CODEWITHSUNDEEP
c++macrosconcatenationc-preprocessor
Ilya Muradyan
Ilya Muradyan

Reputation: 442

Concatenation macro in C++

I have the following definition in my C++ file:

#define SIZE 32

And I want to generate following two lines:

typedef uint32_t bui
typedef uint64_t lui

The first line can be generated by:

#define PASTER(x,y) x ## y ## _t
#define EVALUATOR(x,y)  PASTER(x,y)
#define NAME(fun, size) EVALUATOR(fun, size)

typedef NAME(uint,SIZE) bui

But I cant't generate the second line with

typedef NAME(uint,SIZE*2) lui

Is it possible to do it without defining #define DOUBLE_SIZE 64, using only SIZE macro?

Upvotes: 0

Views: 169

Answers (1)

Richard Hodges
Richard Hodges

Reputation: 69922

Prefer templates over macros where possible (almost always):

#include <cstdlib>
#include <cstdint>

struct Signed {};
struct Unsigned{};


namespace detail {
    template<class SignedNess, std::size_t Bits> struct make_int ;
    template<> struct make_int<Signed, 64> { using type = std::int64_t; };
    template<> struct make_int<Signed, 32> { using type = std::int32_t; };
    template<> struct make_int<Signed, 16> { using type = std::int16_t; };
    template<> struct make_int<Signed, 8> { using type = std::int8_t; };
    template<> struct make_int<Unsigned, 64> { using type = std::uint64_t; };
    template<> struct make_int<Unsigned, 32> { using type = std::uint32_t; };
    template<> struct make_int<Unsigned, 16> { using type = std::uint16_t; };
    template<> struct make_int<Unsigned, 8> { using type = std::uint8_t; };
}

template<class Signedness, std::size_t Bits> using make_int = typename detail::make_int<Signedness, Bits>::type;

int main()
{
    make_int<Signed, 16> x = 5;
}

Upvotes: 1

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