是否可以在 C++ 中开发静态 for 循环?
这样的东西有可能存在吗?
Is it possible for something like this to exist?
template<int Channel>
void deduce_mask(Matrix const &src, int mask[])
{
//I hope i could become a constant and the compiler would unroll the loop at compile time
for(int i = Channel; i != -1; --i)
{
//mapper is a helper class which translate two and three dimension into one dimension index
//constexpr makes it possible to find out the index at compile time
mask[mapper(0, 1, i)] = src(row - 1, col)[i];
mask[mapper(1, 1, i)] = src(row, col)[i];
mask[mapper(2, 1, i)] = src(row + 1, col)[i];
}
}
代替
template<int Channel>
class deduceMask
{
public:
static void deduce_mask(matrix const &src, int mask[]);
};
template<int Channel>
void deduce_mask(matrix const &src, int mask[])
{
mask[mapper(0, 1, Channel)] = src(row - 1, col)[Channel];
mask[mapper(1, 1, Channel)] = src(row, col)[Channel];
mask[mapper(2, 1, Channel)] = src(row + 1, col)[Channel];
deduceMask<Channel - 1>::deduce_mask(src, mask);
}
template<>
class deduceMask<-1>
{
public:
static void deduce_mask(matrix const &src, int mask[])
{
}
};
当我希望编译器在编译时找出结果时，第二个解决方案是我能想到的唯一解决方案.我是否有一种简单的方法可以像元编程解决方案一样使i"成为常量值?对我来说，一个简单的 for 循环比元编程版本更容易使用.
The second solution is the only solution I could come up of when I want the compiler to figure out the result at compile time. Do I have an easy way to make the "i" become constant value likethe metaprogramming solution? For me, a simple for loop is much more easier to work with rather than the metaprogramming version.
推荐答案
C++ 中的模板元编程是纯函数式编程，而在纯函数式编程中，您无法使用 for 或 while 之类的循环，也无法使用根本没有任何可变数据.你所拥有的只是递归.为了更容易地使用递归，您需要稍微提高抽象级别.您拥有的递归代码很好，但迭代和工作可以分开:
Template metaprogramming in C++ is pure functional programming, and in pure functional programming you don't get to use loops like for or while and you don't get to have any mutable data at all. All you have is recursion. To make working with recursion easier, you need to rise abstraction level a bit. The recursive code that you have is fine, but the iteration and work can be split apart:
template <int First, int Last>
struct static_for
{
template <typename Fn>
void operator()(Fn const& fn) const
{
if (First < Last)
{
fn(First);
static_for<First+1, Last>()(fn);
}
}
};
template <int N>
struct static_for<N, N>
{
template <typename Fn>
void operator()(Fn const& fn) const
{ }
};
既然你有了这个元函数，你可以像这样编写你的 deduce_mask 函数:
Now that you have this meta-function, you can write your deduce_mask function like this:
template<int Channel>
void deduce_mask(Matrix const &src, int mask[])
{
static_for<0, Channel>()([&](int i)
{
mask[mapper(0, 1, i)] = src(row - 1, col)[i];
mask[mapper(1, 1, i)] = src(row, col)[i];
mask[mapper(2, 1, i)] = src(row + 1, col)[i];
});
}
带有/Ob1 命令行开关的 Visual C++ 2012 将此代码编译为:
Visual C++ 2012 with /Ob1 command line switch compiles this code into this:
push 0
call <lambda_7588286c1d4f3efe98a2e307bd757f8e>::operator() (010C1270h)
push 1
call <lambda_7588286c1d4f3efe98a2e307bd757f8e>::operator() (010C1270h)
push 2
call <lambda_7588286c1d4f3efe98a2e307bd757f8e>::operator() (010C1270h)
push 3
call <lambda_7588286c1d4f3efe98a2e307bd757f8e>::operator() (010C1270h)
push 4
call <lambda_7588286c1d4f3efe98a2e307bd757f8e>::operator() (010C1270h)
...
如果你不能使用 lambda 函数，你需要写一个函子.与 lambda 函数相比，Functor 有一个优势――您可以指定调用约定(如果您不介意这样做).如果函子的 operator() 具有 __fastcall 调用约定，那么您将在汇编代码中看到 mov edx, x 而不是 push x.
If you can't use lambda functions, you need to write a functor. Functor has one advantage over lambda function - you can specify a calling convention (if you don't mind doing that). If the operator() of the functor has __fastcall calling convention then you will see mov edx, x instead of push x in the assembler code.
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