Have you met the Perl documentation yet? perlop is where you need to look for information on operators.

Shift Operators (within perlop):

Binary "<<" returns the value of its left argument shifted left by the number of bits specified by the right argument. Arguments should be integers. (See also Integer Arithmetic.)

Binary ">>" returns the value of its left argument shifted right by the number of bits specified by the right argument. Arguments should be integers. (See also Integer Arithmetic.)

Note that both "<<" and ">>" in Perl are implemented directly using "<<" and ">>" in C. If use integer (see Integer Arithmetic) is in force then signed C integers are used, else unsigned C integers are used. Either way, the implementation isn't going to generate results larger than the size of the integer type Perl was built with (32 bits or 64 bits).

The result of overflowing the range of the integers is undefined because it is undefined also in C. In other words, using 32-bit integers, 1 << 32 is undefined. Shifting by a negative number of bits is also undefined.

See also: use integer;. The integer pragma can be lexically scoped, so you're not committed to integer-only math throughout the script.


Dave

In reply to Re: How to do 'unsigned shift right' in perl? by davido
in thread How to do 'unsigned shift right' in perl? by mattdeans

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