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