Is using a locally defined type in a template allowed?

Question

While answering a different question, I could not find a reason for why the following code would be disallowed.

template <typename F> void bar (F *f) { f->a = 0; }

int main () {
    struct Foo { int a; } f = { 3 };
    bar(&f);                         // fail
}

The explanation I provided was that the compiler would have no way to legally express the template expansion, but that is more my intuition than a reason. Is there a definitive explanation as to why the code should not compile?

@DavidRodríguez-dribeas: The problem was I was looking at the C++11 draft standard for a reason this was disallowed, and couldn't find it. However, I was using the compiler in it's default mode, not C++11 mode. The code does compile when I pass the C++11 switch to the compiler. — jxh, Sep 19 '12 at 22:24

score 6 · Accepted Answer · edited May 23 '17 at 12:12

6

That is explicitly disallowed in C++03, but legal code in C++11.

Not sure if this is an exact duplicate of Using local classes with STL algorithms, as the standard has changed since I asked that question.

edited May 23 '17 at 12:12

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answered Sep 19 '12 at 21:54

David Rodríguez - dribeas

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Is using a locally defined type in a template allowed?

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