i = i * f *vs* i *= f (Not a bug) (Yes, a bug!)

[Peter S. Shenkin]

Re: Bennett Todd's assertion that the problem is not a bug:

cf K&R p 41:  "In general, if... a "binary operator" has operands of different
types, the "lower" type is *promoted* to the "higher" before the
operation proceeds....

This, coupled with their assertion elsewhere that x = x op y "may be taken
as equivalent" (or words to that effect) to x op= y, would seem to indicate
that i *= f and i = i * f should evaluate the expression with i and f as
floats, then cast the result to int.  In any case, they should evaluate
the expression the same way!  If you disagree, quote me chapter and verse.

Net effect:  it's a bug!

Moral: beware implicit type conversions, since they may not do what K&R
says they do.

By the way, I'd be curious to see the results of the program on other
compilers.

{philabs,cmcl2!rocky2}!cubsvax!peters            Peter S. Shenkin 
Dept of Biol. Sci.;  Columbia Univ.;  New York, N. Y.  10027;  212-280-5517
"In accordance with the recent proclivity for clever mottos, this is mine."