Adding the line:

CFLAGS= -dP -dA -S

to the makefile and interpreting gccsign as gccsign.s (assembly output from the
compiler), one can see what is going on (I did not know that gcc uses LISP
internally for optimization -- hehe that makes sense). Of course gcc optimized
the integer operations away and used direct constants in the printf arguments
... but by using variables a=-1 b=1 and then in printf a >> b, one can see that
the compiler uses the assembly instruction sarl to shift arithmetically right. 

Wonderful stuff when shifting a binary value that may be represented in an
integer register right, so one can mask out a lsb bit using & 1. Of course one
would never see it if the counter would count bits independently from the
register contents. On the other hand, if waiting for a zeroed register to exit
the loop ... Truly a millenium (21st) bug. Thanks for pointing this out, it was
worth investigating.

Peter


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