getc() != EOF

[Morris Keesan]

----------------------------

> > >	  while ( (c = getc()) != EOF )
> > 
> > It will work if "c" is declared "int."
> > It will not work if "c" is declared "char."
> > 
> >                                  Ed Nather
> 
> WRONG! The code above will work if c is int or char.
> Char variables are promoted to int in expressions (see C manual)
> and a char -1 is IDENTICAL with an int -1. Unsigned char c could be
> different (Any C implementors there? (kvm?)). 
> 
> Chris Maltby
> University of Sydney

  1) When saying things like "See C manual", it would sure help if people would
     give references -- preferably section numbers or page numbers.
  2) The reference missing above is section 6.6, "Arithmetic conversions", on
     page 184 of Kernighan and Ritchie:

	A GREAT MANY operators cause conversions . . . called the "usual
	arithmetic conversions."

	First, any operands of type char . . . are converted to int.

    (Emphasis mine -- there are some expressions where the usual arithmetic
    conversions don't apply; above, they apply to !=, but NOT to = ).
  3) From section 6.1 of the C manual (page 183 of K&R): 

	Whether or not sign-extension occurs for characters is machine
	dependent . . .  Of the machines treated by this manual, only the
	PDP-11 sign-extends.

     On many machines, char and unsigned char are equivalent.  On these
     machines, (char)-1 and (int)-1 are very different.  C provides no
     way to specify 'signed char' (a shortcoming of the language).
  4) Even on machines that sign-extend characters, the above code is incorrect
     if c is declared "char", because it will halt not only on EOF, but also on
     (char)-1, which is a valid char.
-- 
					Morris M. Keesan
					{decvax,linus,wjh12,ima}!bbncca!keesan
					keesan @ BBN-UNIX.ARPA