Meaning of "expression"

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Sujet : Meaning of "expression"
De : Keith.S.Thompson+u (at) *nospam* gmail.com (Keith Thompson)
Groupes : comp.lang.c
Date : 08. Jun 2026, 22:05:06
Autres entêtes
Organisation : None to speak of
Message-ID : <1107aq2$3grso$2@kst.eternal-september.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
User-Agent : Gnus/5.13 (Gnus v5.13)
Keith Thompson <Keith.S.Thompson+u@gmail.com> writes:
[...]
The actual text of the standard implies that 42 is not an expression.
I rely on the obvious intent to conclude that it is.

I made the above statement to demonstrate that just following the exact
wording of the standard, without thinking about the (sometimes unclear)
intent behind it, can lead to absurd results.

I've discussed this particular glitch before, but it's been a while.

N3220 6.5.1 says:

    An *expression* is a sequence of operators and operands that
    specifies computation of a value, or that designates an object
    or a function, or that generates side effects, or that performs
    a combination thereof.

I believe the wording is unchanged from C90 up to the latest C202y
draft.  Since the word "expression" is in italics, this is the
standard's definition of the word.

This is a flawed definition.  The terms "operator" and "operand"
are defined in 6.4.6:

    *punctuator: one of
        [ ] ( )
    [snip]
   
    A punctuator is a symbol that has independent syntactic and semantic
    significance. Depending on context, it may specify an operation to
    be performed (which in turn may yield a value or a function
    designator, produce a side effect, or some combination thereof) in
    which case it is known as an *operator* (other forms of operator also
    exist in some contexts). An *operand* is an entity on which an
    operator acts.

Consider this expression statement:

    42;

Is `42` an expression?  Clearly it's intended to be, but there is no
operator, and therefore there is no operand, so it doesn't meet the
standard's definition of the word "expression".

For that matter, consider:

    (void)0;

It's "obvious" that `(void)0` is an expression.  It consists of one
operator `(void)` and one operand `0` (I'll ignore the fact that
the definition uses plurals for both), but it does not specify
computation of a value, or designate an object or a function,
or generates side effects, or perform a combination thereof.

The fact that the standard's definition of "expression" is flawed is
not much of a problem in practice.  Virtually everyone, implementers
and programmers, assumes the obvious intent.  Nobody believes that
`42` isn't an expression.  But it is my strongly held opinion that
the wording should be improved in a future edition of the standard.

I think it should say something to the effect that the meaning
of the term "expression" is defined by the grammar.  The current
wording that claims to be the definition of the term could, with
a few tweaks, still be turned into a valid normative statement
*about* expressions.

I have a similar issue with the standard's definition of "value":
"precise meaning of the contents of an object when interpreted as
having a specific type".  It's obvious that the result of evaluating
a non-void expression (such as the infamous `42`) is a "value",
but the definition implies that a "value" can only be the meaning
of the contents of an object.  Nobody is actually misled by the
current definition, but it should be improved.

--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
void Void(void) { Void(); } /* The recursive call of the void */

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