Providing aliases to Parent class data array in C++ - c++

Probably not the wisest choice, but modelling a Vertex<T, N> class to abstract Vertex2<T>, Vertex3<T> and Vertex4<T> implementations, providing basic access. It's structured like this:
template<typename T, unsigned N>
class _Vertex {
const unsigned _size = N;
T _data[N];
public:
inline _Vertex() : _data() {};
inline T const& operator[](int pos) const { return _data[pos]; }
inline T & operator[](int pos) { return _data[pos]; }
};
Say I want to implement Vertex2<T> as a Vertex<T, 2>, and provide aliases like x and y. The most proper way would be adding functions like:
inline T const& x() const { return (*this)[0]; }
inline T & x() { return (*this)[0]; }
This would be repeated for every property, or alias I'd like to add. It's not a bad design, but usage proves tricky as, provided v is of type Vector2<float>, v.x() = 1.0f is not as friendly as v.x = 1.0f.
Is there any way to provide clearer, friendlier alias?
My main thoughts were "abusing" memory layout, providing access to _data[i] accordingly, but I have no idea where to start.
This question is based on my "re-imagination" of a vertex.h header file provided for an assigment, so this makes it related to homework, but I can assure you the end is not. It's just my curiosity holding me off doing my homework!

I put on my robe and Yakk hat.
operator->* is clearly underutilized in C++. We can overload to take a tag type of our choosing in a way that kind of looks like member access. Here's a simplified example:
template <size_t I> struct index_t { };
constexpr index_t<0> x{};
constexpr index_t<1> y{};
struct Vector {
int data[2];
template <size_t I>
int& operator->*(index_t<I> ) { return data[I]; }
int& operator[](size_t idx) { return data[idx]; }
};
With that, we can do:
int main()
{
Vector v{1, 2};
std::cout << v.data[0] << ' ' << v.data[1] << '\n'; // 1 2
v->*x = 42;
v->*y = 17;
std::cout << v.data[0] << ' ' << v.data[1] << '\n'; // 42 17
}
That said, don't do this. C++ doesn't have properties. The right way to do this would just be named functions that return lvalue references:
int& x();
int& y();
It may not be as "nice" as v.x but it gets the job done without having to add extra members to your class.

v.x = 2.f is not the OOP/C++ way of member access. You are best off using get/set function like t = v.x() and v.x( t ).
It is possible to defeat the data encapsulation using the following:
template< class T >
class Vertex2 : private Vertex< T, 2 >
{
public:
T & x;
T & y;
Vertex2(): Vertex< T, 2 >(), x((*this)[0]), y((*this)[1]) {}
};
This doesn't provide a const access but if you want to do v.x = you probably don't care too much about such things. Unfortunately, you are probably stuck with the extra memory of the two references.

Related

Get byte representation of C++ class

I have objects that I need to hash with SHA256. The object has several fields as follows:
class Foo {
// some methods
protected:
std::array<32,int> x;
char y[32];
long z;
}
Is there a way I can directly access the bytes representing the 3 member variables in memory as I would a struct ? These hashes need to be computed as quickly as possible so I want to avoid malloc'ing a new set of bytes and copying to a heap allocated array. Or is the answer to simply embed a struct within the class?
It is critical that I get the exact binary representation of these variables so that the SHA256 comes out exactly the same given that the 3 variables are equal (so I can't have any extra padding bytes etc included going into the hash function)
Most Hash classes are able to take multiple regions before returning the hash, e.g. as in:
class Hash {
public:
void update(const void *data, size_t size) = 0;
std::vector<uint8_t> digest() = 0;
}
So your hash method could look like this:
std::vector<uint8_t> Foo::hash(Hash *hash) const {
hash->update(&x, sizeof(x));
hash->update(&y, sizeof(y));
hash->update(&z, sizeof(z));
return hash->digest();
}
You can solve this by making an iterator that knows the layout of your member variables. Make Foo::begin() and Foo::end() functions and you can even take advantage of range-based for loops.
If you can increment it and dereference it, you can use it any other place you're able to use a LegacyForwardIterator.
Add in comparison functions to get access to the common it = X.begin(); it != X.end(); ++it idiom.
Some downsides include: ugly library code, poor maintainability, and (in this current form) no regard for endianess.
The solution to the latter downside is left as an exercise to the reader.
#include <array>
#include <iostream>
class Foo {
friend class FooByteIter;
public:
FooByteIter begin() const;
FooByteIter end() const;
Foo(const std::array<int, 2>& x, const char (&y)[2], long z)
: x_{x}
, y_{y[0], y[1]}
, z_{z}
{}
protected:
std::array<int, 2> x_;
char y_[2];
long z_;
};
class FooByteIter {
public:
FooByteIter(const Foo& foo)
: ptr_{reinterpret_cast<const char*>(&(foo.x_))}
, x_end_{reinterpret_cast<const char*>(&(foo.x_)) + sizeof(foo.x_)}
, y_begin_{reinterpret_cast<const char*>(&(foo.y_))}
, y_end_{reinterpret_cast<const char*>(&(foo.y_)) + sizeof(foo.y_)}
, z_begin_{reinterpret_cast<const char*>(&(foo.z_))}
{}
static FooByteIter end(const Foo& foo) {
FooByteIter fbi{foo};
fbi.ptr_ = reinterpret_cast<const char*>(&foo.z_) + sizeof(foo.z_);
return fbi;
}
bool operator==(const FooByteIter& other) const { return ptr_ == other.ptr_; }
bool operator!=(const FooByteIter& other) const { return ! (*this == other); }
FooByteIter& operator++() {
ptr_++;
if (ptr_ == x_end_) {
ptr_ = y_begin_;
}
else if (ptr_ == y_end_) {
ptr_ = z_begin_;
}
return *this;
}
FooByteIter operator++(int) {
FooByteIter pre = *this;
(*this)++;
return pre;
}
char operator*() const {
return *ptr_;
}
private:
const char* ptr_;
const char* const x_end_;
const char* const y_begin_;
const char* const y_end_;
const char* const z_begin_;
};
FooByteIter Foo::begin() const {
return FooByteIter(*this);
}
FooByteIter Foo::end() const {
return FooByteIter::end(*this);
}
template <typename InputIt>
char checksum(InputIt first, InputIt last) {
char check = 0;
while (first != last) {
check += (*first);
++first;
}
return check;
}
int main() {
Foo f{{1, 2}, {3, 4}, 5};
for (const auto b : f) {
std::cout << (int)b << ' ';
}
std::cout << std::endl;
std::cout << "Checksum is: " << (int)checksum(f.begin(), f.end()) << std::endl;
}
You can generalize this further by making serialization functions for all data types you might care about, allowing serialization of classes that aren't plain-old-data types.
Warning
This code assumes that the underlying types being serialized have no internal padding, themselves. This answer works for this datatype because it is made of types which themselves do not pad. To make this work for datatypes that have datatypes that have padding, this method would need to be recursed all the way down.
Just cast a pointer to object to a pointer to char. You can iterate through the bytes by increment. Use sizeof(foo) to check overflow.
As long as you're able to make your class an aggregate, i.e. std::is_aggregate_v<T> == true, you can actually sort-of reflect the members of the structure.
This allows you to easily hash the members without actually having to name them. (also you don't have to remember updating your hash function every time you add a new member)
Step 1: Getting the number of members inside the aggregate
First we need to know how many members a given aggregate type has.
We can check this by (ab-)using aggregate initialization.
Example:
Given struct Foo { int i; int j; };:
Foo a{}; // ok
Foo b{{}}; // ok
Foo c{{}, {}}; // ok
Foo d{{}, {}, {}}; // error: too many initializers for 'Foo'
We can use this to get the number of members inside the struct, by trying to add more initializers until we get an error:
template<class T>
concept aggregate = std::is_aggregate_v<T>;
struct any_type {
template<class T>
operator T() {}
};
template<aggregate T>
consteval std::size_t count_members(auto ...members) {
if constexpr (requires { T{ {members}... }; } == false)
return sizeof...(members) - 1;
else
return count_members<T>(members..., any_type{});
}
Notice that i used {members}... instead of members....
This is because of arrays - a structure like struct Bar{int i[2];}; could be initialized with 2 elements, e.g. Bar b{1, 2}, so our function would have returned 2 for Bar if we had used members....
Step 2: Extracting the members
Now that we know how many members our structure has, we can use structured bindings to extract them.
Unfortunately there is no way in the current standard to create a structured binding expression with a variable amount of expressions, so we have to add a few extra lines of code for each additional member we want to support.
For this example i've only added a max of 4 members, but you can add as many as you like / need:
template<aggregate T>
constexpr auto tie_struct(T const& data) {
constexpr std::size_t fieldCount = count_members<T>();
if constexpr(fieldCount == 0) {
return std::tie();
} else if constexpr (fieldCount == 1) {
auto const& [m1] = data;
return std::tie(m1);
} else if constexpr (fieldCount == 2) {
auto const& [m1, m2] = data;
return std::tie(m1, m2);
} else if constexpr (fieldCount == 3) {
auto const& [m1, m2, m3] = data;
return std::tie(m1, m2, m3);
} else if constexpr (fieldCount == 4) {
auto const& [m1, m2, m3, m4] = data;
return std::tie(m1, m2, m3, m4);
} else {
static_assert(fieldCount!=fieldCount, "Too many fields for tie_struct! add more if statements!");
}
}
The fieldCount!=fieldCount in the static_assert is intentional, this prevents the compiler from evaluating it prematurely (it only complains if the else case is actually hit)
Now we have a function that can give us references to each member of an arbitrary aggregate.
Example:
struct Foo {int i; float j; std::string s; };
Foo f{1, 2, "miau"};
// tup is of type std::tuple<int const&, float const&, std::string const&>
auto tup = tie_struct(f);
// this will output "12miau"
std::cout << std::get<0>(tup) << std::get<1>(tup) << std::get<2>(tup) << std::endl;
Step 3: hashing the members
Now that we can convert any aggregate into a tuple of its members, hashing it shouldn't be a big problem.
You can basically hash the individual types like you want and then combine the individual hashes:
// for merging two hash values
std::size_t hash_combine(std::size_t h1, std::size_t h2)
{
return (h2 + 0x9e3779b9 + (h1<<6) + (h1>>2)) ^ h1;
}
// Handling primitives
template <class T, class = void>
struct is_std_hashable : std::false_type { };
template <class T>
struct is_std_hashable<T, std::void_t<decltype(std::declval<std::hash<T>>()(std::declval<T>()))>> : std::true_type { };
template <class T>
concept std_hashable = is_std_hashable<T>::value;
template<std_hashable T>
std::size_t hash(T value) {
return std::hash<T>{}(value);
}
// Handling tuples
template<class... Members>
std::size_t hash(std::tuple<Members...> const& tuple) {
return std::apply([](auto const&... members) {
std::size_t result = 0;
((result = hash_combine(result, hash(members))), ...);
return result;
}, tuple);
}
template<class T, std::size_t I>
using Arr = T[I];
// Handling arrays
template<class T, std::size_t I>
std::size_t hash(Arr<T, I> const& arr) {
std::size_t result = 0;
for(T const& elem : arr) {
std::size_t h = hash(elem);
result = hash_combine(result, h);
}
return result;
};
// Handling structs
template<aggregate T>
std::size_t hash(T const& agg) {
return hash(tie_struct(agg));
}
This allows you to hash basically any aggregate struct, even with arrays and nested structs:
struct Foo{ int i; double d; std::string s; };
struct Bar { Foo k[10]; float f; };
std::cout << hash(Foo{1, 1.2f, "miau"}) << std::endl;
std::cout << hash(Bar{}) << std::endl;
full example on godbolt
Footnotes
This only works with aggregates
No need to worry about padding because we access the members directly.
You have to add a few more ifs into tie_struct if you need more than 4 members
The provided hash() function doesn't handle all types - if you need e.g. std::array, std::pair, etc... you need to add overloads for those.
It's a lot of boilerplate code, but it's insanely powerful.
You can also use Boost.PFR for the aggregate-to-tuple part, if you are allowed to use boost

Pointer generalization of index

Referencing with indexes compared to referencing with pointers has several advantages, e.g. indexes often survive array reallocation. But pointers/references are often more convenient to use than indexes. In the C++ STL library, the iterators were designed as generalizations of the pointer and now I wonder whether we could do something similar for indexes: little classes that carry the index as local data, but behave like (convenient) pointers. Something like:
#include <iostream>
#include <vector>
#include <compare>
template<class T, class I, T** ppdata>
struct index_ptr_T
{
explicit index_ptr_T(I index = 0)
: m_index(index)
{}
index_ptr_T(T* p)
: m_index(p - *ppdata)
{}
T* operator->() const
{
return *ppdata + m_index;
}
T& operator*() const
{
return *(*ppdata + m_index);
}
friend auto operator<=>(const index_ptr_T<T, I, ppdata>&, const index_ptr_T<T, I, ppdata>&) = default;
I m_index;
};
struct s_t // a test class
{
inline static s_t* pdata{};
using index_ptr_t = index_ptr_T<s_t, short, &pdata>;
int m_i;
index_ptr_t m_next;
};
int main()
{
std::vector<s_t> data(2);
s_t::pdata = &data[0];
s_t::index_ptr_t pa = &data[0];
*pa = { 10, 0 };
s_t::index_ptr_t pb = &data[1];
pb->m_next = pa;
std::cout << pb->m_next->m_i << std::endl;
data.reserve(data.capacity() + 1);
s_t::pdata = &data[0];
std::cout << pb->m_next->m_i << std::endl;
std::cout << "sizeof(s_t::index_ptr_t) = " << sizeof(s_t::index_ptr_t) << std::endl;
}
Does something like this already exist in library form? Or did someone already publish about it? I googled, but couldn’t find what I was looking for. The index_ptr_T class above is quite incomplete, many operators are missing. I would like to use a proven library.
Note 1: I've !!EDITED!! above example to clarify a little bit more the intended use.
Note 2: s_t::pdata = &data[0] must also be done whenever the data is realocated/moved. E.g. when the data is in a vector after elements have been added.
Your proposed design is rather limited, thanks to templating on the array object. A slightly more reasonable design would look like this:
template<typename ContainerType, typename IndexType>
struct index_ptr_T
{
index_ptr_T(ContainerType & container, IndexType index)
: m_container(container), m_index(index) {}
auto* operator->()
{
return &m_container[m_index];
}
auto & operator*()
{
return m_container[m_index];
}
ContainerType & m_container;
IndexType m_index;
};
int main()
{
std::vector<int> v{1, 2, 3};
int i = 1;
index_ptr_T p{v, i};
v.reserve(v.capacity() + 1);
std::cout << *p;
}
Or, more succinctly, with no separate utility:
int main()
{
std::vector<int> v{1, 2, 3};
int i = 1;
auto p = [&v, i]() -> auto & { return v[i]; };
v.reserve(v.capacity() + 1);
std::cout << p();
}
I'm not aware of anything like this in existing frameworks. As you can see, there's just not much in the concept. It's just a bound indexing operation. Making it into a utility would make code harder to read, without making it easier to write.

Member field aliases in derived class (no accessor functions)

Toy example:
template<typename T, std::size_t N>
class static_vector
{
public:
T& operator[](std::size_t i) { return m_elements[i]; }
T const& operator[](std::size_t i) const { return m_elements[i]; }
private:
std::array<T, N> m_elements;
};
template<typename T>
class vector3
: public static_vector<T, 3>
{
public:
using vector_type = static_vector<T, 3>;
// x = vector_type::operator[](0);
// y = vector_type::operator[](1);
// z = vector_type::operator[](2);
};
Let vector3<float> pos;. I want to access pos[0] via pos.x. Clearly, if pos is declared to be const, I want pos.x to be read-only.
Is this possible?
Let me stress the fact that I don't want to use accessor functions of the form
T& x() { return (*this)[0]; }
T const& x() const { return (*this)[0]; }
There is no zero cost way to do this using the exact syntax you want.
Relaxing either cost (compile, maintenance, memory usage and runtime) or syntax (which your () is an example of) gets you what you want.
I added a comment to your question but I thought I'd add an answer with some code to clarify. Be warned, what follows is not a good idea.
You can use simple pointer arithmetic to interpret members of a struct as if they were elements in an array. Because the type of the struct members and the type of elements in the pseudo array are the same, we're safe to reinterpret one as the other with the caveat that there is no padding in between the struct members.
The C++ standard gives no method for defining padding in a struct so you will have to rely on compiler specific directives. I believe however that both MSVC and GCC support #pragma pack.
#pragma pack(push, 1)
template <typename T>
struct Vec3
{
T x;
T y;
T z;
T& operator[](size_t i) { return *(&x + i); }
const T& operator[](size_t i) const { return *(&x + i); }
};
#pragma pack(pop)
So why isn't this a good solution?
You are relying on compiler specific directives making your code less portable.
You need to explicitly declare each member which means you'll need separate templates for Vec2, Vec3 and Vec4.
1 byte alignment isn't supported on all architectures again making your code less portable.
Even on architectures where unaligned memory access is supported (such as x86,) it comes with a performance penalty.

Virtually turn vector of struct into vector of struct members

I have a function that takes a vector-like input. To simplify things, let's use this print_in_order function:
#include <iostream>
#include <vector>
template <typename vectorlike>
void print_in_order(std::vector<int> const & order,
vectorlike const & printme) {
for (int i : order)
std::cout << printme[i] << std::endl;
}
int main() {
std::vector<int> printme = {100, 200, 300};
std::vector<int> order = {2,0,1};
print_in_order(order, printme);
}
Now I have a vector<Elem> and want to print a single integer member, Elem.a, for each Elem in the vector. I could do this by creating a new vector<int> (copying a for all Elems) and pass this to the print function - however, I feel like there must be a way to pass a "virtual" vector that, when operator[] is used on it, returns this only the member a. Note that I don't want to change the print_in_order function to access the member, it should remain general.
Is this possible, maybe with a lambda expression?
Full code below.
#include <iostream>
#include <vector>
struct Elem {
int a,b;
Elem(int a, int b) : a(a),b(b) {}
};
template <typename vectorlike>
void print_in_order(std::vector<int> const & order,
vectorlike const & printme) {
for (int i : order)
std::cout << printme[i] << std::endl;
}
int main() {
std::vector<Elem> printme = {Elem(1,100), Elem(2,200), Elem(3,300)};
std::vector<int> order = {2,0,1};
// how to do this?
virtual_vector X(printme) // behaves like a std::vector<Elem.a>
print_in_order(order, X);
}
It's not really possible to directly do what you want. Instead you might want to take a hint from the standard algorithm library, for example std::for_each where you take an extra argument that is a function-like object that you call for each element. Then you could easily pass a lambda function that prints only the wanted element.
Perhaps something like
template<typename vectorlike, typename functionlike>
void print_in_order(std::vector<int> const & order,
vectorlike const & printme,
functionlike func) {
for (int i : order)
func(printme[i]);
}
Then call it like
print_in_order(order, printme, [](Elem const& elem) {
std::cout << elem.a;
});
Since C++ have function overloading you can still keep the old print_in_order function for plain vectors.
Using member pointers you can implement a proxy type that will allow you view a container of objects by substituting each object by one of it's members (see pointer to data member) or by one of it's getters (see pointer to member function). The first solution addresses only data members, the second accounts for both.
The container will necessarily need to know which container to use and which member to map, which will be provided at construction. The type of a pointer to member depends on the type of that member so it will have to be considered as an additional template argument.
template<class Container, class MemberPtr>
class virtual_vector
{
public:
virtual_vector(const Container & p_container, MemberPtr p_member_ptr) :
m_container(&p_container),
m_member(p_member_ptr)
{}
private:
const Container * m_container;
MemberPtr m_member;
};
Next, implement the operator[] operator, since you mentioned that it's how you wanted to access your elements. The syntax for dereferencing a member pointer can be surprising at first.
template<class Container, class MemberPtr>
class virtual_vector
{
public:
virtual_vector(const Container & p_container, MemberPtr p_member_ptr) :
m_container(&p_container),
m_member(p_member_ptr)
{}
// Dispatch to the right get method
auto operator[](const size_t p_index) const
{
return (*m_container)[p_index].*m_member;
}
private:
const Container * m_container;
MemberPtr m_member;
};
To use this implementation, you would write something like this :
int main() {
std::vector<Elem> printme = { Elem(1,100), Elem(2,200), Elem(3,300) };
std::vector<int> order = { 2,0,1 };
virtual_vector<decltype(printme), decltype(&Elem::a)> X(printme, &Elem::a);
print_in_order(order, X);
}
This is a bit cumbersome since there is no template argument deduction happening. So lets add a free function to deduce the template arguments.
template<class Container, class MemberPtr>
virtual_vector<Container, MemberPtr>
make_virtual_vector(const Container & p_container, MemberPtr p_member_ptr)
{
return{ p_container, p_member_ptr };
}
The usage becomes :
int main() {
std::vector<Elem> printme = { Elem(1,100), Elem(2,200), Elem(3,300) };
std::vector<int> order = { 2,0,1 };
auto X = make_virtual_vector(printme, &Elem::a);
print_in_order(order, X);
}
If you want to support member functions, it's a little bit more complicated. First, the syntax to dereference a data member pointer is slightly different from calling a function member pointer. You have to implement two versions of the operator[] and enable the correct one based on the member pointer type. Luckily the standard provides std::enable_if and std::is_member_function_pointer (both in the <type_trait> header) which allow us to do just that. The member function pointer requires you to specify the arguments to pass to the function (non in this case) and an extra set of parentheses around the expression that would evaluate to the function to call (everything before the list of arguments).
template<class Container, class MemberPtr>
class virtual_vector
{
public:
virtual_vector(const Container & p_container, MemberPtr p_member_ptr) :
m_container(&p_container),
m_member(p_member_ptr)
{}
// For mapping to a method
template<class T = MemberPtr>
auto operator[](std::enable_if_t<std::is_member_function_pointer<T>::value == true, const size_t> p_index) const
{
return ((*m_container)[p_index].*m_member)();
}
// For mapping to a member
template<class T = MemberPtr>
auto operator[](std::enable_if_t<std::is_member_function_pointer<T>::value == false, const size_t> p_index) const
{
return (*m_container)[p_index].*m_member;
}
private:
const Container * m_container;
MemberPtr m_member;
};
To test this, I've added a getter to the Elem class, for illustrative purposes.
struct Elem {
int a, b;
int foo() const { return a; }
Elem(int a, int b) : a(a), b(b) {}
};
And here is how it would be used :
int main() {
std::vector<Elem> printme = { Elem(1,100), Elem(2,200), Elem(3,300) };
std::vector<int> order = { 2,0,1 };
{ // print member
auto X = make_virtual_vector(printme, &Elem::a);
print_in_order(order, X);
}
{ // print method
auto X = make_virtual_vector(printme, &Elem::foo);
print_in_order(order, X);
}
}
You've got a choice of two data structures
struct Employee
{
std::string name;
double salary;
long payrollid;
};
std::vector<Employee> employees;
Or alternatively
struct Employees
{
std::vector<std::string> names;
std::vector<double> salaries;
std::vector<long> payrollids;
};
C++ is designed with the first option as the default. Other languages such as Javascript tend to encourage the second option.
If you want to find mean salary, option 2 is more convenient. If you want to sort the employees by salary, option 1 is easier to work with.
However you can use lamdas to partially interconvert between the two. The lambda is a trivial little function which takes an Employee and returns a salary for him - so effectively providing a flat vector of doubles we can take the mean of - or takes an index and an Employees and returns an employee, doing a little bit of trivial data reformatting.
template<class F>
struct index_fake_t{
F f;
decltype(auto) operator[](std::size_t i)const{
return f(i);
}
};
template<class F>
index_fake_t<F> index_fake( F f ){
return{std::move(f)};
}
template<class F>
auto reindexer(F f){
return [f=std::move(f)](auto&& v)mutable{
return index_fake([f=std::move(f),&v](auto i)->decltype(auto){
return v[f(i)];
});
};
}
template<class F>
auto indexer_mapper(F f){
return [f=std::move(f)](auto&& v)mutable{
return index_fake([f=std::move(f),&v](auto i)->decltype(auto){
return f(v[i]);
});
};
}
Now, print in order can be rewritten as:
template <typename vectorlike>
void print(vectorlike const & printme) {
for (auto&& x:printme)
std::cout << x << std::endl;
}
template <typename vectorlike>
void print_in_order(std::vector<int> const& reorder, vectorlike const & printme) {
print(reindexer([&](auto i){return reorder[i];})(printme));
}
and printing .a as:
print_in_order( reorder, indexer_mapper([](auto&&x){return x.a;})(printme) );
there may be some typos.

C++ Overload index operator for more than one array/vector

I'm not an advanced programmer. How can I overload the [] operator for a class that has two (or more) array/vector type variables?
class X
{
protected:
std::vector<double> m_x, m_y;
public:
double& operator[](const short &i) { return ???; }
};
What should I use for ???, or how can I do it (maybe adding other definitions?) to be able to call either variable?
Additional question: will this allow other classes of type class derived : public X access m_x and m_y for writing?
UPDATE:
Thank you everyone who answered, but I'm afraid that if I draw the line then the answer to my first question is no, and to the second yes. The longer version implies either an extra struct, or class, or plain setters/getters, which I wanted to avoid by using a simple function for all.
As it stands, the current solution is a (temporary) reference to each variable, in each class to avoid the extra X:: typing (and keep code clear), since m_x would have existed, one way or another.
you can write just a function for this, like:
double &get(unsigned int whichVector, unsigned int index)
{
return (whichVector == 0 ? m_x[index] : m_y[index]);
}
or use operator():
struct A
{
std::vector<int> a1;
std::vector<int> a2;
int operator()(int vec, int index)
{
return (vec == 0 ? a1[index] : a2[index]);
}
};
A a;
auto var = a(0, 1);
but still, this is kinda strange :) probably you should just give a const ref outside, like:
const std::vector<double> &getX() const { return m_x; }
and second question: protected will be convert into private in public inheritance (child/derived will have access to these memebers)
Assuming you want m_x and m_y indexed against the same parameter and a single return value:
struct XGetter
{
double& x;
double& y;
};
XGetter operator[](const short &i) { return { m_x[i], m_y[i] }; }
And the const overload:
struct XGetterReadOnly
{
double x;
double y;
};
XGetterReadOnly operator[](const short &i) const { return { m_x[i], m_y[i] }; }
The compiler will make a good job of optimizing away the intermediate classes XGetter and XGetterReadOnly where appropriate which maybe hard to get your head round if you're a new to C++.
If using mixin doesn't make you uncomfortable you could use tag dispatching like:
#include <utility>
#include <vector>
#include <iostream>
template <size_t I>
struct IndexedVector {
std::vector<double> v;
IndexedVector():v(10){}
};
template <size_t I>
struct tag {
int i;
};
template <size_t S, class = std::make_index_sequence<S>>
struct MixinVector;
template <size_t S, size_t... Is>
struct MixinVector<S, std::index_sequence<Is...>>: IndexedVector<Is>... {
template <size_t I>
double &operator[](tag<I> i) {
return IndexedVector<I>::v[i.i];
}
};
int main() {
MixinVector<2> mv;
mv[tag<0>{0}] = 1.0;
std::cout << mv[tag<0>{0}] << std::endl;
}
To use std::index_sequence you need however compiler supporting c++14 (you could though implement it yourself in c++11). The approach is easily expandable to any number of vectors by simple MixinVector template parameter modification.
There are many broken things, either at conceptual and design level.
Are you able to point your finger simultaneously against two distinct things? No? That's why you cannot use one index to address two distinct vector retaining their distinction.
You can do many things: whatever way to "combine" two value int one is good
by a syntactic point of view:
return m_x[i]+m_y[x] or return sin(m_x[i])*cos(m_y[i]) or return whatever_complicated_expression_you_like_much
But what's the meaning of that? The point is WHY THERE ARE TWO VECTOR IN YOUR CLASS? What do you want them to represent? What do you mean (semantically) indexing them both?
Something I can do to keep their distinction is
auto operator[](int i) const
{ return std::make_pair(m_x[i],m_y[i]); }
so that you get a std::pair<double,double> whose fist and second members are m_x[i] and m_y[i] respectively.
Or ... you can return std::vector<double>{m_x[i],m_y[i]};
About your other question: Yes, inheriting as public makes the new class able to access the protected parts: that's what protected is for.
And yes, you cam R/W: public,protected and private are about visibility, not readability and writeability. That's what const is about.
But again: what does your class represent? without such information we cannot establish what make sense and what not.
Ok, stated your comment:
you need two different funcntions: one for read (double operator[](unsigned) const) and one for write (double& operator[](unsigned) const)
If you know vectors have a known length -say 200-, that you can code an idex transforamtion like i/1000 to identify the vector and i%1000 to get the index,so that 0..199 addres the first, 1000..1199 address the second 2000..2199 address the third... etc.
Or ... you can use an std::pair<unsigned,unsigend> as the index (like operator[](const std::pair<unsigned,unsigned>& i), using i.first to identify the vector, and i.second to index into it, and then call x[{1,10}], x[{3,30}] etc.
Or ... you can chain vetor together as
if(i<m_x.size()) return m_x[i]; i-=m_x:size();
if(i<m_y.size()) return m_y[i]; i-=m_y:size();
if(i<m_z.size()) return m_z[i]; i-=m_z:size();
...
so that you index them contiguously.
But you can get more algorithmic solution using an array of vectors instead of distinct vector variables
if you have std::array<std::vector<double>,N> m; instead of m_x, m_y and m_z the above code can be...
for(auto& v: m)
{
if(i<v.size()) return v[i];
i-=v.size();
}
You can return a struct has two double
struct A{
double& x;
double& y;
A(A& r) : x(r.x), y(r.y){}
A(double& x, double& y) : x(x), y(y){}
};
class X
{
protected:
std::vector<double> m_x, m_y;
public:
A operator[](const short &i) {
A result(m_x[i], m_y[i]);
return result;
}
};
Thank for editing to #marcinj