Custom variable scope - java

Is it allowed to just place some curly braces without any if/for/etc statements to limit the variable scope?
An example:
public void myMethod()
{
...
{
int x;
x = 5;
}
...
}
I may want to do this, so I know for sure I won't access/change the variable outside the scope and that it will be destroyed beforehand

Yes, it's allowed. Just try and see for youtself

The curly braces { .. } limit the scope of variables to the block.
However, changes can be made to global variables falling into the scope of { .. } block.
int x = -1;
double y = 5;
{
x = 10;
y = 100;
char c = 'A';
}
System.out.println(x + " " + y); // 10 100.0
System.out.println(c); // Compile time error and out of scope
{
c = 'B'; // Compile time error and out of scope
}

Related

Calling function inside another function [duplicate]

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.

Why can't I access a variable declared in a switch?

this is my code :
there is a probléme "can't acces to variable j"
public static void main(String args[]) {
char digit = 'a';
for (int i = 0; i < 10; i++){
switch (digit){
case 'x' : { int j = 0; System.out.println(j); }
default : { int j = 100; System.out.println(j); }
}
}
int i = j;
System.out.println(i);
}
Each variable has scope. Scope is a restriction regarding where some variable can be accessed.
When you declare a variable in any type of block {}, that variable can only be accessed within that block of code.
You'll have to declare (and possibly initialize) the variable outside the block so its scope is greater, either at the method level or as a static variable (or instance if you were working with instance methods).

java cannot find symbol error: beginner

When compiled this static method comes up with the error that the int array variable, coord, cannot be found. I declared it within the method and it is of the type int[] and I can't figure out why it won't work. I have a feeling that it has to do with the method being static, but changing it to static was the only way I found to make that method work in the first place.
I feel like this is probably really simple for anybody but me especially when all I could find on this subject were much more complicated coding issues.
In case this helps.. this method is supposed to return the (x,y) coordinates for a move location. Sorry for probably not inputting the code correctly. First time doing this. Thanks in advance for any help
CODE:
public static int[] getMove(String player)
{
boolean done = false;
while(!done)
{
Scanner in = new Scanner(System.in);
System.out.println("Input row for " + player);
int x = in.nextInt() - 1;
System.out.println("Input column for " + player);
int y = in.nextInt() - 1;
int[] coord = {x,y};
if(getLoc(coord[0], coord[1]).equals("x") || getLoc(coord[0], coord[1]).equals("o") || coord[0] < 0 || coord[0] > ROWS || coord[1] < 0 || coord[1] > COLUMNS)
{
System.out.println("Invalid coordinates... please retry");
}
else
{
done = true;
}
}
return coord;
}
What you are missing is the scope of the variable. A variable declared in parent block is accessible in child blocks, but not the other way around.
public void someMethod()
{
int x=1;
while(x<10)
{
x++; //x is accessible here, as it is defined in parent block.
int j = 0; //This variable is local to while loop and will cause error if used inside method
j++;
}
System.out.println(j);// The outer block does not know about the variable j!!
}
Now in your case,
Notice where you have defined coors, and in what all places you are using it.
Try to figure where should you define coors variable.
That is because the array coord is local to the while loop. And therefore its not visible outside its scope. Move the declaration of coord outside the while and it should work.
int[] coord = new int[2];
while(!done){
...
...
coord[0] = x;
coord[1] = y;
...
}

Flow analysis in javac - variable a might not have been initialized

In the following code I'd expect that it should not be necessary to initialise variables a and b in last else block, however the compiler does not like it.
import java.util.Random;
public class Foo {
private void foo () {
double a,b;
boolean c;
double r = (new Random()).nextDouble();
if(r < 0.25) {
a = 1;
b = 2;
c = true;
} else if(r >= 0.25 && r < 0.75) {
a = 3;
b = 3;
c = true;
} else {
// why is it necessary to init a and b here?
// given that c is set to false
c = false;
}
if(c) {
double k = a + b;
}
}
}
With the code above, the compiler does complain.
bash-3.2$ javac Foo.java
Foo.java:25: variable a might not have been initialized
double k = a + b;
^
Foo.java:25: variable b might not have been initialized
double k = a + b;
^
2 errors
I would have thought that the compiler could do the static analysis to figure out that k will not be evaluated if c is set to false. So my question is why does the compiler demand that I initialise a and b?
The compiler isn't smart enough to understand that going through the else block will set c to false, and that the next if block thus won't ever be executed. The static analysis is more limited than what you expect, which also makes the comilation faster/
And it's probably a good thing, because changing the code of the else block would suddenly make the next if block not compilable, which would be annoying.

Java: Confused about the variable initialization concept

I'm new to Java and I'm having a bit of trouble understanding the concept of the declaration and the initialization of variables.
For example, when I do:
public class Foo {
public static void main (String[] args) {
int x, y;
for (x = 0 ; x < 10 ; x++) {
y = x + 1;
}
System.out.println(x);
System.out.println(y);
}
}
It does not compile and says that "variable y might not have been initialized."
However, it does not have any trouble if I tell it to just print out the x value after the loop. Of course it would work if I simply declared it in the beginning (saying int y = 0; or something like that), but I wanted to know why x is printed but not y.
Thanks in advance!
Edit:
I understand that the compiler doesn't actually check inside the loop to see if the variable would be initialized or not so it just says it might not have been initialized, but then why does the following code work? Does the compiler check the if loop but not the for loop?
public class Foo {
public static void main (String[] args) {
int x = 0, y;
if (x == 0) {
y = 1;
}
else {
y = 2;
}
System.out.println(y);
}
}
Edit 2:
It looks like it gives me the same error if I actually give another condition for the else part so that it would be:
if (x == 0) {
y = 1;
}
else if (x == 1) {
y = 2;
}
So I guess the other example worked since y was initialized in both the if and the else part, which means the y would always be initialized no matter what the condition given is. Now I really get it. Thank you!!
local variables don't have a default value and you need to initialize them. You are certainly setting the value of x (x=0), but the compiler doesn't check if the loop body will be actually entered. So give y a value of 0.
This is wrong:
public class Foo {
public static void main (String[] args) {
int = x, y; // Wrong
for (x = 0 ; x < 10 ; x++) {
y = x + 1;
}
System.out.println(x);
System.out.println(y);
}
}
This is correct:
public class Foo {
public static void main (String[] args) {
int x, y; // Declaration only: x and y are uninitialized
This is also correct:
public class Foo {
public static void main (String[] args) {
int x = 1, y = 10; // Declaration + initialization
'Hope that helps...
If you look in your code; you have initialized x to 0 in for loop and then incrementing it with x++. But you are initializing Y inside loop which may or may not execute at runtime (nothing to do with compile time). In Java, you have to initialize local variable before using it and if you are not doing so compiler will prompt the error. And that is why x is printed and not Y
It cannot be determined until run-time whether the for loop will be run even once. Therefore that initialization doesn't count (i.e., the compiler cannot depend on it, so it errors).
It cannot be determined until run-time which of the two -- the if or the else clause -- will fire. However, at compile-time we know that one OR the other will fire, so if you initialize in both, the compilation error goes away.
x gets initialized in the for loop (first argument)
Consider what could happen if your loop was
for (x = 0 ; x < someOtherVariable ; x++) {
and the value of someOtherVariable happened to be zero. The loop wouldn't run at all, so y would never be initialized.
Of course, the loop you wrote will always run ten times since the lower and upper bound are both hard-coded, but the compiler (apparently) doesn't analyze the code enough to prove that. It sees a variable that's only initialized within a loop, and following the general rule that a loop might not run at all, it complains that the variable might not be initialized.
BTW, int = x, y; isn't valid Java syntax: the equals sign doesn't belong there.
It depends on two things:
executes no matter what (here "x")
executes based on some constraint (here "y")
The compiler repels ambiguity in any case so it throws error on witnessing a variable initialization in a block which is conditionally executed.

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