conditional declaration of type using decltype

Question

I thought that conditional type could be declared using decltype in template functions. But it seems not. Could anyone point out what's wrong with my test code?

#include <boost/type_index.hpp>
using boost::typeindex::type_id_with_cvr;

#define print_type(var) do { \
  std::cout << type_id_with_cvr<decltype(var)>().pretty_name() << std::endl; \
} while(0)

template <typename T1, typename T2>
auto max(T1 a, T2 b) -> decltype(a < b ? b : a) {
  decltype(a < b ? b : a) c = a < b ? b : a;
  print_type(c);
  return a < b ? b : a;
}

int main() {
  int i = 10;
  double d = 3.3;

  decltype(i < d? d : i) r = i < d? d : i;
  print_type(r); // -> double
  std::cout << r << std::endl; // 10
}

max66 · Accepted Answer · 2018-01-01 04:15:26Z

4

I suppose that your intention with

decltype( a < b ? a : b )

was to obtain the type of b when a < b and the type of a otherwise.

That is: I suppose that your intention was to obtain a type determined run time according to the runt time values of a and b.

This is impossible in C++ because the type of a variable must be decided compile time.

With that decltype() you obtain the type of the ternary operator

a < b ? a : b

that doesn't depend from the values of a and b but only from their types.

So, in the case

decltype(i < d? d : i)

where i is an int and d is a double, you obtain a double and the values of i and d are irrelevant.

edited Jan 1, 2018 at 4:15

answered Jan 1, 2018 at 3:59

max66

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6 Comments

Daniel Lee Over a year ago

For what ground c++(decltype) decide to use double rather than int?

StoryTeller - Unslander Monica Over a year ago

@DanielLee - For it being the common type between double and int.

Qaz Over a year ago

@DanielLee, double is the common type of int and double. Mainly because converting int to double commonly either preserves the value or approximates it reasonably. In fact, it's common for conversions from a 32-bit int to a double to be non-narrowing, as IEEE-754 doubles can store up to ±2⁵³ without loss of precision. On the flip side, converting from double to int misses out on many, many values. C++ simply requires the conditional expression to resolve to a compile-time-known type, so it chooses the one that doesn't miss out on tons of values from the other.

Alex Huszagh Over a year ago

std::conditional exists.... Yes, it must be decided at compile-time, but there's template magic to let you do stuff.

StoryTeller - Unslander Monica Over a year ago

@DanielLee - There's no magic. C++ is a standardized language. You may read how the common type is deduced yourself. And FYI, you can only tag one person in a comment. You tagged me first, so only I was notified.

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