c++ - Calling template function with conditionals -

given function type template :

template<typename typea, typename typeb, typename typec>  void foo (typea a, typeb b, typec c) { ...; }  

then hope call function approach shown follows:

int main (void) {     int ta = 32;      int tb = 64;     int tc = 32;      float *array_a;      double *array_b;      float *array_c;       foo<(ta == 32 ? float : double), (tb == 32 ? float : double), (tc == 32 ? float : double)>(array_a, array_b, array_c);       return 0; } 

of course, code results in compile error...

however, wonder whether there convenient way check ta's, tb's, , tc's value , call function foo accordingly...

first of all, choosing type use instantiate template based on value of variable , conditional operator syntactically wrong. language doesn't allow type chosen method.

second, can let compiler deduce type. can use:

foo(array_a, array_b, array_c); 

the compiler deduce typea float*, typeb double*, , typec float*.

using

foo<float, double, float>(array_a, array_b, array_c); 

is not correct since types used instantiate template don't match argument types.

third, if want able derive type based on value, value has const or constexpr. can use:

template <int n> struct type_chooser { using type = double; }; template <> struct type_chooser<32> { using type = float; };  int main () {     const int ta = 32;      const int tb = 64;     const int tc = 32;      using typea = typename type_chooser<ta>::type;     using typeb = typename type_chooser<tb>::type;     using typec = typename type_chooser<tc>::type;      foo<typea, typeb, typec>(10, 20, 30);       return 0; }