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From: Keith Thompson <Keith.S.Thompson+u@gmail.com>
Newsgroups: comp.lang.c
Subject: Re: What is your opinion about unsigned int u = -2 ?
Date: Fri, 02 Aug 2024 18:31:25 -0700
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Thiago Adams <thiago.adams@gmail.com> writes:
[...]
> It is interesting to compare constexpr with the existing constant
> expression in C that works with integers.Compilers extend to work with
> unsigned long long.
> constexpr works with the sizes as defined , for instance char.

I'm not sure what you mean by "Compilers extend to work with
unsigned long long.".

Preprocessing expressions (used in #if conditions) are evaluated in
intmax_t or uintmax_t, but that's not the case for constant expressions
in general.  For example, in :

    static const signed char c = -2;

the initializer is required to be a constant expression.  -2 is of type
int, which is implicitly converted to char.  There are no constants of
types narrower than int, but you can write:

    static const signed char c = (signed char)-2;

where `(signed char)-2` is a constant expression of type signed char.

-- 
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
void Void(void) { Void(); } /* The recursive call of the void */