Question

我有binop结构，表示相同类型的2个表达式的二进制运算。我有两种不同类型的表达式：arithmetic_expr和logical_expr定义为boost :: variant。

目标是让binop有两个字段：rhs和lhs这些表达式的具体类型。怎么做到这一点？

这是我到目前为止所提出的：

template <typename expr_type, typename tag> struct binop;

struct op_plus  {};
typedef boost::variant<
            int,
            boost::recursive_wrapper<binop<arithmetic_expr, op_plus>>  // <-- fails here 'arithmetic_expr': undeclared identifier
    > arithmetic_expr;


struct op_and {};
typedef boost::variant<
            bool,
            boost::recursive_wrapper<binop<logical_expr, op_and>>
    > logical_expr;


template <typename expr_type, typename tag>
struct binop
{
    explicit binop(const expr_type& l, const expr_type& r) : lhs(l), rhs(r) { }
    expr_type lhs, rhs;
};

用例示例如下：

(((1 + 2) /* arithmetic_expr */ + 3) /* arithmetic_expr */ AND (4 + 5) /* arithmetic_expr */) /* logical_expr */

Answer 1

这样的事情？

#include <iostream>
#include <utility>
#include <boost/variant.hpp>

// identity evaluation
int evaluate(int i) { return i; }

// introduce the concep of binary op
template<class ActualOp> struct binary_op;

using add = binary_op<std::plus<int>>;
using subtract = binary_op<std::minus<int>>;
using times = binary_op<std::multiplies<int>>;

// our variant
using expression = boost::variant
<
int,
boost::recursive_wrapper<add>,
boost::recursive_wrapper<subtract>,
boost::recursive_wrapper<times>
>;

// overload for evaluating the variant
struct compute : boost::static_visitor<int>
{
    template<class T> int operator()(const T& t) const { return evaluate(t); }
};


int evaluate(const expression& e)
{
    return boost::apply_visitor(compute(), e);
};

// evaluate the binary op
template<class ActualOp>
struct binary_op
{
    binary_op(expression l, expression r)
    : _l(std::move(l)), _r(std::move(r))
    {}

    ActualOp _op;
    expression _l, _r;
};
template<class Op>
int evaluate(const binary_op<Op>& o) {
    return o._op(evaluate(o._l), evaluate(o._r));
}

int main()
{
    using namespace std;


    expression e = add(6, times(3,subtract(7,2)));

    cout << evaluate(e) << endl;
    return 0;
}

预期产出：

提升recursive_wrapper递归

1 个答案: