Re: Java vs Ada 95 (Was Re: Once again, Ada absent from DoD SBIR solicitation)

[dewar@merv.cs.nyu.edu (Robert Dewar)]

Robert Duff said

"How about portability of arithmetic?  In Java, int is always exactly 32
bits 2's complement.  In Ada, Integer is whatever the machine supports.
And if I say "type T is range 1..100;" I might get 32-bit arithmetic, or
I might get 8-bit arithmetic, or who-knows-what.  In Java, 64-bit
integers are supported, and no more.  In Ada, 64-bit integers are
supported by GNAT, but not by other compilers.  A compiler could support
more, or less, according to its whim.  Is "type T is range 1..10**10;"
legal?  It is on *some* Ada compilers, but not others.

On the other hand, at least Ada notifies you of overflows -- in Java, it
just silently gets the wrong answer.  (This is the C and C++ culture --
Java inherits much more than just syntax from that culture.)"


That seems a bit confused. You get portable arithmetic in Ada by specifying
the range. That gives you completely portable arithmetic, certainly up to
32 bits in practice. If you say type T is range 1 .. 100, then it does not
matter if you get 8 bit or 32 bit arithmetic if you avoid overflowing this
range. As you note, in Java overflows are ignor3ed in any case. So to get
EXACTLY the equivalent of the Java int semantics, we do

  type int is range - 2 ** 31 .. + 2 ** 31 - 1;

sure, a compiler can give you more than 32 bits, but providing you avoid
overflow that is irrelevant, and if you have overflow, then Java arithemtic
would also be undefined.