🚀 go-pugleaf

In article <4142@ism780c.UUCP> marv@ism780c.UUCP (Marv Rubinstein) writes:
>   unsigned char uc;
>	    char c;
>
>     uc > -1  /* evaluates to 0  (for all uc) because -1 is converted to
>		 unsigned */
>
>      c > -1  /* evaluates to 1 (for all c) because c is converted to
>		 an *integer* whose range is 0..255 and the comparison is
>		 a signed compare */
>
>Now the question, is this C?

Yes, it is correct.  The results are a consequence of the rule that after
automatic widening of types has occurred, if either operand is unsigned,
then the expression is evaluated with unsigned operands.  When I was first
learning C, I made the very silly mistake of writing

    char ch;
    while ((ch = getchar) != EOF)

which would work on some compilers and not on others.  Naturally, the diffi-
culty was that in some cases 'char' by default was unsigned, and so the
comparison was done using unsigned arithmetic.

---
Kenneth R. Ballou		...!ucbvax!cartan!brahms!ballou
Dept. of Mathematics
University of California
Berkeley, California  94720