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Java << operator (In this piece of code)

So I was reading the example source code from this Android page on ViewGroups,
and I came across these lines:
// Report our final dimensions.
setMeasuredDimension(resolveSizeAndState(maxWidth, widthMeasureSpec, childState),
resolveSizeAndState(maxHeight, heightMeasureSpec,
childState << MEASURED_HEIGHT_STATE_SHIFT));
So I hope to learn:
What does the << operator exactly do in Java?
What is happening in last line of the aforementioned snippet?
Thanks.

It isn't an operand, it is an operator. A bit-wise shift operator, to be exact.
Given x and y as operands, x << y shifts the bits of the x value y bits to the left. It is basically the same as multiplying by 2 to the power y.

<< is one of bit shifting operator.
It is quite offen we use higher 16 bits of one single 32 bits int for one thing and lower 16 bits of another thing.
childState << MEASURED_HEIGHT_STATE_SHIFT means childState is passing a height (in its lower 16 bits) which is expecting to be in higher 16 bits of the int passing to resolveSizeAndState().

What does the << operator exactly do in Java?
It's moving bits to the left. As mentioned in previous answers:
1111 1110 << 2
1111 1000 // Have added 0 from right
What is happening in last line of the aforementioned snippet?
Code is changing state. It's sometimes used to represent states. e.g.
state1 = 1 << 0;
state2 = 1 << 1;
state3 = 1 << 2;
so you have 3 unique states.

"<<" is a bit operator.
I quote the explaination from tutorial for Java
The signed left shift operator "<<" shifts a bit pattern to the left, and the signed right shift operator ">>" shifts a bit pattern to the right.
And also I quote an example from here
int a = 60; /* 60 = 0011 1100 */
int c = 0;
c = a << 2; /* 240 = 1111 0000 */
so you can see a << n is almost a*(2^n)

This is Bitwise and Bit Shift Operators. More info can be found at http://docs.oracle.com/javase/tutorial/java/nutsandbolts/opsummary.html
For Ex:
1111 1110 << 2
1111 1000 // Have added 0 from right
0001 1111 >> 3
0000 0011 // Will preserve MSB and shift it right
If you don't want the first bit to be preserved, you use (in Java, Scala, C++, C afaik, and maybe more) a triple-sign-operator:
1100 1100 >>> 1
0110 0110

Shift Operator in Java [duplicate]

This question already has answers here:
Why does Java mask shift operands with 0x1F?
(2 answers)
Closed 9 years ago.
How does the shift operator << work when the value of the shift bits is greater than the total number of bits for the datatype?
For example,
int i = 2;
int j = i<<34;
System.out.println(j);
The size of an integer is 32 bits, however we are shifting 34 bits. How does this work?

I'm not sure the source on this, but according to WikiPedia (emphasis mine):
If the promoted type of the left-hand operand is int, only the five
lowest-order bits of the right-hand operand are used as the shift
distance. It is as if the right-hand operand were subjected to a
bitwise logical AND operator & with the mask value 0x1f (0b11111).[4]
The shift distance actually used is therefore always in the range 0 to
31, inclusive.
Edit: It looks like the WikiPedia entry basically lifted the information straight out of the Java Specification.

When you shift an integer with the << or >> operator and the shift distance is greater than or equal to 32, you take the shift distance mod 32 (in other words, you mask off all but the low order 5 bits of the shift distance).
This can be very counterintuitive. For example (i >> 32) == i, for every integer i. You might expect it to shift the entire number off to the right, returning 0 for positive inputs and -1 for negative inputs, but it doesn't; it simply returns i, because
(i << (32 & 0x1f)) == (i << 0) == i.
Getting back to your original problem, (i << 33) == (i << (33 & 0x1f))
== (i << 1). You can do the whole thing in binary if you like. 270 in binary is : 0000 0000 0000 0000 0000 0001 0000 1110 Shifting right by 1,
you get: 0000 0000 0000 0000 0000 0000 1000 0111 which is 135.
But a better way to do this problem in your head is to dispense with the binary entirely.
The value of i >> s is floor(i / 2<sup>s</sup>) (where s has already been masked off so it's less than 32). So, 270 << 1 = floor(270/2) = 135.
http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.19

If you try 1 << 34, you end up with 4. The runtime basically has an implicit mod NUMBER_OF_BITS on the right-hand operand of the shift. In the previous example, it's 1 << (34 % 32), which becomes 1 << 2, which is 4.

What does ">>>" in java mean?

im trying to translate this code to python, but im having a hard time doing so, don't worry about the index values and variable names, I just want to know what the ">>>" part does exactly, perhaps a python equivalent, but an explanation would be great:
target[0] = (char)(source[sourceIndex] >>> 2);
target[1] = (char)((source[sourceIndex] & 3) << 4 | source[sourceIndex + 1] >>> 4);
target[2] = (char)((source[sourceIndex + 1] & 0xf) << 2 | source[sourceIndex + 2] >>> 6);
target[3] = (char)(source[sourceIndex + 2] & 0x3f);
Any help would be appreciated

It's an "unsigned right shift".
So, if your number (x) is 11110000 (in binary).
x >>> 1 will be 01111000 (in binary).
This is opposed to x >> 1 which will result in 11111000 (in binary).
The >> tries to preserve the "sign bit" but the >>> does not.
Note: I've assumed a 8-bit integer (or a byte in Java). The same thing holds for 2-byte and 4-byte integers.

The "<<<" and ">>" are bit shift operators. Specifically,
The signed left shift operator "<<" shifts a bit pattern to the left, and the signed right shift operator ">>" shifts a bit pattern to the right. The bit pattern is given by the left-hand operand, and the number of positions to shift by the right-hand operand. The unsigned right shift operator ">>>" shifts a zero into the leftmost position, while the leftmost position after ">>" depends on sign extension.
—— from The Java™ Tutorials - Bitwise and Bit Shift Operators

Thats the unsigned right shift operator. It's a bitwise operator that shifts a zero into the leftmost bit of your operand. Here - http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html.

"<<" operator in java

Fallowing statement is from Character class of java:
(1 << Character.PARAGRAPH_SEPARATOR)) >> type
PARAGRAPH_SEPARATOR is a byte and type is an integer.
The operators in this sentence, what do they do? how and where I can use those operators?
Here is the oracles java.lang.Character doc. Nearly all the methods in the class uses those operators.

They are bit-shift operators. << shifts the bits "left" (towards the most-significant bit), and vice-versa for >>. Shifting left or right by n bits is pretty much the same as multiplying or dividing, respectively, by 2n.
See #axtavt's comment for an explanation of how these operators are being used in this context.

These are the bitwise shift operators.
If you left shift the following byte:
00000001
you would get:
00000010
I.e. the pattern has "shifted" to the left and zeros fill in on the right. So if you apply the right shift operator >> on that result, you'll get the original byte again.
You'll notice that the decimal values of these numbers are 1 and 2. If you shift left once again you'll get:
00000100 = 4
So you see that shifting to the left multiplies the number by two (given that it doesn't overflow), while right shifting divides by two. This happens very efficiently in most computers. So that's one example of how you might use these operators in a practical way.

Bitwise operators: http://download.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

<< is the left shift operator: It shifts the binary number stored in the computer left. For example, 9 in binary is 1001. 9 << 2 makes 100100 in binary (36), because it shifts it left and adds 0s at the end. 1 << n is the same thing as Math.pow(2, n) except it is way faster and better in general, as well as returning int, not double.
>> is right shift. It shifts it right, and discards empty bits. 13 is 1101 in binary, so 13 >> 1 is 110 in binary, or 6 normally.

It works as a highly optimized multiple comparison. The signed left shift operator "<<" shifts a bit pattern to the left, and the signed right shift operator ">>" shifts a bit pattern to the right. The bit pattern is given by the left-hand operand, and the number of positions to shift by the right-hand operand. The unsigned right shift operator ">>>" shifts a zero into the leftmost position, while the leftmost position after ">>" depends on sign extension.
for more detail,You can visit below link .
http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

What does >> do in Java?

Okay, I tried looking up what >>, or shift means, but it's way over my head as this site explains it: http://www.janeg.ca/scjp/oper/shift.html
What would the explanation be, if talking to a kid?

Computers are binary devices. Because of this, numbers are represented by a sequence of 1s and 0s.
Bitshifting is simply moving those sequences of 1s and 0s left or right.
So all the >> operator does is shift the bits towards the right one bit.
Consider the number 101:
// Assuming signed 8-bit integers
01100101 // How 101 is represented in binary
00110010 // After right shifting one bit, this represents 50
The least significant bit in this case was truncated. Obviously the devil's in the details, but that's all there is really to it.
The << operator does the opposite operation:
// Assuming signed 8-bit integers
01100101 // How 101 is represented in binary
11001010 // After left shifting one bit, this represents -54
// Assuming unsigned 8-bit integers
01100101 // How 101 is represented in binary
11001010 // After left shifting one bit, this represents 202
In this case, the most significant bit was truncated since I used only 8-bits. If the number had more bits, however:
// Assuming signed 16-bit integers
00000000 01100101 // How 101 is represented in binary
00000000 11001010 // After left shifting one bit, this represents 202
00000001 10010100 // After left shifting one bit again, this represents 404
So you may get different numbers depending on how many bits and the data types associated with those bits you're dealing with.
Addendum: If you're wondering how binary works, think about how the decimal number system works. Consider the number 5287. It can be written like this:
5287
But you can also write it out like this:
5287 = (5 * 1000) + (2 * 100) + (8 * 10) + (7 * 1)
Which you can then write out like this:
5287 = (5 * 10^3) + (2 * 10^2) + (8 * 10^1) + (7 * 10^0)
The above equation explains why the decimal number system is sometimes called the base-10 system. The decimal number system employs the use of 10 digits (0-9). Notice how the exponents correspond to digit position.
The binary number system, or the base-2 system, is the exact same thing but with the number two as the base of the exponents, and employing only two digits: 0 and 1.
5287 = 00010100 10100111 (base 2)
= (0 * 2^15) + (0 * 2^14) + (0 * 2^13) + (1 * 2^12)
+ (0 * 2^11) + (1 * 2^10) + (0 * 2^9) + (0 * 2^8)
+ (1 * 2^7) + (0 * 2^6) + (1 * 2^5) + (0 * 2^4)
+ (0 * 2^3) + (1 * 2^2) + (1 * 2^1) + (1 * 2^0)

Can I assume the kid I'm talking to knows a bit about binary? :)
All numbers can be represented in some kind of binary, like so:
Base 10 : Base 2
1 : 0001
2 : 0010
3 : 0011
4 : 0100
5 : 0101
6 : 0110
7 : 0111
8 : 1000
...
and so on.
The shift operators basically move all of the bits (1s or 0s) across one position. So, for example:
000111 >> 1
shifts all the bits in 000111 right by one number to produce this:
000011
000111 << 1
shifts all those bits left by one, to produce this:
001110
If you shift by more than one, then it just moves the bits further.
Now, depending on what language you're using and the kind of numbers you're working with, it can be a little bit more complicated than that. For example, if you are working in a language where the "most significant bit" (the one furthest to the left in a number) represents whether the number is signed or not, then the language will have to take that into account.
Mathematically speaking, if you take an integer (and ignore the risk of overflows, which are caused by the computer running out of space to store bits,) shift left by 1 (<< 1) is the equivalent of multiplying by 2, and shift right by 1 is the equivalent of dividing by 2. (Think a bit about what a "place value" in binary maths is worth, and that'll make sense)

>> the SHIFT RIGHT operator
Example:
class X
{
public static void main(String args[])
{
System.out.println("20>>2 = "+20>>2);
}
}
Output : 20>>2 = 5
Explanation:
Binary value of 20 is: 00000000000000000000000000010100
shift all bits 2 positions to right 00000000000000000000000000000101
It will give 5 ( 1*2^2 + 0*2^1 + 1*2^0 )

I once wrote an JApplet (bitorgel) and put it on my web page, where one can play around with bit operators. You can try it live, or download the source. AFAIK they work the same in C, C++ and Java - probably in C# too.