Here’s my current favourite Java puzzler. How can you get your code to do this?
Integer b = 2;
Integer c = 1;
System.out.println("b+c : " + (b+c) ); // output: 'b+c : 4' !!There are no tricks with Sytem.out.println() i.e. you would be able to see the same value in a debugger. Clue: You need to add a few lines of code somewhere before this in your program. Scroll down for the solution.
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public static void main(String... args)throws Exception{
Integer a = 2;
Field valField = a.getClass().getDeclaredField("value");
valField.setAccessible(true);
valField.setInt(a, 3);
Integer b = 2;
Integer c = 1;
System.out.println("b+c : " + (b+c) ); // b+c : 4
}As you can see (and I would encourage you to go to the source code for Integer) there is a static cache (look for the Inner class IntegerCache) where the value of the Integer is mapped to its corresponding int value. The cache will store all numbers from -128 to 127 although you can tune this using the property java.lang.Integer.IntegerCache.high.
| Reference: | My favourite Java puzzler 2 + 1 = 4 from our JCG partner Daniel Shaya at the Rational Java blog. |
This would of been really interesting if it was primitive values that were used.
Daniel, that was quite… puzzling :)
Have you read the Java Puzzlers book? There are many brain teasers in it. For example can you make this code become an infinite loop:
while (i == i + 1) {
}
Quite puzzling, huh? :)
If you’d like the answer and a few more puzzlers, I invite you to read more here: http://methodicalprogrammer.com/blog/5-puzzles-that-prove-you-need-to-read-java-puzzlers
in my case output was 3 only. i am using jdk 904