使用模板的 C++ 惰性求值
C++ lazy evaluation with templates
我写了一个c++程序来实现惰性二进制操作。
该程序应该按如下方式工作：创建一个带运算的二叉树，每次我们在二叉运算中使用一棵树和其他东西时，我们把两个东西都作为左和右 children 的新的操作。当我调用 operator T 时，它需要每个 child 的另一个 operator T 递归，以防它也是一个运算符，或者将它传递给与模板参数关联的函数（F).
代码
#include <iostream>
#include <cassert>
#include <vector>
static const auto &add = [](auto a, auto b) {return a + b;};
static const auto &sub = [](auto a, auto b) {return a - b;};
static const auto &mul = [](auto a, auto b) {return a * b;};
static const auto &frac = [](auto a, auto b) {return a / b;};
template <class A, class B, class F> class BinaryOP;
template <class A, class B> using Addition = BinaryOP <A, B, decltype(add)>;
template <class A, class B> using Subtraction = BinaryOP <A, B, decltype(sub)>;
template <class A, class B> using Multiplication = BinaryOP <A, B, decltype(mul)>;
template <class A, class B> using Division = BinaryOP <A, B, decltype(frac)>;
static const auto make_binop = [](auto a, auto b, auto &F) { return BinaryOP<decltype(a), decltype(b), decltype(F)>(a, b, F); };
static const auto make_add = [](auto a, auto b) { return make_binop(a, b, add); };
static const auto make_sub = [](auto a, auto b) { return make_binop(a, b, sub); };
static const auto make_mul = [](auto a, auto b) { return make_binop(a, b, mul); };
static const auto make_div = [](auto a, auto b) { return make_binop(a, b, frac); };
template <class T>
struct is_recursable {
private:
typedef char yes;
typedef short no;
template <typename C> static yes test(decltype(C::RECURSE_ME)*);
template <typename C> static no test(...);
public:
enum { value = sizeof(test<T>(0)) == sizeof(yes) };
};
template <class A, class B, class F>
class BinaryOP {
protected:
typedef BinaryOP <A, B, F> THIS_T;
const A lhs;
const B rhs;
F func;
public:
static const bool RECURSE_ME;
constexpr BinaryOP(const A &lhs, const B &rhs, F &func):
lhs(lhs), rhs(rhs), func(func)
{}
virtual ~BinaryOP()
{}
template <class T>
constexpr operator T() const {
auto a = is_recursable<A>::value ? T(lhs) : lhs;
auto b = is_recursable<B>::value ? T(rhs) : rhs;
return func(a, b);
}
template <class R>
constexpr Addition <THIS_T, R> operator+(const R &b) {
return make_add(*this, b);
}
template <class R>
constexpr Subtraction <THIS_T, R> operator-(const R &b) {
return make_sub(*this, b);
}
template <class R>
constexpr Multiplication <THIS_T, R> operator*(const R &b) {
return make_mul(*this, b);
}
template <class R>
constexpr Division <THIS_T, R> operator/(const R &b) {
return make_div(*this, b);
}
friend std::ostream &operator<<(std::ostream &os, const THIS_T &tree) {
os << "{" << tree.lhs << ":" << is_recursable<A>::value << ", " << tree.rhs << ":" << is_recursable<B>::value << "}";
return os;
}
void printrec() const {
printf("1: {");
if(is_recursable<A>::value == 1)
/* printf("1 %s", typeid(B).name()); */
lhs.printrec();
else
printf("0");
printf(", ");
if(is_recursable<B>::value == 1)
/* printf("1 %s", typeid(B).name()); */
rhs.printrec();
else
printf("0");
printf("}");
}
};
class Matrix {
typedef std::vector <float> vec_t;
typedef std::vector <std::vector <float> > mat_t;
mat_t mat;
public:
Matrix(mat_t mat):
mat(mat)
{}
Matrix(size_t h, size_t w):
mat(mat_t(h, vec_t(w)))
{
for(size_t i = 0; i < h; ++i)
for(size_t j = 0; j < w; ++j)
mat[i][j] = (i == j);
}
size_t height() const { return mat.size(); }
size_t width() const { return (height() == 0) ? 0 : mat.front().size(); }
Matrix operator*(const Matrix &other) {
Matrix res(height(), other.width());
assert(width() == other.height());
for(size_t i = 0; i < height(); ++i) {
for(size_t j = 0; j < other.width(); ++j) {
float sum = 0.;
for(size_t k = 0; k < width(); ++k) {
sum += mat[i][k] * other.mat[k][j];
}
res.mat[i][j] = sum;
}
}
return res;
}
Matrix operator*(const float &scalar) {
Matrix res(height(), width());
for(size_t i = 0; i < height(); ++i) for(size_t j = 0; j < width(); ++j) res.mat[i][j] = mat[i][j] * scalar;
return res;
}
friend std::ostream &operator<<(std::ostream &os, const Matrix &m) {
for(size_t i = 0; i < m.height(); ++i) {
for(size_t j = 0; j < m.width(); ++j) {
os << m.mat[i][j] << " ";
}
os << std::endl;
}
return os;
}
};
int main() {
auto tree = make_add(2, 3) + make_sub(5, 6) * make_add(3, 0) / 3; // = 4
std::cout << tree << std::endl;
std::cout << int(tree) << std::endl;
Matrix m(10, 10), n(10, 10);
auto mattree = make_mul(m, n) + 6;
std::cout << mattree << std::endl;
std::cout << "operator is recursable: " << is_recursable<Addition<float, Matrix> >::value << std::endl;
std::cout << "matrix is recursable: " << is_recursable<Matrix>::value << std::endl;
std::cout << "float is recursable: " << is_recursable<float>::value << std::endl;
std::cout << "mattree is recursable: " << is_recursable<decltype(mattree)>::value << std::endl;
mattree.printrec();
/* std::cout << (mattree.operator Matrix()) << std::endl; */
}
正在编译
c++ -flto -O2 -std=c++1y binary_operator.cpp
铿锵
binary_operator.cpp:93:10: error: member reference base type 'const int' is not a structure or union
rhs.printrec();
~~~^~~~~~~~~
binary_operator.cpp:158:11: note: in instantiation of member function 'BinaryOP<BinaryOP<Matrix, Matrix, const <lambda at binary_operator.cpp:7:26> &>, int, const <lambda at binary_operator.cpp:5:26> &>::printrec' requested here
mattree.printrec();
^
binary_operator.cpp:87:11: error: no member named 'printrec' in 'Matrix'
lhs.printrec();
~~~ ^
binary_operator.cpp:87:11: note: in instantiation of member function 'BinaryOP<Matrix, Matrix, const <lambda at binary_operator.cpp:7:26> &>::printrec' requested here
lhs.printrec();
^
binary_operator.cpp:158:11: note: in instantiation of member function 'BinaryOP<BinaryOP<Matrix, Matrix, const <lambda at binary_operator.cpp:7:26> &>, int, const <lambda at binary_operator.cpp:5:26> &>::printrec' requested here
mattree.printrec();
^
binary_operator.cpp:93:11: error: no member named 'printrec' in 'Matrix'
rhs.printrec();
~~~ ^
3 errors generated.
make: *** [binary_operator] Error 1
gcc
binary_operator.cpp: In instantiation of 'void BinaryOP<A, B, F>::printrec() const [with A = BinaryOP<Matrix, Matrix, const<lambda(auto:5, auto:6)>&>; B = int; F = const<lambda(auto:1, auto:2)>&]':
binary_operator.cpp:158:20: required from here
binary_operator.cpp:93:11: error: request for member 'printrec' in '((const BinaryOP<BinaryOP<Matrix, Matrix, const<lambda(auto:5, auto:6)>&>, int, const<lambda(auto:1, auto:2)>&>*)this)->BinaryOP<BinaryOP<Matrix, Matrix, const<lambda(auto:5, auto:6)>&>, int, const<lambda(auto:1, auto:2)>&>::rhs', which is of non-class type 'const int'
rhs.printrec();
~~~~^~~~~~~~
binary_operator.cpp: In instantiation of 'void BinaryOP<A, B, F>::printrec() const [with A = Matrix; B = Matrix; F = const<lambda(auto:5, auto:6)>&]':
binary_operator.cpp:87:7: required from 'void BinaryOP<A, B, F>::printrec() const [with A = BinaryOP<Matrix, Matrix, const<lambda(auto:5, auto:6)>&>; B = int; F = const<lambda(auto:1, auto:2)>&]'
binary_operator.cpp:158:20: required from here
binary_operator.cpp:87:11: error: 'const class Matrix' has no member named 'printrec'
lhs.printrec();
~~~~^~~~~~~~
binary_operator.cpp:93:11: error: 'const class Matrix' has no member named 'printrec'
rhs.printrec();
~~~~^~~~~~~~
shell returned 1
但是如果我通过更改注释来更改 printrec（参见代码），我将得到：
{{2:0, 3:0}:1, {{{5:0, 6:0}:1, {3:0, 0:0}:1}:1, 3:0}:1}
4
{{1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
:0, 1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
:0}:1, 6:0}
operator is recursable: 1
matrix is recursable: 0
float is recursable: 0
mattree is recursable: 1
1: {1 i, 0}
显然，当我尝试递归地评估（或打印）树时，它无法编译并出现错误，因为它无法在叶子中找到相同的方法（例如 printrec）(non-operator) class，但它似乎（参见 is_recursive）知道 compile-time 中有哪些类型。
if(is_recursable<A>::value == 1)
/* printf("1 %s", typeid(B).name()); */
lhs.printrec();
else
printf("0");
在 constexpr if 存在之前，您无法执行此操作。原因是编译器必须假设两个分支都是可能的，即使一个或另一个不可到达——是的，似乎它应该能够分辨，但语言不允许这样做。您需要使用标签调度或其他一些编译时构造。
见this answer
我写了一个c++程序来实现惰性二进制操作。
该程序应该按如下方式工作：创建一个带运算的二叉树，每次我们在二叉运算中使用一棵树和其他东西时，我们把两个东西都作为左和右 children 的新的操作。当我调用 operator T 时，它需要每个 child 的另一个 operator T 递归，以防它也是一个运算符，或者将它传递给与模板参数关联的函数（F).
代码
#include <iostream>
#include <cassert>
#include <vector>
static const auto &add = [](auto a, auto b) {return a + b;};
static const auto &sub = [](auto a, auto b) {return a - b;};
static const auto &mul = [](auto a, auto b) {return a * b;};
static const auto &frac = [](auto a, auto b) {return a / b;};
template <class A, class B, class F> class BinaryOP;
template <class A, class B> using Addition = BinaryOP <A, B, decltype(add)>;
template <class A, class B> using Subtraction = BinaryOP <A, B, decltype(sub)>;
template <class A, class B> using Multiplication = BinaryOP <A, B, decltype(mul)>;
template <class A, class B> using Division = BinaryOP <A, B, decltype(frac)>;
static const auto make_binop = [](auto a, auto b, auto &F) { return BinaryOP<decltype(a), decltype(b), decltype(F)>(a, b, F); };
static const auto make_add = [](auto a, auto b) { return make_binop(a, b, add); };
static const auto make_sub = [](auto a, auto b) { return make_binop(a, b, sub); };
static const auto make_mul = [](auto a, auto b) { return make_binop(a, b, mul); };
static const auto make_div = [](auto a, auto b) { return make_binop(a, b, frac); };
template <class T>
struct is_recursable {
private:
typedef char yes;
typedef short no;
template <typename C> static yes test(decltype(C::RECURSE_ME)*);
template <typename C> static no test(...);
public:
enum { value = sizeof(test<T>(0)) == sizeof(yes) };
};
template <class A, class B, class F>
class BinaryOP {
protected:
typedef BinaryOP <A, B, F> THIS_T;
const A lhs;
const B rhs;
F func;
public:
static const bool RECURSE_ME;
constexpr BinaryOP(const A &lhs, const B &rhs, F &func):
lhs(lhs), rhs(rhs), func(func)
{}
virtual ~BinaryOP()
{}
template <class T>
constexpr operator T() const {
auto a = is_recursable<A>::value ? T(lhs) : lhs;
auto b = is_recursable<B>::value ? T(rhs) : rhs;
return func(a, b);
}
template <class R>
constexpr Addition <THIS_T, R> operator+(const R &b) {
return make_add(*this, b);
}
template <class R>
constexpr Subtraction <THIS_T, R> operator-(const R &b) {
return make_sub(*this, b);
}
template <class R>
constexpr Multiplication <THIS_T, R> operator*(const R &b) {
return make_mul(*this, b);
}
template <class R>
constexpr Division <THIS_T, R> operator/(const R &b) {
return make_div(*this, b);
}
friend std::ostream &operator<<(std::ostream &os, const THIS_T &tree) {
os << "{" << tree.lhs << ":" << is_recursable<A>::value << ", " << tree.rhs << ":" << is_recursable<B>::value << "}";
return os;
}
void printrec() const {
printf("1: {");
if(is_recursable<A>::value == 1)
/* printf("1 %s", typeid(B).name()); */
lhs.printrec();
else
printf("0");
printf(", ");
if(is_recursable<B>::value == 1)
/* printf("1 %s", typeid(B).name()); */
rhs.printrec();
else
printf("0");
printf("}");
}
};
class Matrix {
typedef std::vector <float> vec_t;
typedef std::vector <std::vector <float> > mat_t;
mat_t mat;
public:
Matrix(mat_t mat):
mat(mat)
{}
Matrix(size_t h, size_t w):
mat(mat_t(h, vec_t(w)))
{
for(size_t i = 0; i < h; ++i)
for(size_t j = 0; j < w; ++j)
mat[i][j] = (i == j);
}
size_t height() const { return mat.size(); }
size_t width() const { return (height() == 0) ? 0 : mat.front().size(); }
Matrix operator*(const Matrix &other) {
Matrix res(height(), other.width());
assert(width() == other.height());
for(size_t i = 0; i < height(); ++i) {
for(size_t j = 0; j < other.width(); ++j) {
float sum = 0.;
for(size_t k = 0; k < width(); ++k) {
sum += mat[i][k] * other.mat[k][j];
}
res.mat[i][j] = sum;
}
}
return res;
}
Matrix operator*(const float &scalar) {
Matrix res(height(), width());
for(size_t i = 0; i < height(); ++i) for(size_t j = 0; j < width(); ++j) res.mat[i][j] = mat[i][j] * scalar;
return res;
}
friend std::ostream &operator<<(std::ostream &os, const Matrix &m) {
for(size_t i = 0; i < m.height(); ++i) {
for(size_t j = 0; j < m.width(); ++j) {
os << m.mat[i][j] << " ";
}
os << std::endl;
}
return os;
}
};
int main() {
auto tree = make_add(2, 3) + make_sub(5, 6) * make_add(3, 0) / 3; // = 4
std::cout << tree << std::endl;
std::cout << int(tree) << std::endl;
Matrix m(10, 10), n(10, 10);
auto mattree = make_mul(m, n) + 6;
std::cout << mattree << std::endl;
std::cout << "operator is recursable: " << is_recursable<Addition<float, Matrix> >::value << std::endl;
std::cout << "matrix is recursable: " << is_recursable<Matrix>::value << std::endl;
std::cout << "float is recursable: " << is_recursable<float>::value << std::endl;
std::cout << "mattree is recursable: " << is_recursable<decltype(mattree)>::value << std::endl;
mattree.printrec();
/* std::cout << (mattree.operator Matrix()) << std::endl; */
}
正在编译
c++ -flto -O2 -std=c++1y binary_operator.cpp
铿锵
binary_operator.cpp:93:10: error: member reference base type 'const int' is not a structure or union
rhs.printrec();
~~~^~~~~~~~~
binary_operator.cpp:158:11: note: in instantiation of member function 'BinaryOP<BinaryOP<Matrix, Matrix, const <lambda at binary_operator.cpp:7:26> &>, int, const <lambda at binary_operator.cpp:5:26> &>::printrec' requested here
mattree.printrec();
^
binary_operator.cpp:87:11: error: no member named 'printrec' in 'Matrix'
lhs.printrec();
~~~ ^
binary_operator.cpp:87:11: note: in instantiation of member function 'BinaryOP<Matrix, Matrix, const <lambda at binary_operator.cpp:7:26> &>::printrec' requested here
lhs.printrec();
^
binary_operator.cpp:158:11: note: in instantiation of member function 'BinaryOP<BinaryOP<Matrix, Matrix, const <lambda at binary_operator.cpp:7:26> &>, int, const <lambda at binary_operator.cpp:5:26> &>::printrec' requested here
mattree.printrec();
^
binary_operator.cpp:93:11: error: no member named 'printrec' in 'Matrix'
rhs.printrec();
~~~ ^
3 errors generated.
make: *** [binary_operator] Error 1
gcc
binary_operator.cpp: In instantiation of 'void BinaryOP<A, B, F>::printrec() const [with A = BinaryOP<Matrix, Matrix, const<lambda(auto:5, auto:6)>&>; B = int; F = const<lambda(auto:1, auto:2)>&]':
binary_operator.cpp:158:20: required from here
binary_operator.cpp:93:11: error: request for member 'printrec' in '((const BinaryOP<BinaryOP<Matrix, Matrix, const<lambda(auto:5, auto:6)>&>, int, const<lambda(auto:1, auto:2)>&>*)this)->BinaryOP<BinaryOP<Matrix, Matrix, const<lambda(auto:5, auto:6)>&>, int, const<lambda(auto:1, auto:2)>&>::rhs', which is of non-class type 'const int'
rhs.printrec();
~~~~^~~~~~~~
binary_operator.cpp: In instantiation of 'void BinaryOP<A, B, F>::printrec() const [with A = Matrix; B = Matrix; F = const<lambda(auto:5, auto:6)>&]':
binary_operator.cpp:87:7: required from 'void BinaryOP<A, B, F>::printrec() const [with A = BinaryOP<Matrix, Matrix, const<lambda(auto:5, auto:6)>&>; B = int; F = const<lambda(auto:1, auto:2)>&]'
binary_operator.cpp:158:20: required from here
binary_operator.cpp:87:11: error: 'const class Matrix' has no member named 'printrec'
lhs.printrec();
~~~~^~~~~~~~
binary_operator.cpp:93:11: error: 'const class Matrix' has no member named 'printrec'
rhs.printrec();
~~~~^~~~~~~~
shell returned 1
但是如果我通过更改注释来更改 printrec（参见代码），我将得到：
{{2:0, 3:0}:1, {{{5:0, 6:0}:1, {3:0, 0:0}:1}:1, 3:0}:1}
4
{{1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
:0, 1 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 1
:0}:1, 6:0}
operator is recursable: 1
matrix is recursable: 0
float is recursable: 0
mattree is recursable: 1
1: {1 i, 0}
显然，当我尝试递归地评估（或打印）树时，它无法编译并出现错误，因为它无法在叶子中找到相同的方法（例如 printrec）(non-operator) class，但它似乎（参见 is_recursive）知道 compile-time 中有哪些类型。
if(is_recursable<A>::value == 1)
/* printf("1 %s", typeid(B).name()); */
lhs.printrec();
else
printf("0");
在 constexpr if 存在之前，您无法执行此操作。原因是编译器必须假设两个分支都是可能的，即使一个或另一个不可到达——是的，似乎它应该能够分辨，但语言不允许这样做。您需要使用标签调度或其他一些编译时构造。
见this answer