C++ convert integer to string at compile time

user_213296 0
Viewed 13444

I want to do something like this:

template<int N>
char* foo() {
  // return a compile-time string containing N, equivalent to doing
  // ostringstream ostr; 
  // ostr << N;
  // return ostr.str().c_str();
}

It seems like the boost MPL library might allow this but I couldn't really figure out how to use it to accomplish this. Is this possible?

boost-mpl templates boost c++ string
6 Answers

This is now possible in C++17 and beyond without any external dependencies, with the conversion done entirely at compile-time. I've packaged the code in a short header file, check out constexpr-to-string on GitHub. The code supports different bases and character widths as well.

Examples of use:

const char *number = to_string<2147483648999954564, 16>; // produces "1DCD65003B9A1884"
puts(number);
puts(to_string<-42>); // produces "-42"
puts(to_string<30, 2>); // produces "11110"

// Below requires C++20
puts(f_to_string<3.1415926>); // Defaults to 5-point precision: "3.14159"
puts(f_to_string<{3.1415926, 7}>); // Specify precision: "3.1415926"
user_110862680

Another useful option:

template <int i, bool gTen>
struct UintToStrImpl
{
   UintToStrImpl<i / 10, (i > 99)> c;
   const char c0 = '0' + i % 10;
};

template <int i>
struct UintToStrImpl <i, false> 
{ 
   const char c0 = '0' + i; 
};

template <int i, bool sign>
struct IntToStrImpl
{
   UintToStrImpl<i, (i > 9)> num_;
};

template <int i>
struct IntToStrImpl <i, false>
{
   const char sign = '-';
   UintToStrImpl<-i, (-i > 9)> num_;
};

template <int i>
struct IntToStr
{
   IntToStrImpl<i, (i >= 0)> num_;
   const char end = '\0';
   const char* str = (char*)this;
};

std::cout << IntToStr<-15450>().str;
user_107336310
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