Many Happy Returns

[Martin Dickopp]

Correct, so far.


Why? Both operands have type int.

You are correct, of course. I have been severely confused.

Martin

 

[Rocke Robertson]

Perhaps I should have explained in more detail. The ~ operator
implements one's complement arithmetic by inverting all the bits. Thus
you get "all-bits-one" for ~0, because "all-bits-zero" is 0.
But! If the number is signed, two's complement arithmetic comes into
play *after* that. On a two's complement machine, any number whose
leftmost bit is 1 is taken as negative, and "all-bits-one" is -1 on
such a machine. Therefore, on a two's complement machine, ~0 == -1.

Boy, I'm sure glad someone understands this stuff cause this last post by Joona just completely lost
me....

 

[Christopher Benson-Manica]

Perhaps I should have explained in more detail. The ~ operator
implements one's complement arithmetic by inverting all the bits. Thus
you get "all-bits-one" for ~0, because "all-bits-zero" is 0.

Is 0 indeed equivalent to 0x00000000? (I ask because I don't know...)

 

[Andreas Kahari]

Is 0 indeed equivalent to 0x00000000? (I ask because I don't know...)

For ints, yes.

The following FAQ entry is relevant to this:
http://www.eskimo.com/~scs/C-faq/q7.31.html

 

[Hallvard B Furuseth]

Is 0 indeed equivalent to 0x00000000? (I ask because I don't know...)

Yes. Both mean 0.

BTW, in case you were thinking that 0x00000000 is a special way to say
'all bits zero': Not so. (void*)0x00000000 means the same as (void*)0,
whose representation need not be all bits zero. The _integer_ 0 does
have all bits zero, though. Possibly except padding bits.

 

[Mark Gordon]

Is 0 indeed equivalent to 0x00000000? (I ask because I don't know...)

How could it not? The base of numbers in the source code does not affect
the value so 0==0x000000000000000000000000000000 and 15==0xf etc.

Or are you asking if the representation of 0 is specified by the
standard as being all bits zero?