| 1 | // This is a silly example, but it demonstrates that the type of a lazy default
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| 2 | // probably should not be evaluated when the parameter is given explicitly
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| 3 | // (since the type of the default is never needed in that case).
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| 4 |
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| 5 | #include <boost/parameter.hpp>
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| 6 |
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| 7 | BOOST_PARAMETER_NAME(param);
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| 8 |
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| 9 | template <typename T> struct empty {};
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| 10 |
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| 11 | struct my_default_computer {
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| 12 | #if 0
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| 13 | // What I need to do now (broken):
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| 14 | typedef empty<int>::type result_type;
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| 15 | #else
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| 16 | // What I want to do:
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| 17 | template <typename T>
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| 18 | struct result {typedef typename empty<T>::type type;};
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| 19 | #endif
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| 20 | };
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| 21 |
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| 22 | int main(int, char**) {
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| 23 | (_param = 0)[_param || my_default_computer()];
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| 24 | return 0;
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| 25 | }
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