sizeof() confusion

[Dan Merget]

In comp.lang.c, dhesi%cirrusl at oliveb.ATC.olivetti.com (Rahul Dhesi) writes:

> >>        printf("sizeof c       = %d\n", sizeof c);
> >>        printf("sizeof 'c'     = %d\n", sizeof 'c');

> >Were you suprised that "'c'" was 4?
> >You shouldn't be.  That expression evaluates to an int, not a char.
                      ^^^^
> If characters are promoted to ints in expressions, then why isn't
> |sizeof c| equivalent to |sizeof (int) c|?  The confusion arises
> because the term "expression" is defined differently in the definition
> of C and in colloquoal conversation.
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The confusion has nothing to do with the definition of "expression", and
nothing is getting promoted.  The poster did not say "All chars are promoted
to ints in expressions"; he said "That expression ["'c'"] evaluates to an int".

Remember, a "character" in c is simply the integer which corresponds to the
ASCII (or EBCDIC, etc) representation of that character.  When storing an array
of characters, you should use the 8-bit "char" type.  However, when you are
dealing with an individual character, you will get better milage out of the
"int" type.  In recognition of this fact, A CHARACTER BETWEEN TWO SINGLE
QUOTES IS AN **INT**!  EVEN THOUGH THEY CALL IT A "CHARACTER CONSTANT", IT IS
**NOT** AN 8-BIT CHAR!  (Exception: some compiler optimizations)
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    Dan Merget
    danm at hpnmdla.HP.COM