if scope, so you can't it outside.
To use variables outside if scope, take a look at the example below.
Note that warnings are different than errors. Compiler WILL compile your code even if it contains warnings like (variable not used) but NOT errors.
example :
//define var
int x;
if(test){
//initialize it here
x = 1;
} else {
//or here
x = 2;
}
//use it here
System.out.println(x);
I'm trying to make a method to add SetOnMousePressed-functions to multiple nodes, and i've tried using a couple different loops etc, but i always end up with the error "Local variable x defined in an enclosing scope must be final or effectively final." This is as far as i've gotten:
public static int playerSelectingCategory(int intScorecard[][], Rectangle[][] scoreboardBackground, int categoryCounter, int nrOfPlayers, boolean limitCheck)
{
int counter = 0;
int y = 0;
for(int x = 0; x<YatzyConstants.getNrCategories(); x++)
{
if(counter < nrOfPlayers)
{
if(y < YatzyConstants.getNrCategories())
{
scoreboardBackground[counter][y].setOnMousePressed(e ->
{
scoreboardBackground[counter][y].setFill(javafx.scene.paint.Color.ALICEBLUE);
});
y++;
}
counter++;
}}
return intScorecard[counter][y];
}
I originally declared everything one by one, but i have to think that there should be a more effective way of doing it. Any help is appreciated, really hit a brick wall here.
Since y is modified by your code, it's not efectively final and therefore cannot be accessed from a anonymus class or lambda expression. The same is true for counter.
I'd recommend storing scoreboardBackground[counter][y] in a variable that is (effectively) final (that is unless the array is modified and you want to color the Rectangle at the position where the original rect was stored...):
final Rectangle rect = scoreboardBackground[counter][y];
rect.setOnMousePressed(e -> {
rect.setFill(javafx.scene.paint.Color.ALICEBLUE);
});
Or alternatively simply use the Node that is the source of the event:
final EventHandler<MouseEvent> handler = event -> {
((Shape) event.getSource()).setFill(javafx.scene.paint.Color.ALICEBLUE);
};
for(int x = 0; x<YatzyConstants.getNrCategories(); x++)
{
if(counter < nrOfPlayers)
{
if(y < YatzyConstants.getNrCategories())
{
scoreboardBackground[counter][y].setOnMousePressed(handler);
y++;
}
....
I wrote a method for checking if a number is a prime:
static boolean isPrime(int x) {
for (int i = 2; i <= Math.sqrt(x); i++) {
if (x % i == 0)
return false;
}
return true;
}
In a collection of exercises we're learning from, the solution is:
static boolean isPrime(int x) {
boolean hasDivisors = false;
for (int i = 2; i <= Math.sqrt(x); i++) {
if (x % i == 0) {
hasDivisors = true;
break;
}
}
return !hasDivisors;
}
In my case, If i find a divisor, I return that the number is not a prime (return false) and that replaces the need for a break in second method. The only other obvious reason is that the second method only uses a single return statement.
Is there a reason for this (speed/memory wise)?
It's a matter of style, mostly. Some coding conventions dictate that a method have only a single return statement. This makes a lot of sense in languages where you have to free resources explicitly, but doesn't have any functional impact in Java.
Personally, I prefer to just return as soon as you know the result (like in the first snippet), but again, it's a matter of personal style.
Both solutions work and both are valid for all the reasons you have already identified. Good analysis. As others have noted, the differences are purely stylistic. Here is an interesting discussion about the single return statement style in java.
Performance-wise, I wouldn't expect a significant difference between the 2 approaches. But if you want to be certain, perform a benchmark test. If you want a real performance boost, you can eliminate the expensive call to Math.sqrt() by replacing:
for (int i = 2; i <= Math.sqrt(x); i++) {
with
for (int i = 2; i*i <= x; i++) {
int x = 1;
Consumer<Object> f = (i) -> {
int x = 1; // invalid
};
vs.
Consumer<Object> f = (i) -> {
int x = 1;
};
int x = 1; // valid
Imagine those two blocks inside a method. Why is the second block valid?
This is very similar to normal Java scopes:
int i;
{
int i; // invalid
}
vs.
{
int i; // valid
}
int i;
In the first block you are "shadowing" the original x variable. You will not be able to access the first x anymore by creating a new x.
The second block is okay, because the second x is created on a moment when the first x does not longer exists (it is out of scope).
Basically: in the first case, you are trying to have two variables called x at the same time. In the second case, you are creating two x variables after each other: their lifetimes will not overlap.