string literal parameter not accepted to a constexpr function

Question

Call to the extract function below does not work for me on g++ 4.9.0 (20130421). The error I get is that s1 is not a constant expression. If i can be initialized as constexpr then j and k should too. Is that wrong?

#include <tuple>

template <unsigned N1, unsigned N2> 
constexpr bool strmatch(const char (&s1)[N1], const char (&s2)[N2], unsigned i = 0)
{
  return (s1[i]==s2[i]) ? 
            (s1[i]=='\0') ? 
               true
               : strmatch(s1, s2, i+1) 
            : false;
}

template<unsigned N>
constexpr int extract(const std::tuple<int, int> & t1, const char (&array)[N]) {
  return std::get<strmatch(array, "m0")>(t1);
}

int main(void)
{
  constexpr int i = strmatch("m0", "m0");  // OK
  constexpr int j = extract(std::make_tuple(10, 20), "m0"); 
  constexpr int k = extract(std::make_tuple(10, 20), "m1");

  return 0;
}

I'm 75% sure this is a compiler bug. Function invocation substitution ought to clean up any reference binding issues. — aschepler
– aschepler, Commented May 11, 2013 at 0:15

Richard Smith · Accepted Answer · 2013-05-14 05:00:14Z

1

Your code is ill-formed. The problem is that array is not a core constant expression, so can't be used in the template argument in the call to std::get:

template<unsigned N>
constexpr int extract(const std::tuple<int, int> & t1, const char (&array)[N]) {
  return std::get<strmatch(array, "m0")>(t1);
}

Remember that constexpr functions can be called at runtime: this code would use the value of a runtime parameter to this function (array) during translation (in the evaluation of the call to strmatch).

answered May 14, 2013 at 5:00

Richard Smith

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4 Comments

Sumant Over a year ago

That's pretty lame! I can assign the result of strmatch to a constexpr integer (i) and use it as a template argument for std::get. I should be able to do just that without having to use a named variable. Isn't that the idea behind function invocation substitution?

Richard Smith Over a year ago

Function invocation substitution is just a formalism to describe how the function call is evaluated. To reiterate, your function extract<3> can be called at runtime, which would in general require template instantiation to happen at runtime if this were allowed.

Sumant Over a year ago

Right. But as long as all the invocations (in a translation unit) of extract<N> can be evaluated at compile-time, the code should work, right?

Richard Smith Over a year ago

No, the function must still work at runtime even if you never happen to call it that way. constexpr is not like template instantiation -- it does not let you generate code (which would be necessary if you could use parameters to constexpr functions in template arguments).

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