Re: Elements of constant array as template arguments.

From:
"Victor Bazarov" <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Tue, 2 May 2006 17:45:42 -0400
Message-ID:
<e38ju8$gkg$1@news.datemas.de>
Andrzej 'Foxy' D. wrote:

I cannot figure out, why is the following code incorrect:

const int array[] = { 1 };
const int var = 1;

template<int I>
struct Blah {};

int main() {
       Blah<var> b1;
       Blah<array[0]> b2; // ERROR!
}

I know that template arguments can only be constant expressions. But
I don't understand why isn't 'array[0]' treated as a constant
expression.


Indexing is a run-time thing. 'array' is converted to a pointer and
then indexing is performed on that pointer. It does not qualify as
a compile-time constant expression which is defined in subclause 5.19.
Pointers just cannot be part of a constant expression except as
arguments of 'sizeof'.

It seems to me that if an array is const and the index is
a constant expression (e.g. 0), then the whole expression is
constant. I can't imagine, how a compiler can NOT know, what's the
value of array[0] during compile-time.


The langauge just doesn't specify it that way. As to why it is so,
ask in 'comp.std.c++', they could explain the rationale behind
different parts of the Standard.

V
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