I have a matrix class defined like so:
template <typename T, unsigned int N, unsigned int M>
class TMatrixNxM //Rows x columns
{
public:
TMatrixNxM(T = T(0)); //Default constructor
TMatrixNxM(const std::array<std::array<T, M>, N>&); //Construct from array
TMatrixNxM(std::initializer_list<std::initializer_list<T>>); //Initializer
//...
private:
std::array<std::array<T, M>, N> data; //ROW-MAJOR
};
Now, in the code which uses matrices I have:
Math::Matrix3x3 b({{1, 2, 3}, {4, 5, 6}, {7, 8, 9}});
(Note: Matrix3x3 is a typedef for TMatrixNxM< float, 3, 3> and also, it is in a Math namespace)
Until now, it worked, because I didn't always have that array constructor, only initializer list one. But now, however, the compiler doesn't even finish compiling, it crashes! (I get "stopped working" popup and I have to close it, I am using MS VS Express 2013)
If I do it like this:
Math::Matrix3x3 b = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
Then it works fine. This is what I assume:
When I do it like that, then there is no ambiguity since you can only call the initializer_list constructor that way. With the first approach the compiler may be confused, since array is an aggregate type which means the initialization starts with double braces like: {{...}}, but also since I have an initializer list of initializer lists I have to use double braces. But should it really be a problem, since I never actually do double braces, those are single-braced lists inside a larger single-braces list?
What is really happening here and how do I solve this problem?
Thank you for your time!
EDIT
Compiler doesn't crash anymore if I make the constructor take the array by const pointer (since I'm never really going to straight-out plop the array in the constructor call, I have initializer list for that):
TMatrixNxM(const std::array<std::array<T, M>, N>*);
However can someone explain what was an actual problem before, was my assumption right?
Here's a minimal compilable (or, well, not) code that you can use to test:
#include <array>
#include <initializer_list>
template <typename T, unsigned int N, unsigned int M>
class TMatrixNxM //Rows x columns
{
public:
TMatrixNxM(T = T(0));
TMatrixNxM(const std::array<std::array<T, M>, N>&);
TMatrixNxM(std::initializer_list<std::initializer_list<T>>);
private:
std::array<std::array<T, M>, N> data; //ROW-MAJOR
};
template <typename T, unsigned int N, unsigned int M>
TMatrixNxM<T, N, M>::TMatrixNxM(T par_value)
{
std::array<T, M> temp;
temp.fill(par_value);
data.fill(temp);
}
template <typename T, unsigned int N, unsigned int M>
TMatrixNxM<T, N, M>::TMatrixNxM(const std::array<std::array<T, M>, N> &par_values)
{
data = par_values;
}
template <typename T, unsigned int N, unsigned int M>
TMatrixNxM<T, N, M>::TMatrixNxM(std::initializer_list<std::initializer_list<T>> par_values)
{
int i = 0;
for(std::initializer_list<T> row : par_values)
{
int j = 0;
for(T value : row)
{
data[i][j] = value;
++j;
}
++i;
}
}
int main()
{
TMatrixNxM<float, 3, 3> b({{1, 2, 3}, {4, 5, 6}, {7, 8, 9}});
return 0;
}
If you comment out the constructor declaration/definition with arrays then it should compile and execute fine. As stated previously, I am using compiler from MS VS Express 2013.
Well, looks like it just doesn't work on Microsoft's compiler. I tried to compile the same code on my Linux machine that uses GCC and it compiles with no problems. And it is indeed calling the initializer list constructor as it should. If I make an array in a variable and pass it, it also compiles fine, and it calls the array constructor:
std::array<float, 3> a = {{1, 2, 3}};
std::array<std::array<float, 3>, 3> b = {{a, a, a}};
TMatrixNxM<float, 3, 3> mat(b);
It also compiles and calls the array constructor if I initialize the array directly:
TMatrixNxM<float, 3, 3> mat({{ {{1, 2, 3}}, {{4, 5, 6}}, {{7, 8, 9}} }});
Related
I am trying to implement matrix-like class, using an std::array to actually store the data. All of the data is known at compile-time.
I want to be able to use initializer-lists to initialize the Matrix. Something along the lines of
Matrix m = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
I also want to be able to set the dimensions of my matrix during instantiation.
My current code is similar to the following:
template<int m, int n>
class Matrix {
public:
using initializer_list2d = std::initializer_list< std::initializer_list<float> >;
using array2d = std::array< std::array<float, m>, n >;
consteval Matrix(initializer_list2d initList) {
// static_assert to check initList length [...]
int i = 0;
for (auto &row : initList) {
// static_assert to check row length [...]
std::copy(row.begin(), row.end(), data_[i].begin());
i++;
}
}
// Definitions of operators and methods [...]
private:
array2d data_;
};
In order to use this though, I have to set the dimensions through the template:
Matrix<3, 3> m = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
It seems to me, that since the constructor is consteval (and all of its parameters have to be known at compile-time anyway) it should somehow bet possible to deduce the dimensions through the initializer-list alone.
Using a dynamic data type (e.g. vector) is not possible for my application, since this program has to be able to run without access to the heap.
Is there some way to achieve this?
Thanks for any help in advance!
Before C++17 there is no deduction of template arguments for a class template, so this is not possible.
From C++17, you can write this deduction guide:
template<int m, int n>
Matrix(float const (&)[n][m]) -> Matrix<m, n>;
and add an extra pair of braces when constructing the Matrix:
Matrix m =
{
{
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
}
};
Here's a demo.
I found another interesting solution, using variadic templates:
template<int m, class T, class... U>
struct Matrix {
Matrix(const T (&t)[m], const U (& ...u)[m]) {
std::copy(t, t+m, arr[0]);
int i = 0;
(std::copy(u, u+m, arr[++i]), ...);
}
T (arr[m])[sizeof...(U) + 1];
};
It even allows for getting rid of the second brace required in cigien's answer and doesn't need a template deduction guide (Although it still needs C++17 in order for int m to be automatically deduced):
Matrix mat = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
Also, the compiler seems to be able to optimze this quite a bit, as seen here
This is the code I found online.
template<class T, unsigned ... RestD> struct array;
template<class T, unsigned PrimaryD >
struct array<T, PrimaryD>
{
typedef T type[PrimaryD];
type data;
T& operator[](unsigned i) { return data[i]; }
};
template<class T, unsigned PrimaryD, unsigned ... RestD >
struct array<T, PrimaryD, RestD...>
{
typedef typename array<T, RestD...>::type OneDimensionDownArrayT;
typedef OneDimensionDownArrayT type[PrimaryD];
type data;
OneDimensionDownArrayT& operator[](unsigned i) { return data[i]; }
};
int main()
{
array<int, 2, 3>::type a4 = { { 1, 2, 3}, { 1, 2, 3} };
array<int, 2, 3> a5{ { { 1, 2, 3}, { 4, 5, 6} } };
std::cout << a5[1][2] << std::endl;
array<int, 3> a6{ {1, 2, 3} };
std::cout << a6[1] << std::endl;
array<int, 1, 2, 3> a7{ { { { 1, 2, 3}, { 4, 5, 6 } } }};
std::cout << a7[0][1][2] << std::endl;
}
Could you explain what this code does exactly? I understand that recursion is used in some form here to create a multidimensional array, but I am a little confused on how this process works.
I am also confused about this line:
array<int, 2, 3>::type a4 = { { 1, 2, 3}, { 1, 2, 3} };
What is the ::type?
This is whats called template specialization.
The first thing you must know is that anything defined with template is evaluated at compile time.
Looking at your code, there are two main cases we want to consider:
When the array is of one-dimension
When the array has more than one dimension
The following describes the one-dimensional case:
template<class T, unsigned PrimaryD >
struct array<T, PrimaryD>
{
typedef T type[PrimaryD];
type data;
T& operator[](unsigned i) { return data[i]; }
};
As you can see the above struct is specialized to only accept two template parameters, the first one is T the data type of the array, and the second is PrimaryD which simply means how long should the array be.
Now for the 2nd specialization:
template<class T, unsigned PrimaryD, unsigned ... RestD >
struct array<T, PrimaryD, RestD...>
{
// if the template contains array<int, 1, 2, 3, 4>
// then RestD... will unpack the values 2, 3, 4
typedef typename array<T, RestD...>::type
OneDimensionDownArrayT;
typedef OneDimensionDownArrayT type[PrimaryD];
type data;
// This will return an array type
// you can keep on making a call to operator[] until the type of the OneDimensionDownArrayT refers to the previous specialization
OneDimensionDownArrayT& operator[](unsigned i) { return
data[i];
}
};
Now for your example:
array<int, 2, 3>::type a4 = { { 1, 2, 3}, { 1, 2, 3} }
What the compiler will do is sort of match the template parameters namely, int, 2, and 3 which evaluate to the 2nd specialization.
Let's follow it through:
array<int, 2, 3> which is a struct contains the following concrete types:
OneDimensionalT which is equal to array<int, 3>
and an operator[] function that returns an array<int, 3>.
OneDimensionalT is equal to array<int, 3> because if you unpack the RestD... values you will only get 3, the 2 is already consumed by the first template parameter, PrimaryD.
Adding a 3rd dimension follows the same logic, keep on defining smaller dimension struct array types recursively using only the n-1 dimension values of the parameter pack RestD...
Oh and btw, the line on the top is just to declare the base type so we can specialize it.
Can I initialize an array using the std::initializer_list object instead of brace-enclosed initializer?
As known, we can do this: http://en.cppreference.com/w/cpp/language/aggregate_initialization
unsigned char b[5]{"abc"};
// equivalent to unsigned char b[5] = {'a', 'b', 'c', '\0', '\0'};
int ar[] = {1,2,3};
std::array<int, 3> std_ar2{ {1,2,3} }; // std::array is an aggregate
std::array<int, 3> std_ar1 = {1, 2, 3};
But I can't initialize an array by std::initializer_list il;:
http://ideone.com/f6aflX
#include <iostream>
#include <initializer_list>
#include <array>
int main() {
int arr1[] = { 1, 2, 3 }; // OK
std::array<int, 3> arr2 = { 1, 2, 3 }; // OK
std::initializer_list<int> il = { 1, 2, 3 };
constexpr std::initializer_list<int> il_constexpr = { 1, 2, 3 };
//int arr3[] = il; // error
//int arr4[] = il_constexpr; // error
//std::array<int, 3> arr5 = il; // error
//std::array<int, 3> arr6 = il_constexpr; // error
return 0;
}
But how can I use std::initializer_list il; to initialize an array?
Other answered correctly said this is not possible upfront. But with little helpers, you can get pretty close
template<typename T, std::size_T N, std::size_t ...Ns>
std::array<T, N> make_array_impl(
std::initializer_list<T> t,
std::index_sequence<Ns...>)
{
return std::array<T, N>{ *(t.begin() + Ns) ... };
}
template<typename T, std::size_t N>
std::array<T, N> make_array(std::initializer_list<T> t) {
if(N > t.size())
throw std::out_of_range("that's crazy!");
return make_array_impl<T, N>(t, std::make_index_sequence<N>());
}
If you are open to more work arounds, you can put this into a class to catch statically-known length violations for the cases where you pass a braced init list. But be warned that most people who read this code will head-desk
template<typename T, std::size_t N>
struct ArrayInitializer {
template<typename U> struct id { using type = U; };
std::array<T, N> t;
template<typename U = std::initializer_list<T>>
ArrayInitializer(typename id<U>::type z)
:ArrayInitializer(z, std::make_index_sequence<N>())
{
if(N > z.size())
throw std::out_of_range("that's crazy!");
}
template<typename ...U>
ArrayInitializer(U &&... u)
:t{ std::forward<U>(u)... }
{ }
private:
template<std::size_t ...Ns>
ArrayInitializer(std::initializer_list<T>& t,
std::index_sequence<Ns...>)
:t{ *(t.begin() + Ns) ... }
{ }
};
template<typename T, std::size_t N>
std::array<T, N> f(ArrayInitializer<T, N> ai) {
return std::move(ai.t);
}
int main() {
f<int, 5>({1, 2, 3, 4, 5}); // OK
f<int, 5>({1, 2, 3, 4, 5, 6}); // "too many initializers for array<int, 5>"
std::initializer_list<int> il{1, 2, 3, 4, 5};
f<int, 5>(il); // ok
}
Note that both the non-static case at the top of the answer and the "head-desk" case do only check whether you provide too few initializing elements, and errors out then, for the initializer_list case. If you provide too many for the initializer_list case, the trailing elements are just ignored.
As far I know, no: you can't initialize a std::array with a std::initializer_list.
The problem is that std::array is intended as a lightweight replacement (a wrapper) for the classic C-style array. So light that is without constructors, so only implicit constructor can be used.
The construction with aggregate initialization (via implicit constructor) is possible because it's possible for the C-style array.
But std::initializer_list is a class, more complicated than an aggregate inizialization.
You can initialize, by example, a std::vector with a std::initializer_list but only because there is an explicit constructor, for std::vector, that receive a std::initializer_list. But std::vector is a heavier class.
The only solution that I see is a 2 step way: (1) construction and (2) copy of the std::initializer_list values. Something like
std::array<int, 3> arr5;
auto ui = 0U;
auto cit = il.cbegin();
while ( (ui < arr5.size()) && (cit != il.cend()) )
arr5[ui++] = *cit++;
p.s.: sorry for my bad English.
The problem with std::array is that it is required to be an aggregate type, hence it does not have constructors.
Hence only aggregate initialization or trivial copy are possible.
std::initializer_list is a class other than std::array, so a (missing) implicit conversion is required.
See http://en.cppreference.com/w/cpp/language/aggregate_initialization
and http://en.cppreference.com/w/cpp/container/array
for reference.
I'm trying to initialize a class Vec with a brace-enclosed initializer list, to be called like in this minimal example:
int main()
{
Vec<3> myVec1{{1,2,3}}; // or: myVec1({1,2,3});
Vec<3> myVec2;
myVec2 = {1, 2, 3};
Vec<3> myVec3 = {1,2,3};
return 0;
}
All these initializations (and the assignment) should work. (Plus custom default constructor. Using an aggregate class is thus impossible.)
While I could just use a std::initializer_list like such:
template <unsigned C>
struct Vec
{
Vec(){}
Vec(std::initializer_list<int> list)
{
//does not work for obvious reasons:
//static_assert(list.size() == C, "");
}
};
I can't statically ensure the number of parameters to be equal to my template parameter, i.e., an initialization with {1, 2, 3, 4} would only fail during runtime.
Browsing SO, I came up with the following:
template <unsigned C>
struct Vec
{
Vec(){}
Vec(const unsigned(&other)[C]){}
Vec& operator=(const unsigned(&other)[C]){return *this;}
};
This works fine for the assignment and () or {} initialization (as for myVec1) - but it fails for the initialization using = (as for myVec3).
(GCC gives error "could not convert '{1, 2, 3}' from '' to 'Vec<3u>'")
I don't get why one of the initializations should work, but not the other.
Any other ideas how I can use the brace-enclosed initializer list, but also ensure the correct length at compile time?
Thanks :)
Initializer lists are more appropriate for situations where the size of the list is dynamic. In your case, where the size of Vec is a template parameter (i.e. static), you're probably better off with variadic parameters:
template <unsigned C>
struct Vec
{
Vec(){}
template<typename ... V>
Vec(V ... args)
{
static_assert(sizeof...(V) == C, "");
//...
}
};
then these will work:
Vec<3> myVec2;
myVec2 = {1, 2, 3};
Vec<3> myVec3 = {1,2,3};
But this one won't, as it explicitly requires an std::initializer_list:
Vec<3> myVec1{{1,2,3}};
You need two pair of brackets:
Vec<3> myVec3 = {{1,2,3}};
When you do:
Vec<3> myVec3 = {1, 2, 3};
// It is the same as (except that explicit constructor are not considered):
Vec<3> myVec3 = Vec<3>{1, 2, 3};'
So basically, the compiler will look for a constructor for Vec<3> which either takes an std::initializer_list or a constructor that takes 3 arguments, all of which being constructible from int.
If you want a std::array like structure, you need to make your type an aggregate, meaning that it should not have user-defined constructor:
template <unsigned N>
struct Vec {
int data[N];
};
void f() {
Vec<3> v1{{1, 2, 3}};
Vec<3> v2 = {{1, 2, 3}};
}
I have this template matrix struct (I provided a constructor which takes std::initializer_list):
template<int rows, int cols, typename scalar = float>
struct matrix;
with a product operator defined outside the matrix struct, like this:
template<int n, int m, int p, typename scalar>
matrix<n, m, scalar> operator*(const matrix<m, p, scalar>& left, const matrix<p, n, scalar>& left);
and then declared as a friend inside the struct. So if I instantiate two matrices:
matrix<2, 3> A = { 1, 2, 3, 4, 5, 6 };
matrix<3, 2> B = { 7, 8, 9, 10, 11, 12 };
and I want to create a matrix C = A * B, I have to write:
matrix<2, 2> C = A * B;
And that's fine, but is there a way to omit the <2, 2> template? I believe that it can be deducted at compile time (because auto works fine):
auto C = A * B; // no errors
I'd like to write just matrix instead of auto, is it possible?
No, you cannot (if you don't have some non-template base matrix). matrix is not a type, it's template and you should specify template parameters. auto is simplest thing, that you can do. Or, instead of auto you can use decltype
decltype(A * B) C = A * B;