Silvio Mayolo Silvio Mayolo - 1 year ago 85
C++ Question

Use compile-time constant if known

I have a value of some mystery type

(for this example, we can assume
is an integral type). I would like to use this value as a template argument in some template function (for this example, the argument to
). The trouble is that
might not be a constant type. If it isn't a constant type, I would like to default to
in my template function. If it is constant, however, I would like to use the value itself, since it is known.

Currently, I have the following (non-compiling) code.

#include <type_traits>

template <typename T>
struct CompileTimeStuff {
constexpr static int value(T arg) { return 0; }

template <typename T>
struct CompileTimeStuff<const T> {
constexpr static int value(const T arg) { return (int)arg; }

template <typename T>
constexpr int magic_function(T arg) {
return CompileTimeStuff<T>::value(arg);

const int cvalue = 7;
int ivalue = 7;

// This should be 7, since cvalue is constant and thus known at compile-time.
typedef std::integral_constant<int, magic_function(cvalue)> type1;
// Since ivalue is non-constant, this should take the default value of 0.
typedef std::integral_constant<int, magic_function(ivalue)> type2;

gives the following error.

templatestuff.cpp:28:58: error: the value of ‘ivalue’ is not usable in a constant expression
typedef std::integral_constant<int, magic_function(ivalue)> type2;

The compiler doesn't like
being used as a template argument, even though I never directly use its value in the instantiation.

Answer Source

Yes, you can do this. Using Johannes Schaub's trick, we can do the following:

template<typename T> 
constexpr typename std::remove_reference<T>::type makeprval(T && t) {
  return t;

#define constexpr_or_zero(e) (noexcept(makeprval(e)) ? (e) : 0)

const int cvalue = 7;
int ivalue = 7;

using type1 = std::integral_constant<int, constexpr_or_zero(cvalue)>;
using type2 = std::integral_constant<int, constexpr_or_zero(ivalue)>;

Here's a demo.

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