Const and non-const c++ constructor - c++

Is there a way in C++03 (or earlier) to write a class that can either store a const or non-const pointer, and handles access appropriately? Take the usage of the non-functional "SometimesConst" class as an example:
class SometimesConst
{
public:
SometimesConst(int * buffer) : buffer(buffer) {} // Needs const qualifier?
int* get() { return buffer; } // Needs const qualifier?
void increment() { counter++; }
private:
int * buffer; // Needs const qualifier?
int counter;
};
void function(int * n, const int * c)
{
// These are both okay
SometimesConst wn(n);
SometimesConst wc(c);
// Reading the value is always allowed
printf("%d %d", wn.get()[0], wc.get()[0]);
// Can increment either object's counter
wn.increment();
wc.increment();
// Can set non-const pointer
wn.get()[0] = 5;
// Should generate a compiler error
wc.get()[0] = 5;
}
Creating a const SometimesConst would not allow modification of the counter property of the object. Can a class be designed that has compile-time const safety for input objects, only if they are passed in as const?

No, not the way you are wanting to use it. The only way to have different behavior at compile time is to have different types. However, you can make that fairly easy to use:
#include <stdio.h>
template <typename T>
class SometimesConst
{
public:
SometimesConst(T* buffer) : buffer(buffer) { }
T* get() { return buffer; }
void increment() { ++counter; }
private:
T *buffer;
int counter;
};
typedef SometimesConst<const int> IsConst;
typedef SometimesConst<int> IsNotConst;
void function(int * n, const int * c)
{
IsNotConst wn(n);
IsConst wc(c);
// Reading the value is always allowed
printf("%d %d", wn.get()[0], wc.get()[0]);
// Can increment either object's counter
wn.increment();
wc.increment();
// Can set non-const pointer
wn.get()[0] = 5;
// Should generate a compiler error
wc.get()[0] = 5;
}

The language already mostly lets you do this with a simple class; with the way const cascades to access to members (combined with mutable for the counter member, which you've indicated should always be mutable), you can provide both read-only and read-write access to a buffer quite easily:
class C
{
public:
C(int* buffer) : buffer(buffer) {}
const int* get() const { return buffer; }
int* get() { return buffer; }
void increment() const { counter++; }
private:
int* buffer;
mutable int counter;
};
void function(int* n)
{
// These are both okay
C wn(n);
const C wc(n);
// Reading the value is always allowed
printf("%d %d", wn.get()[0], wc.get()[0]);
// Can increment either object's counter
wn.increment();
wc.increment();
// Can set non-const pointer
wn.get()[0] = 5;
// Generates a compiler error
wc.get()[0] = 5;
}
What you can't do with this is neatly arrange for the class to be instantiated with either a int* or a const int*; the two lead to totally different semantics for your class, so you should split it into two if you really need that.
Fortunately, templates make this easy:
template <typename T>
class C
{
public:
C(T* buffer) : buffer(buffer) {}
const T* get() const { return buffer; }
T* get() { return buffer; }
void increment() const { counter++; }
private:
T* buffer;
mutable int counter;
};
Now a C<int> is as above, but a C<const int> only provides read-only access to the buffer, even when the C<const int> object itself is not marked as const:
void function(int* n1, const int* n2)
{
C<int> a(n1);
C<const int> b(n2);
const C<int> c(n1);
const C<const int> d(n2);
// Reading the value is always allowed
printf("%d %d %d %d",
a.get()[0], b.get()[0],
c.get()[0], d.get()[0]
);
// Incrementing the counter is always allowed
a.increment();
b.increment();
c.increment();
d.increment();
// Can set non-const pointer
a.get()[0] = 5;
// Cannot set const pointer, or const/non-const pointer behind const object
//b.get()[0] = 5;
//c.get()[0] = 5;
//d.get()[0] = 5;
}
Live demo

I think that there is a design problem if you want to store two different things which must be handled in different ways in one class. But yes, you can do it:
struct X{};
class A
{
public:
A(const X*) { cout << "const" << endl; }
A(X*) { cout << "non const" << endl; }
};
int main()
{
const X x1;
X x2;
A a1(&x1);
A a2(&x2);
}
the output is expected:
const
non const

Related

How else can I access a bit field members?

So I understand that you can't have pointers to bit-fields because the pointers can only distinguish addresses to the byte level, not bit level. References to bit-fields are also not allowed. Are there any other ways that I would be able to reference the members of the bit field indirectly? Ideally I would be able to access them following using array syntax similar to the mys1array line below. I know arrays of references are illegal but perhaps someone has some sage knowledge out there about some other mechanisms which could achieve a similar goal.
typedef struct{
unsigned short a : 5;
unsigned short b : 3;
unsigned short c : 8;
}myStruct;
class myClass{
public:
myStruct s1;
//unsigned short &mys1array[] = {&s1.a, &s1.b ,&s1.c};
};
You could use an array of function pointers that are initialized by lambdas to access each element of the bitfield with the different functions.
class myClass {
public:
myStruct s1;
static constexpr unsigned short (*accessors)(myStruct const &s)[] = {
+[](myStruct const &s) -> unsigned short { return s.a; }
// ...
};
};
With this you have to pass an instance of myStruct to the functions. Another method is using std::function and use capturing lambdas:
class myClass {
public:
myStruct s1;
std::function<unsigned short()> accessors[3];
myClass(myStruct s)
: s1(s),
accessors{
[this]() -> unsigned short { return this->s1.a; },
// ...
}
{}
// ...
};
Don't forget that with this, you have to implement copy and move constructors and assignment operators, as the lambda captures this.
You can only access bit fields through the class. You can get indirection by having a pointer or reference to the enclosing class object.
You could write a custom iterator if you wanted to iterate the bitfields within the class, but implementation of such iterator may require some explicit hard-coding since C++ lacks reflection capabilities to automate it. Here is an incomplete proof-of-concept:
struct myStruct {
unsigned short a : 5;
unsigned short b : 3;
unsigned short c : 8;
struct reference {
myStruct* parent;
unsigned char field;
operator unsigned short() {
switch(field) {
case 0: return parent->a;
case 1: return parent->b;
case 2: return parent->c;
default: assert(false);
}
}
reference& operator=(unsigned short u) {
switch(field) {
case 0: parent->a = u; return *this;
case 1: parent->b = u; return *this;
case 2: parent->c = u; return *this;
default: assert(false);
}
}
void operator++() {
++field;
}
friend auto operator<=>(const reference&, const reference&) = default;
};
struct iterator
{
//TODO add missing member definitions, const overloads etc.
reference current;
reference operator*() {
return current;
}
void operator++() {
++current;
}
friend auto operator<=>(const myStructIterator&, const myStructIterator&) = default;
};
iterator begin() {
return {this, 0};
}
iterator end() {
return {this, 3};
}
};
int main()
{
myStruct s {};
for(int i=3; auto f : s) {
f = i++;
}
for(auto f : s) {
std::cout << f << '\n';
}
}
The reference class is sufficient to represent indirection for the bit fields and the iterator allows treating the fields as an iterable range.

C++ Force const-ness of lvalue in initializer expression

I would like the compiler to enforce const-ness of an lvalue (non-reference) but don't know if this is possible in C++. An example:
int foo() { return 5; }
int main() {
// Is there anything I can add to the declaration of foo()
// that would make the following cause a compile-error?
int a = foo();
// Whereas this compiles fine.
const int a = foo();
}
This is not really possible with something like an int because you need to give access to read the int and if they can read the int then they can copy it into a non-const int.
But from your comments it sounds like what you have in reality is not an int but a more complex user defined type, some sort of container perhaps. You can easily create an immutable container. This container could be a wrapper, or alternative implementation of your existing container. It then doesn't matter if the caller uses a const or non-const variable it is still immutable.
class MyClass {
std::vector<int> data;
public:
MyClass(size_t size) : data(size) {}
int& operator[](size_t index) { return data[index]; }
int operator[](size_t index) const { return data[index]; }
size_t size() const { return data.size(); }
};
class MyClassImmutable {
MyClass mc;
public:
MyClassImmutable(MyClass&& mc) : mc(std::move(mc)){}
int operator[](size_t index) const { return mc[index]; }
size_t size() const { return mc.size(); }
const MyClass& get() const { return mc; }
};
MyClassImmutable foo() {
MyClass mc(100);
mc[10] = 3;
return mc;
}
void func(const MyClass& mc);
int main() {
MyClassImmutable mci = foo();
std::cout << mci[10] << "\n"; // Can read individual values
//mci[10] = 4; // Error immutable
func(mc.get()); // call function taking a const MyClass&
}
Live demo.
Of course there is nothing to stop the caller from copying each and every value from your immutable container and inserting them into a mutable container.
Edit: An alternative approach might be to return a smart pointer-to-const. The only downside is you have to pay for a dynamic memory allocation:
std::unique_ptr<const MyClass> foo() {
auto mc = std::make_unique<MyClass>(100);
(*mc)[10] = 3;
return mc;
}
void func(const MyClass& mc);
int main() {
auto mc = foo();
std::cout << (*mc)[10] << "\n"; // Can read individual values
//(*mc)[10] = 4; // Error const
func(*mc); // can pass to a function taking a const MyClass&
}
It's not possible. foo() has no way of knowing about the type of the left hand side of the assignment, because when the assignment itself happens, foo() is already evaluated. The best you could hope for is to change the return value, to try and cause a type-based error on the initialization:
#include <type_traits>
struct my_int {
const int m;
template<typename T, typename std::enable_if<std::is_const<T>::value, T>::type* = nullptr>
constexpr operator T() const {return m;}
};
constexpr my_int foo() { return {5};}
int main() {
const int a = foo();
int b = foo();
}
Live example
But this will also not work, because the typename in the template will never be substitued by a const-qualified type (in this specific case, it will be int for both lines in main()).
As the following is possible
const int x = 4;
int y = x;
the C++ language will not provide such a mechanism.
Remains making a int const by a macro mechanism.
#define int_const_foo(var) const int var = ___foo()
int_const_foo(a);
Drawback: foo cannot be hidden, and the syntax is no longer C style.

C++ - How to update pointer (or members) between instances of same class

I have a simple class which consists of a void pointer and an int (this is some sort of a boost::Variant educational project).
I also have a working copy constructor and a destructor.
But what grinds my gears is, how I would accomplish something like this:
Container cont1("some value"); //simple construction
Container cont2;
cont2.createLink(cont1); //this should initialize members with a reference (or something alike)
std::cout<<cont1; //yields "some value"
cont2.set(20); //setting this container should update the original container too, since I initialized with a reference (or smth alike)
std::cout<<cont1; //yields 20
This is the simplified version of the class:
class Container {
public:
Container(){}
Container(const std::string &val){var.type = STRING; var.data = new std::string(val);}
Container(int val){ /* same for int */}
Container(const Container &val){ /* do a memory copy */}
void set(int val){ /* set the value if type matches, otherwise allocate a new pointer */}
void set(const std::string &val){ /* the same as for int */}
void createLink(const Container &val){ /* somehow assign a reference or whatsoever */}
private:
typedef struct VAR {
int type = 0;
void *data = NULL; }
VAR var;
}
If I set the value of cont2 to a string (i.e. the same data type it holds at the moment), everything is fine, because the set would not allocate a new pointer and rather assign a new value.
But how do I make sure the pointer of cont1 updates if I assign a different value to cont2 and therefore have to allocate a new pointer?
Would I need something like shared_pointer?
Thanks for any insight!
EDIT:
I changed to function name to make it more clear what should happen.
There is a solution that only involves straight OO. You could create an interface for your variant type, and use double indirection to the variant instance to allow linked containers to share the same variant instance.
The reason double indirection is required is because of the way you want the set() method to automatically allocate a new variant instance if the new type doesn't match the original type. If we simply shared a pointer to the variant from both containers, then after set() creates a new variant instance, each container would be referring to different instances again.
To get around that, we can use a pointer to a pointer to a variant in the container instead.
Here is a possible way to define your variant interface, and how it could be subclassed:
typedef std::ostream Out;
struct BadType {};
struct Var {
virtual ~Var () = default;
virtual Out & print (Out &os) { return os << "(BadType)"; }
virtual void set (int) { throw BadType(); }
virtual void set (const std::string &) { throw BadType(); }
};
struct VarInteger : Var {
int data;
VarInteger (int v) : data(v) {}
Out & print (Out &os) { return os << data; }
void set (int v) throw() { data = v; }
};
struct VarString : Var {
std::string data;
VarString (const std::string &v) : data(v) {}
Out & print (Out &os) { return os << data; }
void set (const std::string &v) throw() { data = v; }
};
Here is how you could define your pointer to pointer, and how they could be initialized:
typedef std::shared_ptr<Var> VarPtr;
std::shared_ptr<VarPtr> varptr_;
static VarPtr make_var () { return std::make_shared<Var>(); }
static VarPtr make_var (int v) { return std::make_shared<VarInteger>(v); }
static VarPtr make_var (const std::string &v) {
return std::make_shared<VarString>(v);
}
VarPtr & var () { return *varptr_; }
const VarPtr & var () const { return *varptr_; }
Container () : varptr_(std::make_shared<VarPtr>(make_var())) {}
Container (int v) : varptr_(std::make_shared<VarPtr>(make_var(v))) {}
Container (const std::string &v)
: varptr_(std::make_shared<VarPtr>(make_var(v))) {}
And here is how your set() methods and createLink() method could be implemented.
void set (int v) {
try { var()->set(v); }
catch (BadType) { var() = make_var(v); }
}
void set (const std::string &v) {
try { var()->set(v); }
catch (BadType) { var() = make_var(v); }
}
void createLink (const Container &val) { varptr_ = val.varptr_; }
Demo
How about the following. Of course createLink cannot not take a const reference so I made it to take a non-const pointer.
class Container {
const int STRING = 0x0000001;
const int INT = 0x0000002;
const int LINK = 0x8000000;
public:
...
void set(int val){...}
void set(const std::string &val)
{
if (var.type == LINK)
{
reinterpret_cast<Container*>(var.data)->set(val);
}
else
...
}
void createLink(Container* val)
{
var.data = val;
var.type = LINK;
}
private:
typedef struct VAR {
int type = 0;
void *data = NULL;
};
VAR var;
};
There are a some important points to think about - relative lifetimes of the link and the linked is the most obvious one.

How can I make this function act like an l-value?

Why can't I use the function ColPeekHeight() as an l-value?
class View
{
public:
int ColPeekHeight(){ return _colPeekFaceUpHeight; }
void ColPeekHeight( int i ) { _colPeekFaceUpHeight = i; }
private:
int _colPeekFaceUpHeight;
};
...
{
if( v.ColPeekHeight() > 0.04*_heightTable )
v.ColPeekHeight()-=peek;
}
The compiler complains at v.ColPeekHeight()-=peek. How can I make ColPeekHeight() an l-value?
Return the member variable by reference:
int& ColPeekHeight(){ return _colPeekFaceUpHeight; }
To make your class a good one, define a const version of the function:
const int& ColPeekHeight() const { return _colPeekFaceUpHeight; }
when I declare the function with the
two consts
When you want to pass an object into a function that you don't expect it to modify your object. Take this example:
struct myclass
{
int x;
int& return_x() { return x; }
const int& return_x() const { return x; }
};
void fun(const myclass& obj);
int main()
{
myclass o;
o.return_x() = 5;
fun(o);
}
void fun(const myclass& obj)
{
obj.return_x() = 5; // compile-error, a const object can't be modified
std::cout << obj.return_x(); // OK, No one is trying to modify obj
}
If you pass your objects to functions, then you might not want to change them actually all the time. So, to guard your self against this kind of change, you declare const version of your member functions. It doesn't have to be that every member function has two versions! It depends on the function it self, is it modifying function by nature :)
The first const says that the returned value is constant. The second const says that the member function return_x doesn't change the object(read only).
It can be rewritten like:
class View
{
public:
int GetColPeekHeight() const { return _colPeekFaceUpHeight; }
void SetColPeekHeight( int i ) { _colPeekFaceUpHeight = i; }
private:
int _colPeekFaceUpHeight;
};
...
{
cph = v.GetColPeekHeight();
if ( cph > 0.04 * _heightTable )
v.SetColPeekHeight( cph - peek );
}

Member pointer to array element

It's possible to define a pointer to a member and using this later on:
struct foo
{
int a;
int b[2];
};
int main()
{
foo bar;
int foo::* aptr=&foo::a;
bar.a=1;
std::cout << bar.*aptr << std::endl;
}
Now I need to have a pointer to a specific element of an array, so normally I'd write
int foo::* bptr=&(foo::b[0]);
However, the compiler just complains about an "invalid use of non-static data member 'foo::b'"
Is it possible to do this at all (or at least without unions)?
Edit: I need a pointer to a specific element of an array, so int foo::* ptr points to the second element of the array (foo::b[1]).
Yet another edit: I need to access the element in the array by bar.*ptr=2, as the pointer gets used somewhere else, so it can't be called with bar.*ptr[1]=2 or *ptr=2.
However, the compiler just complains about an "invalid use of non-static data member 'foo::b'"
This is because foo::a and foo::b have different types. More specifically, foo::b is an array of size 2 of ints. Your pointer declaration has to be compatible i.e:
int (foo::*aptr)[2]=&foo::b;
Is it possible to do this at all (or at least without unions)?
Yes, see below:
struct foo
{
int a;
int b[2];
};
int main()
{
foo bar;
int (foo::*aptr)[2]=&foo::b;
/* this is a plain int pointer */
int *bptr=&((bar.*aptr)[1]);
bar.a=1;
bar.b[0] = 2;
bar.b[1] = 11;
std::cout << (bar.*aptr)[1] << std::endl;
std::cout << *bptr << std::endl;
}
Updated post with OP's requirements.
The problem is that, accessing an item in an array is another level of indirection from accessing a plain int. If that array was a pointer instead you wouldn't expect to be able to access the int through a member pointer.
struct foo
{
int a;
int *b;
};
int main()
{
foo bar;
int foo::* aptr=&(*foo::b); // You can't do this either!
bar.a=1;
std::cout << bar.*aptr << std::endl;
}
What you can do is define member functions that return the int you want:
struct foo
{
int a;
int *b;
int c[2];
int &GetA() { return a; } // changed to return references so you can modify the values
int &Getb() { return *b; }
template <int index>
int &GetC() { return c[index]; }
};
typedef long &(Test::*IntAccessor)();
void SetValue(foo &f, IntAccessor ptr, int newValue)
{
cout << "Value before: " << f.*ptr();
f.*ptr() = newValue;
cout << "Value after: " << f.*ptr();
}
int main()
{
IntAccessor aptr=&foo::GetA;
IntAccessor bptr=&foo::GetB;
IntAccessor cptr=&foo::GetC<1>;
int local;
foo bar;
bar.a=1;
bar.b = &local;
bar.c[1] = 2;
SetValue(bar, aptr, 2);
SetValue(bar, bptr, 3);
SetValue(bar, cptr, 4);
SetValue(bar, &foo::GetC<0>, 5);
}
Then you at least have a consistent interface to allow you to change different values for foo.
2020 update, with actual solution:
The Standard does currently not specify any way to actually work with the member pointers in a way that would allow arithmetics or anything to get the pointer to the "inner" array element
OTOH, the standard library now has all the necessities to patch the appropriate member pointer class yourself, even with the array element access.
First, the member pointers are usually implemented as "just offsets", although quite scary. Let's see an example (on g++9, arch amd64):
struct S { int a; float b[10]; };
float(S::*mptr)[10] = &S::b;
*reinterpret_cast<uintptr_t *>(&mptr) //this is 4
int S::*iptr = &S::a;
*reinterpret_cast<uintptr_t *>(&iptr) //this is 0
iptr = nullptr;
*reinterpret_cast<uintptr_t *>(&iptr) //this seems to be 18446744073709551615 on my box
Instead you can make a bit of a wrapper (it's quite long but I didn't want to remove the convenience operators):
#include <type_traits>
template<class M, typename T>
class member_ptr
{
size_t off_;
public:
member_ptr() : off_(0) {}
member_ptr(size_t offset) : off_(offset) {}
/* member access */
friend const T& operator->*(const M* a, const member_ptr<M, T>& p)
{ return (*a)->*p; }
friend T& operator->*(M* a, const member_ptr<M, T>& p)
{ return (*a)->*p; }
/* operator.* cannot be overloaded, so just take the arrow again */
friend const T& operator->*(const M& a, const member_ptr<M, T>& p)
{ return *reinterpret_cast<const T*>(reinterpret_cast<const char*>(&a) + p.off_); }
friend T& operator->*(M& a, const member_ptr<M, T>& p)
{ return *reinterpret_cast<T*>(reinterpret_cast<char*>(&a) + p.off_); }
/* convert array access to array element access */
member_ptr<M, typename std::remove_extent<T>::type> operator*() const
{ return member_ptr<M, typename std::remove_extent<T>::type>(off_); }
/* the same with offset right away */
member_ptr<M, typename std::remove_extent<T>::type> operator[](size_t offset) const
{ return member_ptr<M, typename std::remove_extent<T>::type>(off_)+offset; }
/* some operators */
member_ptr& operator++()
{ off_ += sizeof(T); return *this; };
member_ptr& operator--()
{ off_ -= sizeof(T); return *this; };
member_ptr operator++(int)
{ member_ptr copy; off_ += sizeof(T); return copy; };
member_ptr operator--(int)
{ member_ptr copy; off_ -= sizeof(T); return copy; };
member_ptr& operator+=(size_t offset)
{ off_ += offset * sizeof(T); return *this; }
member_ptr& operator-=(size_t offset)
{ off_ -= offset * sizeof(T); return *this; }
member_ptr operator+(size_t offset) const
{ auto copy = *this; copy += offset; return copy; }
member_ptr operator-(size_t offset) const
{ auto copy = *this; copy -= offset; return copy; }
size_t offset() const { return off_; }
};
template<class M, typename T>
member_ptr<M, T> make_member_ptr(T M::*a)
{ return member_ptr<M, T>(reinterpret_cast<uintptr_t>(&(((M*)nullptr)->*a)));}
Now we can make the pointer to the array element directly:
auto mp = make_member_ptr(&S::b)[2];
S s;
s->*mp = 123.4;
// s.b[2] is now expectably 123.4
Finally, if you really, really like materialized references, you may get a bit haskell-lensish and make them compose:
// in class member_ptr, note transitivity of types M -> T -> TT:
template<class TT>
member_ptr<M,TT> operator+(const member_ptr<T,TT>&t)
{ return member_ptr<M,TT>(off_ + t.offset()); }
// test:
struct A { int a; };
struct B { A arr[10]; };
B x;
auto p = make_member_ptr(&B::arr)[5] + make_member_ptr(&A::a)
x->*p = 432.1;
// x.arr[5].a is now expectably 432.1
typedef int (foo::*b_member_ptr)[2];
b_member_ptr c= &foo::b;
all works.
small trick for member and function pointers usage.
try to write
char c = &foo::b; // or any other function or member pointer
and in compiller error you will see expected type, for your case int (foo::*)[2].
EDIT
I'm not sure that what you want is legal without this pointer. For add 1 offset to your pointer you should get pointer on array from your pointer on member array. But you can dereference member pointer without this.
You can't do that out of the language itself. But you can with boost. Bind a functor to some element of that array and assign it to a boost::function:
#include <boost/lambda/lambda.hpp>
#include <boost/lambda/bind.hpp>
#include <boost/function.hpp>
#include <iostream>
struct test {
int array[3];
};
int main() {
namespace lmb = boost::lambda;
// create functor that returns test::array[1]
boost::function<int&(test&)> f;
f = lmb::bind(&test::array, lmb::_1)[1];
test t = {{ 11, 22, 33 }};
std::cout << f(t) << std::endl; // 22
f(t) = 44;
std::cout << t.array[1] << std::endl; // 44
}
I'm not sure if this will work for you or not, but I wanted to do a similar thing and got around it by approaching the problem from another direction. In my class I had several objects that I wanted to be accessible via a named identifier or iterated over in a loop. Instead of creating member pointers to the objects somewhere in the array, I simply declared all of the objects individually and created a static array of member pointers to the objects.
Like so:
struct obj
{
int somestuff;
double someotherstuff;
};
class foo
{
public:
obj apples;
obj bananas;
obj oranges;
static obj foo::* fruit[3];
void bar();
};
obj foo::* foo::fruit[3] = { &foo::apples, &foo::bananas, &foo::oranges };
void foo::bar()
{
apples.somestuff = 0;
(this->*(fruit[0])).somestuff = 5;
if( apples.somestuff != 5 )
{
// fail!
}
else
{
// success!
}
}
int main()
{
foo blee;
blee.bar();
return 0;
}
It seems to work for me. I hope that helps.