Compile time type conversion check (constexpr and user defined literals)
c++, c++11, constexpr
Solution
Self answer: I found a complete solution, inspired by the comments and other answers to my question, and other questions such as https://stackoverflow.com/a/13384317/1149664.
The solution is to use the template form of user defined literals and sum up the number manually, multiplying the sum based on the already parsed digits by 10.
I wrote a detailed version of this self-answer here: http://scrupulousabstractions.tumblr.com/post/38460349771/c-11-type-safe-use-of-integer-user-defined-literals
template<char... Chars>
int operator"" _steps(){
return {litparser<0,Chars...>::value};
}
Litparser is a little template meta-program which takes a list of characters as arguments expanded from the input characters held by the Chars parameter pack.
typedef unsigned long long ULL;
// Delcare the litparser
template<ULL Sum, char... Chars> struct litparser;
// Specialize on the case where there's at least one character left:
template<ULL Sum, char Head, char... Rest>
struct litparser<Sum, Head, Rest...> {
// parse a digit. recurse with new sum and ramaining digits
static const ULL value = litparser<
(Head <'0' || Head >'9') ? throw std::exception() :
Sum*10 + Head-'0' , Rest...>::value;
};
// When 'Rest' finally is empty, we reach this terminating case
template<ULL Sum> struct litparser<Sum> {
static const ULL value = Sum;
};
Problem
Update: I posted my own answer below And there's a longer version of this matter here: http://scrupulousabstractions.tumblr.com/post/38460349771/c-11-type-safe-use-of-integer-user-defined-literals Question: I've made a simple `constexpr` user defined literal `_X` that gets the value as an unsigned long long (that's how numeral user defined literals work: http://en.cppreference.com/w/cpp/language/user_literal), and then I make sure that the value fits in a signed long long. It all works well (too big values cause compilation error), but only when I explicitly create a variable such as ``` constexpr auto a= 150_X; ``` If instead I write something typical like ``` cout << 150_X << endl;; ``` the tests are not performed at compile time. Are constexpr functions only executed at compile time if they are assigned to a constexpr variable? (I could not find that in the standard) Is it possible to achieve the safe behaviour of `_X` that I'm looking for? Full example: ``` #include<iostream> #include<stdexcept> inline constexpr long long testConv(unsigned long long v) { return (v > 100 ) ? throw std::exception() : v; } // will eventually use actual limit from numeric_limits inline constexpr long long operator "" _X(unsigned long long f) { return testConv(f) ; } int main(){ constexpr auto a= 5_X; std::cout << a << std::endl; std::cout << 200_X << std::endl; // This bad literal is accepted at compile time constexpr auto c=250_X; // This bad literal is not accepted at compile time std::cout << c << std::endl; } ``` oh, for reference: I used gcc4.7.2.