CODEWITHSUNDEEP
javajava-8bit-shift
dragonfly
dragonfly

Reputation: 17773

Shift operators - operands must be convertible to an integer primitive?

I'm preparing myself to a Java exam, and I'm reading "OCA Java SE 8 Programmer Study Guide (Exam 1Z0-808)". In operators section I found this sentence:

Shift Operators: A shift operator takes two operands whose type must be convertible to an integer primitive.

I felt odd to me so I tested it with long:

public class HelloWorld{

     public static void main(String []args){
         long test = 3147483647L;
         System.out.println(test << 1);

     }
}

and it worked, no compiler errors and result is correct. Does the book has a bug or am I misunderstanding the quote from the book?

Upvotes: 10

Views: 351

Answers (2)

Tunaki
Tunaki

Reputation: 137114

The shift operators >> and << are defined in JLS section 15.19. Quoting:

Unary numeric promotion (§5.6.1) is performed on each operand separately. (Binary numeric promotion (§5.6.2) is not performed on the operands.)

It is a compile-time error if the type of each of the operands of a shift operator, after unary numeric promotion, is not a primitive integral type.

When talking about "integer primitive", the book is really talking about "primitive integral type" (defined in JLS section 4.2.1):

The values of the integral types are integers in the following ranges:

  • For byte, from -128 to 127, inclusive
  • For short, from -32768 to 32767, inclusive
  • For int, from -2147483648 to 2147483647, inclusive
  • For long, from -9223372036854775808 to 9223372036854775807, inclusive
  • For char, from '\u0000' to '\uffff' inclusive, that is, from 0 to 65535

Upvotes: 11

Kayaman
Kayaman

Reputation: 73558

They're using integer not in the Java int fashion, but rather as "integer type instead of floating point or other type". Java's long is an integer too, it's just a 64-bit wide integer.

Upvotes: 4

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