Re: syntax error in std::vector<T>::const_iterator where T = (class) U<V>

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Wed, 9 Feb 2011 02:41:55 -0800 (PST)
Message-ID:
<8d90824d-8c7d-4eff-a8a0-04c2e914a2e0@y31g2000prd.googlegroups.com>
On Feb 8, 12:38 pm, "Paul" <pchris...@yahoo.co.uk> wrote:

"James Kanze" <james.ka...@gmail.com> wrote in message

news:751342be-2e1a-4f71-bc0a-275dcfabe41e@o7g2000prn.googlegroups.com...


    [...]

struct A
{
    // 'f' is a function
    static void f() { std::cout << "A\n"; }
};

struct B
{
    // 'f' is a type
    struct f
    {
        f() { std::cout << "B\n"; }
    };
};

int main()
{
    foo<A>();
    foo<B>(); // Fails because B::f is a type}
//---------------------------------------------------------------


That's what the standard requires.

I don't see why the compiler couldn't defer the decision in this
case; I suppose that the only reason it doesn't is to be
orthogonal with cases where it must know.


This doesn't happen on my compiler. MS 32-bit optimising compiler version
14.00......


Not all compilers are up to date concerning this. A lot
(included VC++ through 2005, at least) continue to defer all
template parsing until instantiation, at which point in time,
the meaning of f is unambiguously known. Implementing the
latest rules for name lookup in templates (aka breaking existing
code) isn't always the highest priority of compiler writers.

--
James Kanze

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