如何在 C++ 中使用表达式模板实现符号微分
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通常,您需要一种表示符号的方法(即编码 eg 的表达式模板3 * x * x + 42
),以及可以计算导数的元函数。希望您对 C++ 中的元编程足够熟悉,知道它的含义和含义,但可以给您一个想法:
// This should come from the expression templates
template<typename Lhs, typename Rhs>
struct plus_node;
// Metafunction that computes a derivative
template<typename T>
struct derivative;
// derivative<foo>::type is the result of computing the derivative of foo
// Derivative of lhs + rhs
template<typename Lhs, typename Rhs>
struct derivative<plus_node<Lhs, Rhs> > {
typedef plus_node<
typename derivative<Lhs>::type
, typename derivative<Rhs>::type
> type;
};
// and so on
然后,您将两个部分(表示和计算)捆绑起来,以便使用。例如derivative(3 * x * x + 42)(6)
,可能意味着“计算3 * x * x + 42
x 在 6 处的导数”。
但是,即使您确实知道编写表达式模板需要什么以及用 C++ 编写元程序需要什么,我也不建议您这样做。模板元编程需要大量样板文件并且可能很乏味。相反,我将您引向天才Boost.Proto库,该库专门用于帮助编写 EDSL(使用表达式模板)并在这些表达式模板上进行操作。学习使用它不一定容易,但我发现学习如何在不使用它的情况下实现相同的目标更难。这是一个实际上可以理解和计算的示例程序derivative(3 * x * x + 42)(6)
:
#include <iostream>
#include <boost/proto/proto.hpp>
using namespace boost::proto;
// Assuming derivative of one variable, the 'unknown'
struct unknown {};
// Boost.Proto calls this the expression wrapper
// elements of the EDSL will have this type
template<typename Expr>
struct expression;
// Boost.Proto calls this the domain
struct derived_domain
: domain<generator<expression>> {};
// We will use a context to evaluate expression templates
struct evaluation_context: callable_context<evaluation_context const> {
double value;
explicit evaluation_context(double value)
: value(value)
{}
typedef double result_type;
double operator()(tag::terminal, unknown) const
{ return value; }
};
// And now we can do:
// evalutation_context context(42);
// eval(expr, context);
// to evaluate an expression as though the unknown had value 42
template<typename Expr>
struct expression: extends<Expr, expression<Expr>, derived_domain> {
typedef extends<Expr, expression<Expr>, derived_domain> base_type;
expression(Expr const& expr = Expr())
: base_type(expr)
{}
typedef double result_type;
// We spare ourselves the need to write eval(expr, context)
// Instead, expr(42) is available
double operator()(double d) const
{
evaluation_context context(d);
return eval(*this, context);
}
};
// Boost.Proto calls this a transform -- we use this to operate
// on the expression templates
struct Derivative
: or_<
when<
terminal<unknown>
, boost::mpl::int_<1>()
>
, when<
terminal<_>
, boost::mpl::int_<0>()
>
, when<
plus<Derivative, Derivative>
, _make_plus(Derivative(_left), Derivative(_right))
>
, when<
multiplies<Derivative, Derivative>
, _make_plus(
_make_multiplies(Derivative(_left), _right)
, _make_multiplies(_left, Derivative(_right))
)
>
, otherwise<_>
> {};
// x is the unknown
expression<terminal<unknown>::type> const x;
// A transform works as a functor
Derivative const derivative;
int
main()
{
double d = derivative(3 * x * x + 3)(6);
std::cout << d << '\n';
}
于 2012-05-10T05:38:44.513 回答