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Subject: Re: [Boost-users] sizeof of boost::int8_t
From: Robert Ramey (ramey_at_[hidden])
Date: 2009-12-16 13:12:19

John Maddock wrote:
>> running sizeof on a boost::int8_t returns 1, but what is the type?
>> int/char?
> Strictly speaking you're not allowed to ask that question... it's a
> secret :-)
> The idea is that int8_t is whatever platform-dependent type has
> 8-bits and is signed.
> But since you ask, it's "signed char" on any platform you're ever
> likely to encounter. You should never make any assumptions about the
> actual types of the int#_t typedefs though - some of them really are
> "unmentionable" platform dependent types on some platforms - but take
> a look at boost/cstdint.hpp for the details.
> HTH, John.

This is a subject I accidently revisited recently. I wanted a arithmetic
type that was only 8 bits long. boost::int8_t seemed just the ticket.
But then when one did something like

boost::int8_t i;
std::cout << i;

One usually wouldn't get what he would expect. To me this is a big "hole"
in the whole concept. What I would really like to see is:

int8_t defined with something like BOOST_SERIALIZATION_STRONGTYPEDEF

so that int8_t becomes a true type rather than just a "macro" of some other

To me - we should have a whole new set of (true) numeric types and start to
them. I sort of suspect that C99 had this intention, but by using typedef
implement them, they aren't useful for my needs. Perhaps the "mistake" is
not having typedef create a truely new type. Or maybe a new keyword

I realize that fixing this would be a herculean task at this point so this
isn't a real suggestion
 - its really just a rant. But it makes me feel better.

Robert Ramey

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