Question

一点背景：我有一些奇怪的多个嵌套循环，我将其转换为平面工作队列（基本上将单个索引循环折叠为单个多索引循环）。现在每个循环都是手工编码的。我正在尝试使用lambda表达式来处理任何边界的通用方法：

例如：

// RANGE(i,I,N) is basically a macro to generate  `int i = I; i < N; ++i `
// for (RANGE(lb, N)) {
//      for (RANGE(jb, N)) {
//          for (RANGE(kb, max(lb, jb), N)) {
//              for (RANGE(ib, jb, kb+1)) {

// is equivalent to something like (overload , to produce range)
flat<1, 3, 2, 4>((_2, _3+1), (max(_4,_3), N), N, N)

公寓的内部结构如下：

template<size_t  I1, size_t I2, ...,
         class L1_, class L2, ..._>
 boost::array<int,4> flat(L1_ L1, L2_ L2, ...){
     //boost::array<int,4> current; class or static variable

     // basically, the code below this is going to be done using recursion templates
     // but to do that I need to apply lambda expression to current array
     // to get runtime bounds

     bool advance;
     L2_ l2 = L2.bind(current); // bind current value to lambda
     {
          L1_ l1 = L1.bind(current); //bind current value to innermost lambda
          l1.next();
          advance = !(l1 < l1.upper()); // some internal logic
          if (advance) {
              l2.next();
              current[0] = l1.lower();
          }
     }
     //...,
}

我的问题是，你能给我一些想法如何编写lambda（派生自boost），它可以绑定到索引数组引用，根据lambda表达式返回上限，下限？

非常感谢

bummers，lambda只支持三个占位符。

Answer 1

嗯，这是迄今为止的原型

119 #include <boost/lambda/lambda.hpp>
120
121 namespace generator {
122
123     // there is no _1 because it's innermost
124     // and lambda only has three placeholders
125     boost::lambda::placeholder1_type _2;
126     boost::lambda::placeholder2_type _3;
127     boost::lambda::placeholder3_type _4;
128
129     template<class L, class U>
130     struct range_ {
131         typedef boost::array<int,4> index_type;
132         range_(L lower, U upper) : lower_(lower), upper_(upper) {}
133         size_t lower(const index_type &index) {
134             return lower_(index[1], index[2], index[3]);
135         }
136         size_t upper(const index_type &index) {
137             return upper_(index[1], index[2], index[3]);
138         }
139         L lower_; U upper_;
140     };
141
142     template<class L, class U>
143     range_<L,U> range(L lower, U upper) {
144         return range_<L,U>(lower, upper);
145     }
146
147     template<class R1, class R2, class R3, class R4>
148     struct for_{
149         typedef boost::array<int,4> index_type;
150         index_type index;
151         R1 r1_; R2 r2_; R3 r3_; R4 r4_;
152         for_(R1 r1, R2 r2, R3 r3, R4 r4)
153             : r1_(r1), r2_(r2), r3_(r3), r4_(r4) {}
154         index_type next() {
155             index_type next = index;
156
157             bool advance = false;
158             index[0] += 1;
159
160             advance = !(index[0] < r1_.upper(index));
161             if (advance) {
162                 index[1] += 1;
163                 index[0] = r1_.lower(index);
164             }
165
166             advance = advance && !(index[1] < r2_.upper(index));
167             if (advance) {
168                 index[2] += 1;
169                 index[1] = r2_.lower(index);
170                 index[0] = r1_.lower(index);
171             }
172
173             advance = advance && !(index[2] < r3_.upper(index));
174             if (advance) {
175                 index[3] += 1;
176                 index[2] = r3_.lower(index);
177                 index[1] = r2_.lower(index);
178                 index[0] = r1_.lower(index);
179             }
180
181             //std::cout << next << std::endl;
182             return next;
183
184         }
185     };
186
187     template<class R1, class R2, class R3, class R4>
188     for_<R1, R2, R3, R4> For(R1 r1, R2 r2, R3 r3, R4 r4) {
189         return for_<R1, R2, R3, R4>(r1, r2, r3, r4);
190     }
191
192 }

示例（可能已损坏，需要更多功能）

using namespace generator;
For(range(_2, _3), range(std::max(_3, _4), N), range(N), range(N));

自定义C ++ boost :: lambda表达式帮助

1 个答案: