我想写一个简单的编译时间尺寸分析库。我想创建一个编译选项来删除库中的所有内容,而无需更改代码。所以基本上我做了我自己的基本类型的版本,并且想要将它们替换为实际的基本类型(如果选择该选项的话)。用原始类型替换模板类
这个代码的最小工作示例
#include <iostream>
#include <stdint.h>
#define DEBUG
#ifdef DEBUG
template<int lenght, int time, int mass, int charge, int temperature, int amount, int intensity>
struct Dimensions {
static const int64_t LENGHT = lenght;
static const int64_t TIME = time;
static const int64_t MASS = mass;
static const int64_t CHARGE = charge;
static const int64_t TEMPERATURE = temperature;
static const int64_t AMOUNT = amount;
static const int64_t INTENSITY = intensity;
};
typedef Dimensions< 0, 0, 0, 0, 0, 0, 0 > Adimensional;
typedef Dimensions< 1, 0, 0, 0, 0, 0, 0 > Length;
typedef Dimensions< 0, 1, 0, 0, 0, 0, 0 > Time;
template<typename Dims> class Int32 {
private:
int32_t m_value;
public:
inline Int32() : m_value(0) {}
inline Int32(int32_t value) : m_value(value) {}
inline int32_t value() {
return m_value;
}
};
template<typename Dims>
Int32<Dims> inline operator+(Int32<Dims> &lhs, Int32<Dims> &rhs) {
return Int32<Dims>(lhs.value() + rhs.value());
}
struct Unmatched_dimensions_between_operands;
template<typename DimsLhs, typename DimsRhs>
Unmatched_dimensions_between_operands inline operator+(Int32<DimsLhs> &lhs, Int32<DimsRhs> &rhs);
#else
template<typename Dims> using Int32<Dims> = std::int32_t;
#endif
int main(int argc, char* argv[]) {
Int32<Time> a = 2;
Int32<Time> b = 5;
std::cout << (a + b).value() << "\n";
return 0;
}
当我删除了#define DEBUG
线我得到的编译错误
Error C2988 unrecognizable template declaration/definition 59
是否有替换的Int32
任何模板版本有道具有原始类型的代码?
“没有工作”*什么都不起作用?你得到了什么错误?* –
并且不要椭圆化整个基本展示...... –
但是......'Int32'是一个模板类型? Int32的'Dims'是什么?如果'Int32'是一个简单的(无模板)类型,那么'使用Int32 = std :: int32_t'呢? – max66