[英]c++ template specialization-specific declaration
Basically, what i'm trying to do is build a Vector template, with two arguments: dimensions, and variable type. 基本上,我想做的是构建一个Vector模板,其中包含两个参数:维度和变量类型。
template <typename Type, unsigned ElementCount>
class TVector
{
private:
public:
union
{
struct
{
Type X, Y, Z, W;
};
struct
{
Type Red, Green, Blue, Alpha;
};
Type Values[ElementCount];
};
}
It all works ok, but as you may have noticed, this is only for 4-element vectors, since only the Values variable is dependent on ElementCount. 一切正常,但您可能已经注意到,这仅适用于4元素向量,因为只有Values变量依赖于ElementCount。 ElementCount=1 declares only X and Red, 2 declares Y and Green, and so on... So, what i wanted was to have the other variables being declared also depending on the ElementCount value.
ElementCount = 1仅声明X和Red,2声明Y和Green,依此类推...因此,我想要的是根据ElementCount值声明其他变量。
Is it at all possible? 有可能吗? I don't think so, but wanted to make sure, anyway.
我不这么认为,但无论如何都想确保。
I was thinking of declaring that whole union separately and passing it along as a template parameter, but that's ugly as hell.... 我当时正在考虑分别声明整个联合,并将其作为模板参数传递,但这很丑陋。
EDIT: Dammit. 编辑:该死。 Now i remembered something... What about the constructor-by-value?
现在我想起了什么...按值构造函数呢?
The argument count is dependent of the template parameter ElementCount... How to do this? 参数计数取决于模板参数ElementCount ...如何执行此操作?
You could do it with partial specialization. 您可以通过部分专业化来实现。 Something like
就像是
template<typename T, unsigned N>
struct S
{
union
{
struct
{
T X, Y, Z, W;
};
struct
{
T Red, Green, Blue, Alpha;
};
T Values[N];
};
};
template<typename T>
struct S<T, 1>
{
union
{
struct
{
T X;
};
struct
{
T Red;
};
T Values[1];
};
};
int main()
{
using S1 = S<char, 1>;
using S5 = S<char, 5>;
std::cout << "sizeof(S1) = " << sizeof(S1) << '\n';
std::cout << "sizeof(S5) = " << sizeof(S5) << '\n';
}
It should print 它应该打印
sizeof(S1) = 1 sizeof(S5) = 5
Also, doing eg 另外,例如
S1 s1;
s1.Y = 0;
should give you an error that the structure have no member Y
. 应该给您一个错误,该结构没有成员
Y
Yes it's a lot more to write (or copy-paste) but it should solve your problem. 是的,编写(或复制粘贴)内容要多得多,但它可以解决您的问题。
You don't need to pass in the whole union -- just the struct and the type of the fields. 您不需要传递整个联合-只需传递结构和字段的类型即可。 I've also added a syntactic sugar in operator-> so you don't have to reference the .Names part if you don't wish.
我还在operator->中添加了语法糖,因此,如果您不想的话,不必引用.Names部分。
#include <iostream>
struct PointStruct {
int X;
int Y;
int Z;
};
struct ColorStruct {
int Red;
int Green;
int Blue;
int Alpha;
};
template <typename Type, typename Struct>
class TVector
{
public:
const static int LEN = sizeof(Struct) / sizeof(Type);
union
{
Struct Names;
Type Values[LEN];
};
Struct* operator->() { return &Names; }
const Struct* operator->() const { return &Names; }
};
int main(int argc, const char** argv) {
TVector < int, PointStruct > points;
points.Names.X = 55;
points.Names.Y = -67;
points.Names.Z = 42;
for (int x = 0; x < points.LEN; ++x) {
std::cout << "points[" << x << "]: " << points.Values[x] << std::endl;
}
points.Values[1] = 135;
std::cout << "points->Y: " << points->Y << std::endl;
TVector < int, ColorStruct > colors;
colors.Names.Red = 1;
colors.Names.Green = 2;
colors.Names.Blue = 3;
colors.Names.Alpha = 4;
for (int x = 0; x < colors.LEN; ++x) {
std::cout << "colors[" << x << "]: " << colors.Values[x] << std::endl;
}
}
With g++ 4.3 I get 使用g ++ 4.3我得到
points[0]: 55
points[1]: -67
points[2]: 42
points->Y: 135
colors[0]: 1
colors[1]: 2
colors[2]: 3
colors[3]: 4
I would avoid specialization at this level, rather do it inside the union itself: 我会避免在此级别进行专业化,而应在工会内部进行:
template <typename T> struct X { T x; };
template <typename T> struct XY { T x, y; };
...
template <typename T, int Count>
struct CountToXYZWType {};
template <typename T> struct CountToXYZWType<T,1> { typedef X<T> type; };
template <typename T> struct CountToXYZWType<T,2> { typedef XY<T> type; };
...
template <typename T, int Count>
struct TVector {
union {
typename CountToXYZWType<Count>::type;
// similarly CountToColorType<Count> ...
T values[Count];
}
...
}; };
Ok, first of all, a big thank you for everyone, for all the ideas. 好的,首先,非常感谢大家以及所有的想法。 I finally managed to do what i wanted, using a mix of all you suggested.
最终,我综合使用了您提出的所有建议来完成自己想做的事情。
second, let me explain, then. 其次,让我解释一下。 I wanted to rewrite my math library.
我想重写我的数学库。 I basically had a class for 2d, 3d and 4d vectors each.
我基本上有一个分别用于2d,3d和4d向量的类。 It was stupid, since most of the code is the same, so i decided to try to refactor using templates.
这很愚蠢,因为大多数代码都是相同的,所以我决定尝试使用模板进行重构。
I also had a small class named RGB and another called RGBA for obvious reasons :). 由于明显的原因,我也有一个名为RGB的小类,另一个名为RGBA的小类:)。
And as such, i embarked on this adventure with templates ( i'm not a big fan, but mainly because i can't understand them. they're certainly very useful ). 因此,我开始了使用模板的冒险活动(我不是一个忠实的粉丝,但主要是因为我听不懂它们。它们肯定非常有用)。
So, after several failed tries, I changed approach. 因此,在几次尝试失败之后,我改变了方法。 The final result is a mix of all the above and the kitchen sync.
最终结果是以上所有内容和厨房同步处理的混合。 A small taste, including member functions ( i had some trouble with those ), the union with the named member variables is available here: http://goo.gl/4Zlt20
包括成员函数在内的一些小问题(我对那些函数有些麻烦),可以在此处找到带有命名成员变量的联合: http : //goo.gl/4Zlt20
Now, for one final question, if i may... 现在,对于最后一个问题,我是否可以...
It all works the way i want it, but for some reason, i can't get rid of that "this->" on the inline functions. 一切都按我想要的方式工作,但是由于某种原因,我无法消除内联函数上的“ this->”。 Can anyone explain why?
谁能解释为什么?
Thanks. 谢谢。
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