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From: brianjparker (brianjparker_at_[hidden])
Date: 2002-02-21 16:48:25

--- In boost_at_y..., "Peter Dimov" <pdimov_at_m...> wrote:
> From: "Fernando Cacciola" <fcacciola_at_g...>
> > I don't think that the very small increased complexity of
requiring auto
> > declarations to have a type name are not worth the increased
> > Don't you think it does improves safety a lot?
> No, actually, I don't. Partly because the right time to use auto
> declarations is exactly when the actual type doesn't matter, and
> because I don't see how requiring a dummy identifier would improve
> somehow.

Having the deduced type available is essential for functionality
reasons as well as type safety improvements.

template<typename T1, typename T2>
void func(T1 t1, T2 t2)
     auto L temp = t1 + t2;
     // ...
     vector<L> v;
     // creating this vector cannot be done by typeless auto

(in my preferred syntax, this would be-
template<typename T1, typename T2, typename L>
void func(T1 t1, T2 t2)
     L temp = t1 + t2;
     // ...
     vector<L> v;

A real-world example that I have posted earlier (using the template
syntax above) is-

template<typename R, typename T1, typename T2, typename L>
complex<R> operator*(const complex<T1>& lhs, imaginary<T2> rhs)
L l1 = -( * rhs); // L is deduced here
L l2 = * rhs;

complex<L> ret; = l1; = l2;
return ret;

How would this be implemented using typeless auto alone (without
using a full typeof)?

As far as improved type safety is concerned, a (contrived) example
would be-

template<typename T1, typename T2, typename L>
void func2(T1 t1, T2 t2)
    complex<L> z = func3(t1, t2);
    // compile-time type check that func3 indeed returns a complex
    // ...

Actually, this example shows the limitation of even the type-
annotated "auto" proposal, and is a manifestation of the fact that
the various "auto" proposals are attempting to duplicate the type
deduction mechanisms already existing for template function arguments.

,Brian Parker

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