JAVA - Bit Shifts in Java: Right, Left & Right Zeros with int Variable of Value 256

Question

Hello everyone ;) I learn java on my own from the book Practical JAVA course. My problem is that I stopped on a task which reads Declare an int variable of any name. Assign it the value 256. Do three things: bit shift right two places, bit shift left two places, and bit shift to the right two...

Anonymous · Answer

You write Integer.toBinaryString (a). Here is an example to play with: Code: JavaLog in, to see the code Effect of action: Code: text Expand Select all Copy to clipboard 25511 = 11111  255>>>3 = 31  11111111>>>11 = 11111 Start by understanding offsets without attribution, and you'll understand what's going on. The difference between >> and >> = is that in the latter case, you have an offset with assignment. I recommend it by the way Thinking in Java .

novix8 · Answer

Thanks for the hint

But to be honest, I've been working on it for a while and I'm still not doing it; /

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Yes, I changed the program, but I think something is wrong: / I can't use If's yet, because I haven't figured it out yet. I have to show it the easiest way.

Anonymous · Answer

eh ... Yes, I thought that when I put a training program, you will not understand what's going on.

You probably don't know printf either, but you won't need that for now.

Read yourself what is bit shift .

The entry "Integer.toBinaryString (a)" was only meant to help you convert an integer into binary. My program was supposed to show you why you can't shift 256 256 positions right or left. Well, it's hard ...

Your program after corrections with an offset of 2.

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result:

 Pierwsze przesunięcie będzie wyglądało następująco: 11111111>>2 = 111111
 Kolejne przesunięcie będzie wyglądało tak: 111111>2 = 111111

In Java, we write the names of classes with capital letters, and names of functions / methods in small letters, similarly, we write names of variables with lowercase letters.

novix8 · Answer

Thanks for your patience And I have another question Why was the second variable (b) needed, and why a = 255 and not 256 and b = 2?

Anonymous · Answer

First, let's write the positive part of integers. In Java, int can have values from - (2 ^ 31-1) to (2 ^ 31-1), that is: from -2147483647 to 2147483647. I will not explain how to write negative numbers because it misses the point at this stage . So let's write the positive part of integers: Code: text Expand Select all Copy to clipboard 111 1111 1111 1111 1111 1111 1111 1111 -> (2^31-1) = 2147483647 (integer - część dodatnia) So let's assume that 0 looks like this: Code: text Expand Select all Copy to clipboard 000 0000 0000 0000 0000 0000 0000 0000 -> 0 Bit shift for 255: Code: text Expand Select all Copy to clipboard 000 0000 0000 0000 0000 0000 1111 1111 -> 255 000 0000 0000 0000 0000 0011 1111 1100 -> 255>2 = 63 Bit shift for 256: Code: text Expand Select all Copy to clipboard 000 0000 0000 0000 0000 0001 0000 0000 -> 256 000 0000 0000 0000 0000 0100 0000 0000 -> 256>2 = 64 Offset in your case: Code: text Expand Select all Copy to clipboard 000 0000 0000 0000 0000 0000 1111 1111 -> 255 000 0000 0000 0000 0000 0000 0011 1111 -> 255>>2 = 63 000 0000 0000 0000 0000 0000 1111 1100 -> (255>>2) 256>>2 = 64 000 0000 0000 0000 0000 0001 0000 0000 -> (256>>2) 256>8 = 1 However, if you would take a different number, e.g. 255, then the range of shifts that do not change the value of the number will be smaller. Unless you use longs and then you can move left to the left.

memlon · Answer

This course helped me a lot in learning this language Link .

JAVA - Bit Shifts in Java: Right, Left & Right Zeros with int Variable of Value 256

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