Quiz Time

[Eranga]

Shall we refresh this?

Is it able to compile the following code, and if so what is the output and why?

		int i = 10;
		int j = 10;
		
		boolean b = false;
		
		if(b = i == j) {
			System.out.println("True");
		}
		else {
			System.out.println("False");
		}

[pbrockway2]

The Java compiler sometimes disallows things we might be tempted to write in C, and thereby delivers us from the evil thereof. It's not always possible though.

[vinci123]

answer is "yes"

[Steve11235]

I agree. == has higher precedence as is executed before =.

[Eranga]

Accepted.

Anyone else?

[Tolls]

It's no different to the classic loop for reading files:

String line;
while ((line = reader.readLine()) != null) {
}

[Vinod Mukundan]

It will compile and print "True" as the "==" will be performed first and then "=" giving a true for the condition .... as said its same as any of the checks we do in a for,while or if ..the operation done and the check executed.

[JosAH]

I dusted off this thread; as you all probably know the Math.random() method returns a number in the range [0, 1) (the range includes 0), so for two numbers x and y in that range, the result x^y (^ denotes the power operator here) is in the range [0, 1] (including both 0 and 1); the number 1 is only reached iff y == 0. An observation: x^y > x for all values of x and y in that particular first range. Have a look at the following code snippet:

public class T {

	public static void main(String[] args) {

		int count= 1;
		for (double x= Math.random(); x < 1; count++, x= Math.pow(x, Math.random()))
			System.out.println(count+": "+x);
		System.out.println(count);
		
	}
}

This little method calculates x^y0^y1^y2 ... until the result equals 1; theoreticaly that number is never reached if all y values are unequal to zero. Run the snippet and see for yourself that this program always terminates within 50 iterations or so for al values of y not equal to zero. Why?

kind regards,

Jos

[DarrylBurke]

You already said why.

An observation: x^y > x for all values of x and y

The rest can be blamed on ncg_goldberg

db

[JosAH]

You already said why.

The rest can be blamed on ncg_goldberg

True, but why isn't the limiting value 1-ulp(1)? The result should be less than 1 ... (unless y equals zero).

kind regards,

Jos

[DarrylBurke]

why isn't the limiting value 1-ulp(1)?

That's more CS than I know about, but the documentation for the Math class does mention that

a larger error bound of 1 or 2 ulps is allowed for certain methods.

db

[DarrylBurke]

Not important as far as the result is concerned, but your code doesn't show the value of y, only of x^y

db

[JosAH]

Not important as far as the result is concerned, but your code doesn't show the value of y, only of x^y

But it is important for the final result; you're right about not dispaying the value of y; here's the modified code:

public class T {

	public static void main(String[] args) {

		int count= 1;
		for (double x= Math.random(), y; x < 1; count++, x= Math.pow(x, y)) {
			y= Math.random();
			System.out.println(count+": "+x+" "+y);
		}
		System.out.println(count);
		
	}
}

kind regards,

Jos