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Re: why not to define std::common_type<int32_t, uint32_t>::type as int64_t?

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On 2013-05-29 09:45, Walter Mascarenhas wrote:
>    Probably one reason would be that we would
> need to do it by hand (i.e., it could not
> be inferred from the the ternary operator)
>
>    However, defining std::common_type<int32_t, uint32_t>::type as
> int64_t seems to be more consistent for me.
>
>     Am I missing something?

std::common_type was intentionally defined in terms of a rather simple
rule based on operator?:. This rule *is* consistent, because it reflects
how the core language works. If you don't like it, just define your own
form of common_type that matches better your intentions. You are neither
required to use std::common_type nor do you need to use it directly. For
example std::duration_cast uses a more complex logic to deduce a
reasonable guess for an effective common type:

"Let CF be ratio_divide<Period, typename ToDuration::period>, and CR be
common_type<typename ToDuration::rep, Rep, intmax_t>::type"

You could (for example) provide your own extended common_type_ext trait
that is built on top of std::common_type (or not at all) and that
attempts to impose further constraints such as attempting to return a
type that always has a larger value representation compared to the
operands or that attempts to impose such rules at least for
unsigned/signed type combinations.

HTH & Greetings from Bremen,

Daniel Krügler


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why not to define std::common_type<int32_t, uint32_t>::type as int64_t? Walter Mascarenhas <walter.mascarenhas@googlemail.com> - 2013-05-29 01:45 -0600
  Re: why not to define std::common_type<int32_t, uint32_t>::type as int64_t? Daniel Krügler <daniel.kruegler@googlemail.com> - 2013-05-29 10:43 -0600
  Re: why not to define std::common_type<int32_t, uint32_t>::type as  int64_t? James Kuyper <jameskuyper@verizon.net> - 2013-05-30 13:57 -0600

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