[M3devel] Arithmetic Shift

[dirk muysers]

Sure, but then, since the amount of shift is a variable, it will oblige one
to index into a table of such constants, which is rather clumsy, and
possibly (?) poorly optimised by the compiler.

PROCEDURE ReadInt (VAR x: INTEGER) =
  VAR ch: CHAR;  n, y := 0;
  BEGIN
    Read (ch);
    WHILE ch >= 128 DO
      INC (y, Word.LeftShift (ORD (ch) - 128, n));
      INC (n, 7);
      Read (ch)
    END;
    x := arithmetic-shift (Word.LeftShift (ORD (ch), 25), n - 25) + y
  END ReadInt;

From: Antony Hosking 
Sent: Saturday, July 04, 2015 12:11 PM
To: dirk muysers 
Cc: m3devel at elegosoft.com 
Subject: Re: [M3devel] Arithmetic Shift

In this code you use DIV by a constant power of two.  That will turn into an arithmetic right shift. 

  On Jul 4, 2015, at 7:48 PM, dirk muysers <dmuysers at hotmail.com> wrote:

  Why is there no arithmetic right shift operation in Modula-3 ?

  Oberon software has a clever way to minimally encode
  signed integers independently of their size.

  PROCEDURE WriteInt (i: INTEGER) =
    BEGIN
      WHILE x < –64 OR x > 63 DO
        Write (VAL (i MOD 128 + 128, CHAR));
        i := i DIV 128;
      END;
      Write (VAL (i MOD 128), CHAR));
    END Write;

  Unfortunately it is impossible to implement the corresponding
  decoder in M3, because it requires an arithmetic right shift
  where the sign bit is replicated into the left end.
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