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问题:

I understand what the unsigned right shift operator ">>>" in Java does, but why do we need it, and why do we not need a corresponding unsigned left shift operator?

回答1:

The >>> operator lets you treat int and long as 32- and 64-bit unsigned integral types, which are missing from the Java language.

This is useful when you shift something that does not represent a numeric value. For example, you could represent a black and white bit map image using 32-bit ints, where each int encodes 32 pixels on the screen. If you need to scroll the image to the right, you would prefer the bits on the left of an int to become zeros, so that you could easily put the bits from the adjacent ints:

 int shiftBy = 3;
 int[] imageRow = ...
 int shiftCarry = 0;
 // The last shiftBy bits are set to 1, the remaining ones are zero
 int mask = (1 << shiftBy)-1;
 for (int i = 0 ; i != imageRow.length ; i++) {
     // Cut out the shiftBits bits on the right
     int nextCarry = imageRow & mask;
     // Do the shift, and move in the carry into the freed upper bits
     imageRow[i] = (imageRow[i] >>> shiftBy) | (carry << (32-shiftBy));
     // Prepare the carry for the next iteration of the loop
     carry = nextCarry;
 }

The code above does not pay attention to the content of the upper three bits, because >>> operator makes them

There is no corresponding << operator because left-shift operations on signed and unsigned data types are identical.

回答2:

>>> is also the safe and efficient way of finding the rounded mean of two (large) integers:

int mid = (low + high) >>> 1;

If integers high and low are close to the the largest machine integer, the above will be correct but

int mid = (low + high) / 2;

can get a wrong result because of overflow.

Here's an example use, fixing a bug in a naive binary search.

回答3:

Basically this has to do with sign (numberic shifts) or unsigned shifts (normally pixel related stuff).

Since the left shift, doesn't deal with the sign bit anyhow, it's the same thing (<<< and <<)...

Either way I have yet to meet anyone that needed to use the >>>, but I'm sure they are out there doing amazing things.

As you have just seen, the >> operator automatically fills the high-order bit with its previous contents each time a shift occurs. This preserves the sign of the value. However, sometimes this is undesirable. For example, if you are shifting something that does not represent a numeric value, you may not want sign extension to take place. This situation is common when you are working with pixel-based values and graphics. In these cases you will generally want to shift a zero into the high-order bit no matter what its initial value was. This is known as an unsigned shift. To accomplish this, you will use java’s unsigned, shift-right operator,>>>, which always shifts zeros into the high-order bit.

回答4:

The signed right-shift operator is useful if one has an int that represents a number and one wishes to divide it by a power of two, rounding toward negative infinity. This can be nice when doing things like scaling coordinates for display; not only is it faster than division, but coordinates which differ by the scale factor before scaling will differ by one pixel afterward. If instead of using shifting one uses division, that won't work. When scaling by a factor of two, for example, -1 and +1 differ by two, and should thus differ by one afterward, but -1/2=0 and 1/2=0. If instead one uses signed right-shift, things work out nicely: -1>>1=-1 and 1>>1=0, properly yielding values one pixel apart.

The unsigned operator is useful either in cases where either the input is expected to have exactly one bit set and one will want the result to do so as well, or in cases where one will be using a loop to output all the bits in a word and wants it to terminate cleanly. For example:

void processBitsLsbFirst(int n, BitProcessor whatever)
{
  while(n != 0)
  {
    whatever.processBit(n & 1);
    n >>>= 1;
  }
}

If the code were to use a signed right-shift operation and were passed a negative value, it would output 1's indefinitely. With the unsigned-right-shift operator, however, the most significant bit ends up being interpreted just like any other.

The unsigned right-shift operator may also be useful when a computation would, arithmetically, yield a positive number between 0 and 4,294,967,295 and one wishes to divide that number by a power of two. For example, when computing the sum of two int values which are known to be positive, one may use (n1+n2)>>>1 without having to promote the operands to long. Also, if one wishes to divide a positive int value by something like pi without using floating-point math, one may compute ((value*5468522205L) >>> 34) [(1L<<34)/pi is 5468522204.61, which rounded up yields 5468522205]. For dividends over 1686629712, the computation of value*5468522205L would yield a "negative" value, but since the arithmetically-correct value is known to be positive, using the unsigned right-shift would allow the correct positive number to be used.

回答5:

A normal right shift >> of a negative number will keep it negative. I.e. the sign bit will be retained.

An unsigned right shift >>> will shift the sign bit too, replacing it with a zero bit.

There is no need to have the equivalent left shift because there is only one sign bit and it is the leftmost bit so it only interferes when shifting right.

Essentially, the difference is that one preserves the sign bit, the other shifts in zeros to replace the sign bit.

For positive numbers they act identically.

For an example of using both >> and >>> see BigInteger shiftRight.

回答6:

In the Java domain most typical applications the way to avoid overflows is to use casting or Big Integer, such as int to long in the previous examples.

int hiint = 2147483647;
System.out.println("mean hiint+hiint/2 = " + ( (((long)hiint+(long)hiint)))/2);
System.out.println("mean hiint*2/2 = " + ( (((long)hiint*(long)2)))/2);

BigInteger bhiint = BigInteger.valueOf(2147483647);
System.out.println("mean bhiint+bhiint/2 = " + (bhiint.add(bhiint).divide(BigInteger.valueOf(2))));