问题
Is it possible to have a generic constructor that takes any type of initializer list, even if this has nested lists within?
Say you have the following partial template specialization for a class that takes in its constructor nested initializer lists:
template class ClassA;
template <>
class ClassA<4> {
typedef std::initializer_list<double> list_type;
typedef std::initializer_list<list_type> llist_type;
typedef std::initializer_list<llist_type> lllist_type;
typedef std::initializer_list<lllist_type> initializer_type;
size_t n_[4] = {0};
double* data_;
public:
ClassA(initializer_type l) {
assert(l.size() > 0);
assert(l.begin()->size() > 0);
assert(l.begin()->begin()->size() > 0);
assert(l.begin()->begin()->begin()->size() > 0);
size_t m = n_[0] = l.size();
size_t n = n_[1] = l.begin()->size();
size_t o = n_[2] = l.begin()->begin()->size();
n_[3] = l.begin()->begin()->begin()->size();
data_ = new double[m*n*o*n_[3]];
int i=0, j=0, k=0, p=0;
for (const auto& u : l) {
assert(u.size() == n_[1]);
for (const auto& v : u) {
assert(v.size() == n_[2]);
for (const auto& x : v) {
assert(x.size() == n_[3]);
for (const auto& y : x) {
data_[i + m*j + m*n*k + m*n*o*p] = y;
++p;
}
p = 0;
++k;
}
k = 0;
++j;
}
j = 0;
++i;
}
}
size_t size() const {
size_t n = 1;
for (size_t i=0; i<4; ++i)
n *= n_[i];
return n;
}
friend std::ostream& operator<<(std::ostream& os, const ClassA& a) {
for (int i=0; i<a.size(); ++i)
os<<" "<<a.data_[i];
return os<<endl;
}
};
int main()
{
ClassA<4> TT = { {{{1.}, {7.}, {13.}, {19}}, {{2}, {8}, {14}, {20}}, {{3}, {9}, {15}, {21}}}, {{{4.}, {10}, {16}, {22}}, {{5}, {11}, {17}, {23}}, {{6}, {12}, {18}, {24}}} };
cout<<"TT -> "<<TT<<endl;
return 0;
}
This code prints:
TT -> 1 4 2 5 3 6 7 10 8 11 9 12 13 16 14 17 15 18 19 22 20 23 21 24
Now, I'm trying to generalize the constructor so that I don't have to specialize the class template for each dimension. The problem is that when I replace the constructor with something like:
template <class L>
ClassA(std::initializer_list<L> l) {
cout<<"generic list constructor"<<endl;
}
The clang compiler fails with error:
error: no matching constructor for initialization of 'ClassA<4>
Can someone point out why is this happening? The template matching is not working for initializer lists, probably because this is a new C++ feature? Thank you all...
EDIT
Thanks to the help of @JohannesSchaub-litb and @Daniel Frey, I was able to craft a very generic constructor that takes the initializer_list of any dimension. This is the resulting code:
template <int d, typename T>
class ClassA {
size_t n_[d] = {0};
T* data_;
template <int D, typename U>
struct Initializer_list {
typedef std::initializer_list<typename Initializer_list<D-1,U>::list_type > list_type;
Initializer_list(list_type l, ClassA& a, size_t s, size_t idx) {
a.n_[d-D] = l.size();
size_t j = 0;
for (const auto& r : l)
Initializer_list<D-1, U> pl(r, a, s*l.size(), idx + s*j++);
}
};
template <typename U>
struct Initializer_list<1,U> {
typedef std::initializer_list<T> list_type;
Initializer_list(list_type l, ClassA& a, size_t s, size_t i) {
a.n_[d-1] = l.size();
if (!a.data_)
a.data_ = new T[s*l.size()];
size_t j = 0;
for (const auto& r : l)
a.data_[i + s*j++] = r;
}
};
typedef typename Initializer_list<d,T>::list_type initializer_type;
public:
// initializer list constructor
ClassA(initializer_type l) : data_(nullptr) {
Initializer_list<d, T> r(l, *this, 1, 0);
}
size_t size() const {
size_t n = 1;
for (size_t i=0; i<4; ++i)
n *= n_[i];
return n;
}
friend std::ostream& operator<<(std::ostream& os, const ClassA& a) {
for (int i=0; i<a.size(); ++i)
os<<" "<<a.data_[i];
return os<<endl;
}
};
int main()
{
ClassA<4, double> TT = { {{{1.}, {7.}, {13.}, {19}}, {{2}, {8}, {14}, {20}}, {{3}, {9}, {15}, {21}}}, {{{4.}, {10}, {16}, {22}}, {{5}, {11}, {17}, {23}}, {{6}, {12}, {18}, {24}}} };
cout<<"TT -> "<<TT<<endl;
return 0;
}
Of course the code prints
TT -> 1 4 2 5 3 6 7 10 8 11 9 12 13 16 14 17 15 18 19 22 20 23 21 24
I love this template metaprogramming stuff! Thank you guys for helping figuring this out.
aa
回答1:
I believe what you really want to do is to automatically build the right type
template<int S, typename E>
class make_list_type {
public:
typedef std::initializer_list<
typename make_list_type<S-1, E>::type
> type;
};
template<typename E>
class make_list_type<0, E> {
public:
typedef E type;
};
template<int S>
class ClassA {
typedef typename make_list_type<S, double>::type initializer_type;
public:
ClassA(initializer_type l)
};
As for why your try did not work, see Templates don't always guess initializer list types
回答2:
In general, the answer is (AFAIK): No. But for your specific case, you might use the knowledge that it all ends with double as the leafs:
class arg_type
{
private:
bool is_value;
double d;
std::vector<arg_type> subs;
public:
arg_type(double v) : is_value(true), d(v) {}
arg_type(std::initializer_list<arg_type> l) : is_value(false), subs(l) {}
};
and change your ctor to:
typedef std::initializer_list<arg_type> initializer_type;
ClassA(initializer_type l) {
// ...
}
Hope it helps...
Update: As pointed out by Mankarse (thanks!), the above has undefined behaviour. Here's a version that should fix it without using Boost:
#include <vector>
#include <memory>
#include <iostream>
#include <initializer_list>
class arg_type
{
private:
std::shared_ptr<void> subs; // empty => d is valid
double d;
public:
arg_type(double v) : d(v) {}
arg_type(std::initializer_list<arg_type> l);
void print() const;
};
arg_type::arg_type(std::initializer_list<arg_type> l)
: subs(std::make_shared<std::vector<arg_type>>(l))
{}
void arg_type::print() const
{
if( subs ) {
std::cout << "( ";
for( const auto& e : *std::static_pointer_cast<std::vector<arg_type>>(subs) ) {
e.print();
}
std::cout << ") ";
}
else {
std::cout << d << " ";
}
}
struct MyClass
{
MyClass( std::initializer_list<arg_type> l) {
for( const auto& e : l ){
e.print();
}
}
};
int main()
{
MyClass m { 1, 2, { 3, 4, { 6, 7, { 8 }}}, 5 };
}
If you want to play with it, here's the live example.
来源：https://stackoverflow.com/questions/15848781/constructor-for-nested-initializer-lists