Aldy Hernandez <aldyh@redhat.com> writes:
Hi folks!
Calling print_hex() on a widest_int with the most significant bit turned
on can lead to a leading zero being printed (0x0ffff....). This produces
confusing dumps to say the least, especially when you incorrectly assume
an integer is NOT signed :).
That's the intended behaviour though. wide_int-based types only use as
many HWIs as they need to store their current value, with any other bits
in the value being a sign extension of the top bit. So if the most
significant HWI in a widest_int is zero, that HWI is there to say that
the previous HWI should be zero- rather than sign-extended.
So:
0x0ffffffff -> (1 << 32) - 1 to infinite precision
(i.e. a positive value)
0xffffffff -> -1
Thanks,
Richard