Bitshifting a long in Java

问题

Im sure this is an easy one for whoever sees it first!

Why in Java does code like

   long one = 1 << 0;
   long thirty = 1 << 30;
   long thirtyOne = 1 << 31;
   long thirtyTwo = 1 << 32;

   System.out.println(one+" = "+Long.toBinaryString(1 << 0));
   System.out.println(thirty+" = "+Long.toBinaryString(1 << 30));
   System.out.println(thirtyOne+" = "+Long.toBinaryString(1 << 31));
   System.out.println(thirtyTwo+" = "+Long.toBinaryString(1 << 32));

1 = 1
1073741824 = 1000000000000000000000000000000
-2147483648 = 1111111111111111111111111111111110000000000000000000000000000000
1 = 1

This does not make sense to me. long is a 64 bit number - whereas it seems to act like an int in the above. I know bitshifted bytes undergo int promotion but I dont see whats going on in this case.

Any pointers on whats going on here would be good :)

Thx

EDIT: thanks for all the answers - i realised what was going on as soon as I clicked 'submit' but SO went into readonly mode and I couldnt delete! Many thanks!

回答1:

It's because 1 is an int literal, so << is applied to an integer. The result is cast to a long, but by then it's too late.

If you write 1L << 32, etc., then all will be well. L is used to denote a long literal.

回答2:

First of all:

Java Language Specification paragraph 15.19

Shift Operators

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 & (§15.22.1) with the mask value 0x1f (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive.

If the promoted type of the left-hand operand is long, then only the six 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 & (§15.22.1) with the mask value 0x3f (0b111111). The shift distance actually used is therefore always in the range 0 to 63, inclusive.

So, why does it work as if you are shifting ints instead of longs? Because you are shifting ints! Try using a long literal instead:

long thirtyTwo = 1L << 32;

回答3:

Well you're using int literals, which behave like ints. Use a long literal (i.e. 1L) instead:

System.out.println(1 << 31);
System.out.println(1L << 31);  // <--

-2147483648
2147483648

When you have an expression like 1 << 31, it evaluates to a fixed value irrespective of what it's being assigned to, be it int or long.

回答4:

It acts like an int because you are shifting ints ! To shift a long, you must suffix all your number with l.

long one = 1l << 0;
long thirty = 1l << 30;
long thirtyOne = 1l << 31;
long thirtyTwo = 1l << 32;

System.out.println(one+" = "+Long.toBinaryString(1l << 0));
System.out.println(thirty+" = "+Long.toBinaryString(1l << 30));
System.out.println(thirtyOne+" = "+Long.toBinaryString(1l << 31));
System.out.println(thirtyTwo+" = "+Long.toBinaryString(1l << 32));

Which gives the following result :

1 = 1
1073741824 = 1000000000000000000000000000000
2147483648 = 10000000000000000000000000000000
4294967296 = 100000000000000000000000000000000

回答5:

This is because the number 1 in the line long thirtyTwo = 1 << 32; is an int by default. You need to explicit mark it as a long before you do the shift operation. What happens now is that you do a shift on an int which then after the shift is done is casted from an int to a long.

   long one = 1l << 0;
   long thirty = 1l << 30;
   long thirtyOne = 1l << 31;
   long thirtyTwo = 1l << 32;

回答6:

The problem is that 1 is an integer. It will be converted to a long after the shift. Try

   long one = 1l << 0;
   long thirty = 1l << 30;
   long thirtyOne = 1l << 31;
   long thirtyTwo = 1l << 32;

回答7:

literal 1 is an int. Use 1L (L as long) and retry.

来源：https://stackoverflow.com/questions/22330107/bitshifting-a-long-in-java

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java

bit-shift