🚀 go-pugleaf

In article <4e17uk$g3@kocrsv08.delcoelect.com>,
Richard F. Smiley <c2xrfs@eng.delcoelect.com> wrote:
> Is the program fragment
>    if ( Boolean_A  & =  0x0020 )
> legal?  Specifically, should the three-character sequence  "& ="  be
> treated as representing the two-character operator  "&="?  

No to both questions.

> As I understand the standard (ISO section 5.1.1.2, phase 7), the "&"
> is converted into one token, and the "=" is converted into another. 
> This then precludes treating them as one operator, and the program
> fragment is illegal.  

Correct.

> However, a compiler I respect instead treats this as a legal fragment
> involving the "&=" operator.

This is permitted, provided at least one diagnostic is produced along
the way.

> (The program fragment
> was caused by a bad macro definition.  If the compiler had complained,
> we would have found the problem earlier.)

The &= must exist before the preprocessor works - it can't be generated
by, for example:

    #define EQ =

    ... &EQ

unless you use the ## preprocessor operator, which generates new
tokens.

-- 
Clive D.W. Feather                       | If you lie to the compiler,
cdwf@cityscape.co.uk   (work, preferred) | it will get its revenge.
clive@stdc.demon.co.uk (home)            |   - Henry Spencer