Re: operands of membership tests

[stt@houdini.camb.inmet.com (Tucker Taft)]

Matthew Heaney (mheaney@ni.net) wrote:

: A membership never raises Constraint_Error, but evaluation of its operands
: might.  Suppose I do this

: with Ada.Text_IO; use Ada.Text_IO;
: procedure Test_Fixed is

:    type Fixed is delta 1.0 range 0.0 .. 256.0;
:    for Fixed'Small use 1.0;

: begin

:   Put (Boolean'Image (33_000.0 in Fixed));

: end;

: ...
: Let's suppose 33_000.0 isn't even in the base range of Fixed.

: Can evaluation of the literal 33_000.0 raise Constraint_Error?

Not in this case, because Fixed is a static subtype, so the membership
test is static (in Ada 95, not Ada 83).
On the other hand, if the membership test had not been static,
then 33_000.0 would be illegal at compile-time, per RM95 4.9(35).

: In RM83 11.6 (9), the note states that an implementation is allowed to
: replace the expression with a constant value (in my example, False), even
: though the operand is outside the base range of the type.  What's the story
: in Ada 95?

The compiler is required to evaluate a membership test statically in Ada 95
if the operand and the target subtype are both static (see 4.9(11).

: What's the more general case, for any relational operator?  Under what
: circumstances does evaluation of (numeric) literals raise Constraint_Error? 
: Under what circumstances can an implementation remove an expression it can
: evaluate statically, containing (static) operands that are outside the base
: range of the type?

Evaluation of numeric literals, per se, doesn't raise Constraint_Error.
However, they are illegal if outside the base range of the expected
type when used in a non-static operation.  When used in a static
expression, overflow checks are not performed.  See RM95 4.9(33-35).

: --------------------------------------------------------------------
: Matthew Heaney
: Software Development Consultant
: <mailto:matthew_heaney@acm.org>
: (818) 985-1271

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-Tucker Taft   stt@inmet.com   http://www.inmet.com/~stt/
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