https://stackoverflow.com/a/332086/462608
modifying a formerly const value is only undefined if the original variable is const
...
if you use it to take the const off a reference to something that wasn't declared with const, it is safe.
...
This can be useful when overloading member functions based on const, for instance. It can also be used to add const to an object, such as to call a member function overload.
I am unable to understand the meanings of the above quotes. I request you to give me examples to practically show what these quotes mean.
Regarding your first two quotes:
void do_not_do_this(const int& cref) {
const_cast<int&>(cref) = 42;
}
int main() {
int a = 0;
// "if you use it to take the const off a reference
// to something that wasn't declared with const, it is safe."
do_not_do_this(a); // well-defined
// a is now 42.
// "modifying a formerly const value is only
// undefined if the original variable is const"
const int b = 0;
do_not_do_this(a); // undefined behavoiur
}
Regarding your final quote:
// "This can be useful when overloading member functions based
// on const, for instance. It can also be used to add const
// to an object, such as to call a member function overload."
class A {
const int& get() const
{
// ... some common logic for const and
// non-const overloads.
return a_;
}
int& get() {
// Since this get() overload is non-const, the object itself
// is non-const in this scope. Moreover, the a_ member
// is non-const, and thus casting away the const of the return
// from the const get() (after 'this' has been casted to
// const) is safe.
A const * const c_this = this;
return const_cast<int&>(c_this->get());
}
private:
int a_{0};
}
How about this:
#include <iostream>
void foo(const int& ub, const int& ok)
{
const_cast<int&>(ub) = 0.0; // undefined behaviour - the original object is const
const_cast<int&>(ok) = 1.0; // this is fine - the original object is not const
}
int main()
{
const int ub = 1.0;
int ok = 0.0;
foo(ub, ok);
std::cout << ub << " " << ok << std::ends;
}
Note the output on common compilers is 1 1: the rationale being that the compiler knows that ub cannot change in main so it substitutes 1 for ub in the std::cout call.
Your third paragraph is alluding to a function body of a non-const member function calling the const version as a means of obviating code repetition.
What would be a valid use case for a signature like this?:
T&& foo();
Or is the rvalue ref only intended for use as argument?
How would one use a function like this?
T&& t = foo(); // is this a thing? And when would t get destructed?
For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.
One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have
class foo
{
std::vector<int> bar;
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() { return bar; }
};
If you do
auto vec = foo(10).get_vec();
you have to copy because get_vec returns an lvalue. If you instead use
class foo
{
std::vector<int> bar;
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() & { return bar; }
std::vector<int>&& get_vec() && { return std::move(bar); }
};
Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.
T&& t = foo(); // is this a thing? And when would t get destructed?
An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:
T& foo();
T& t = foo(); // when is t destroyed?
The answer is that t is still valid to use as long as the object is refers to lives.
The same answer still applies to you rvalue reference example.
But... does it make sense to return an rvalue reference?
Sometimes, yes. But very rarely.
consider this:
std::vector<int> v = ...;
// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));
// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(std::move(parameters));
// useful for calling function in generic context without copying
consume(std::get<0>(std::move(parameters)));
So yes there are example. Here, another interesting one:
struct wrapper {
auto operator*() & -> Heavy& {
return heavy;
}
auto operator*() && -> Heavy&& {
return std::move(heavy);
}
private:
Heavy instance;
};
// by value
void use_heavy(Heavy);
// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());
I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:
class Logger
{
public:
void log(const char* msg){
logs.append(msg);
}
std::vector<std::string>&& dumpLogs(){
return std::move(logs);
}
private:
std::vector<std::string> logs;
};
But I admit I made this up now, I never actually used it and it also can be done like this:
std::vector<std::string> dumpLogs(){
auto dumped_logs = logs;
return dumped_logs;
}
According to §7.1.5.1/4:
Except that any class member declared mutable (7.1.1) can be modified, any attempt to modify a const object during its lifetime (3.8) results in undefined behavior.
So my question becomes: when is an object a const object?
In particular, is a const member in a non-const object considered a const object?
class Foo {
const Bar bar;
void replaceBar(Bar bar2) {
*(const_cast<Bar *>&bar) = bar2; // Undefined behavior?
}
}
This comes up because I have an immutable class (all fields are const), but I want to have a move constructor, which technically modifies the value passed in. I'm ok with "cheating" in that case, since it doesn't break logical constness.
The simple rule is: it is ok to cast away constness if the original object is not const. So if you have a non-cont object and, say, you pass the const reference to it to a function, it is legal to cast away constness in the function.
In your example the original object is const, so casting constness away is undefined behaviour.
Let us make this a full example:
struct Bar { int x; };
struct Foo {
const Bar bar;
Foo( int x ):bar(x) {}
void replaceBar(Bar bar2) {
*(const_cast<Bar *>&bar) = bar2; // Undefined behavior?
}
};
now, let us break the world.
int main() {
Foo f(3);
Bar b = {2};
f.replaceBar(b);
std::cout << f.bar.x << "\n";
}
the above can and probably should output 3, because a const object Bar was created with x=3. The compiler can, and should, assume that the const object will be unchanged throughout its lifetime.
Let's break the world more:
struct Bar {
int* x;
Bar(int * p):x(p) {}
~Bar(){ if (x) delete x; }
Bar(Bar&& o):x(o.x){o.x=nullptr;}
Bar& operator=(Bar&& o){
if (x) delete x;
x = o.x;
o.x = nullptr;
}
Bar(Bar const&)=delete;
Bar& operator=(Bar const&)=delete;
};
struct Foo {
const Bar bar;
Foo( int* x ):bar(x) {}
void replaceBar(Bar bar2) {
*(const_cast<Bar *>&bar) = bar2; // Undefined behavior?
}
};
now the same game can result in the compiler deleting something twice.
int main() {
int* p1 = new int(3);
Foo f( p1 );
Bar b( new int(2) );
f.replaceBar(std::move(b));
}
and the compiler will delete p1 once within replaceBar, and should delete it also at the end of main. It can do this, because you guaranteed that f.bar.x would remain unchanged (const) until the end of its scope, then you violated that promise in replaceBar.
Now, this is just things the compiler has reason to do: the compiler can literally do anything once you have modified an object that was declared const, as you have invoked undefined behavior. Nasal demons, time travel -- anything is up for grabs.
Compilers use the fact that some behavior is undefined (aka, not allowed) to optimize.
The following pattern has arisen in a program I'm writing. I hope it's not too contrived, but it manages to mutate a Foo object in the const method Foo::Questionable() const, without use of any const_cast or similar. Basically, Foo stores a reference to FooOwner and vice versa, and in Questionable(), Foo manages to modify itself in a const method by calling mutate_foo() on its owner. Questions follow the code.
#include "stdafx.h"
#include <iostream>
using namespace std;
class FooOwner;
class Foo {
FooOwner& owner;
int data;
public:
Foo(FooOwner& owner_, int data_)
: owner(owner_),
data(data_)
{
}
void SetData(int data_)
{
data = data_;
}
int Questionable() const; // defined after FooOwner
};
class FooOwner {
Foo* pFoo;
public:
FooOwner()
: pFoo(NULL)
{}
void own(Foo& foo)
{
pFoo = &foo;
}
void mutate_foo()
{
if (pFoo != NULL)
pFoo->SetData(0);
}
};
int Foo::Questionable() const
{
owner.mutate_foo(); // point of interest
return data;
}
int main()
{
FooOwner foo_owner;
Foo foo(foo_owner, 0); // foo keeps reference to foo_owner
foo_owner.own(foo); // foo_owner keeps pointer to foo
cout << foo.Questionable() << endl; // correct?
return 0;
}
Is this defined behavior? Should Foo::data be declared mutable? Or is this a sign I'm doing things fatally wrong? I'm trying to implement a kind of lazy-initialised 'data' which is only set when requested, and the following code compiles fine with no warnings, so I'm a little nervous I'm in UB land.
Edit: the const on Questionable() only makes immediate members const, and not the objects pointed to or referenced by the object. Does this make the code legal? I'm confused between the fact that in Questionable(), this has the type const Foo*, and further down the call stack, FooOwner legitimately has a non-const pointer it uses to modify Foo. Does this mean the Foo object can be modified or not?
Edit 2: perhaps an even simpler example:
class X {
X* nonconst_this; // Only turns in to X* const in a const method!
int data;
public:
X()
: nonconst_this(this),
data(0)
{
}
int GetData() const
{
nonconst_this->data = 5; // legal??
return data;
}
};
Consider the following:
int i = 3;
i is an object, and it has the type int. It is not cv-qualified (is not const or volatile, or both.)
Now we add:
const int& j = i;
const int* k = &i;
j is a reference which refers to i, and k is a pointer which points to i. (From now on, we simply combine "refer to" and "points to" to just "points to".)
At this point, we have two cv-qualified variables, j and k, that point to a non-cv-qualified object. This is mentioned in §7.1.5.1/3:
A pointer or reference to a cv-qualified type need not actually point or refer to a cv-qualified object, but it is treated as if it does; a const-qualified access path cannot be used to modify an object even if the object referenced is a non-const object and can be modified through some other access path. [Note: cv-qualifiers are supported by the type system so that they cannot be subverted without casting (5.2.11). ]
What this means is that a compiler must respect that j and k are cv-qualified, even though they point to a non-cv-qualified object. (So j = 5 and *k = 5 are illegal, even though i = 5 is legal.)
We now consider removing the const from those:
const_cast<int&>(j) = 5;
*const_cast<int*>(k) = 5;
This is legal (§refer to 5.2.11), but is it undefined behavior? No. See §7.1.5.1/4:
Except that any class member declared mutable (7.1.1) can be modified, any attempt to modify a const object during its lifetime (3.8) results in undefined behavior.
Emphasis mine.
Remember that i is not const and that j and k both point to i. All we've done is tell the type system to remove the const-qualifier from the type so we can modify the pointed to object, and then modified i through those variables.
This is exactly the same as doing:
int& j = i; // removed const with const_cast...
int* k = &i; // ..trivially legal code
j = 5;
*k = 5;
And this is trivially legal. We now consider that i was this instead:
const int i = 3;
What of our code now?
const_cast<int&>(j) = 5;
*const_cast<int*>(k) = 5;
It now leads to undefined behavior, because i is a const-qualified object. We told the type system to remove const so we can modify the pointed to object, and then modified a const-qualified object. This is undefined, as quoted above.
Again, more apparent as:
int& j = i; // removed const with const_cast...
int* k = &i; // ...but this is not legal!
j = 5;
*k = 5;
Note that simply doing this:
const_cast<int&>(j);
*const_cast<int*>(k);
Is perfectly legal and defined, as no const-qualified objects are being modified; we're just messing with the type-system.
Now consider:
struct foo
{
foo() :
me(this), self(*this), i(3)
{}
void bar() const
{
me->i = 5;
self.i = 5;
}
foo* me;
foo& self;
int i;
};
What does const on bar do to the members? It makes access to them go through something called a cv-qualified access path. (It does this by changing the type of this from T* const to cv T const*, where cv is the cv-qualifiers on the function.)
So what are the members types during the execution of bar? They are:
// const-pointer-to-non-const, where the pointer points cannot be changed
foo* const me;
// foo& const is ill-formed, cv-qualifiers do nothing to reference types
foo& self;
// same as const int
int const i;
Of course, the types are irrelevant, as the important thing is the const-qualification of the pointed to objects, not the pointers. (Had k above been const int* const, the latter const is irrelevant.) We now consider:
int main()
{
foo f;
f.bar(); // UB?
}
Within bar, both me and self point to a non-const foo, so just like with int i above we have well-defined behavior. Had we had:
const foo f;
f.bar(); // UB!
We would have had UB, just like with const int, because we would be modifying a const-qualified object.
In your question, you have no const-qualified objects, so you have no undefined behavior.
And just to add an appeal to authority, consider the const_cast trick by Scott Meyers, used to recycle a const-qualified function in a non-const function:
struct foo
{
const int& bar() const
{
int* result = /* complicated process to get the resulting int */
return *result;
}
int& bar()
{
// we wouldn't like to copy-paste a complicated process, what can we do?
}
};
He suggests:
int& bar(void)
{
const foo& self = *this; // add const
const int& result = self.bar(); // call const version
return const_cast<int&>(result); // take off const
}
Or how it's usually written:
int& bar(void)
{
return const_cast<int&>( // (3) remove const from result
static_cast<const foo&>(*this) // (1) add const to this
.bar() // (2) call const version
);
}
Note this is, again, perfectly legal and well-defined. Specifically, because this function must be called on a non-const-qualified foo, we are perfectly safe in stripping the const-qualification from the return type of int& boo() const.
(Unless someone shoots themselves with a const_cast + call in the first place.)
To summarize:
struct foo
{
foo(void) :
i(),
self(*this), me(this),
self_2(*this), me_2(this)
{}
const int& bar() const
{
return i; // always well-formed, always defined
}
int& bar() const
{
// always well-formed, always well-defined
return const_cast<int&>(
static_cast<const foo&>(*this).
bar()
);
}
void baz() const
{
// always ill-formed, i is a const int in baz
i = 5;
// always ill-formed, me is a foo* const in baz
me = 0;
// always ill-formed, me_2 is a const foo* const in baz
me_2 = 0;
// always well-formed, defined if the foo pointed to is non-const
self.i = 5;
me->i = 5;
// always ill-formed, type points to a const (though the object it
// points to may or may not necessarily be const-qualified)
self_2.i = 5;
me_2->i = 5;
// always well-formed, always defined, nothing being modified
// (note: if the result/member was not an int and was a user-defined
// type, if it had its copy-constructor and/or operator= parameter
// as T& instead of const T&, like auto_ptr for example, this would
// be defined if the foo self_2/me_2 points to was non-const
int r = const_cast<foo&>(self_2).i;
r = const_cast<foo* const>(me_2)->i;
// always well-formed, always defined, nothing being modified.
// (same idea behind the non-const bar, only const qualifications
// are being changed, not any objects.)
const_cast<foo&>(self_2);
const_cast<foo* const>(me_2);
// always well-formed, defined if the foo pointed to is non-const
// (note, equivalent to using self and me)
const_cast<foo&>(self_2).i = 5;
const_cast<foo* const>(me_2)->i = 5;
// always well-formed, defined if the foo pointed to is non-const
const_cast<foo&>(*this).i = 5;
const_cast<foo* const>(this)->i = 5;
}
int i;
foo& self;
foo* me;
const foo& self_2;
const foo* me_2;
};
int main()
{
int i = 0;
{
// always well-formed, always defined
int& x = i;
int* y = &i;
const int& z = i;
const int* w = &i;
// always well-formed, always defined
// (note, same as using x and y)
const_cast<int&>(z) = 5;
const_cast<int*>(w) = 5;
}
const int j = 0;
{
// never well-formed, strips cv-qualifications without a cast
int& x = j;
int* y = &j;
// always well-formed, always defined
const int& z = i;
const int* w = &i;
// always well-formed, never defined
// (note, same as using x and y, but those were ill-formed)
const_cast<int&>(z) = 5;
const_cast<int*>(w) = 5;
}
foo x;
x.bar(); // calls non-const, well-formed, always defined
x.bar() = 5; // calls non-const, which calls const, removes const from
// result, and modifies which is defined because the object
// pointed to by the returned reference is non-const,
// because x is non-const.
x.baz(); // well-formed, always defined
const foo y;
y.bar(); // calls const, well-formed, always defined
const_cast<foo&>(y).bar(); // calls non-const, well-formed,
// always defined (nothing being modified)
const_cast<foo&>(y).bar() = 5; // calls non-const, which calls const,
// removes const from result, and
// modifies which is undefined because
// the object pointed to by the returned
// reference is const, because y is const.
y.baz(); // well-formed, always undefined
}
I refer to the ISO C++03 standard.
IMO, you are not doing anything technically wrong. May-be it would be simpler to understand if the member was a pointer.
class X
{
Y* m_ptr;
void foo() const {
m_ptr = NULL; //illegal
*m_ptr = 42; //legal
}
};
const makes the pointer const, not the pointee.
Consider the difference between:
const X* ptr;
X* const ptr; //this is what happens in const member functions
As to references, since they can't be reseated anyway, the const keyword on the method has no effect whatsoever on reference members.
In your example, I don't see any const objects, so you are not doing anything bad, just exploiting a strange loophole in the way const correctness works in C++.
Without actually getting to whether it is/should/could be allowed, I would greatly advice against it. There are mechanisms in the language for what you want to achieve that don't require writing obscure constructs that will most probably confuse other developers.
Look into the mutable keyword. That keyword can be used to declare members that can be modified within const member methods as they do not affect the perceivable state of the class. Consider class that gets initialized with a set of parameters and performs a complex expensive calculation that may not be needed always:
class ComplexProcessor
{
public:
void setInputs( int a, int b );
int getValue() const;
private:
int complexCalculation( int a, int b );
int result;
};
A possible implementation is adding the result value as a member and calculating it for each set:
void ComplexProcessor::setInputs( int a, int b ) {
result = complexCalculation( a, b );
}
But this means that the value is calculated in all sets, whether it is needed or not. If you think on the object as a black box, the interface just defines a method to set the parameters and a method to retrieve the calculated value. The instant when the calculation is performed does not really affect the perceived state of the object --as far as the value returned by the getter is correct. So we can modify the class to store the inputs (instead of the outputs) and calculate the result only when needed:
class ComplexProcessor2 {
public:
void setInputs( int a, int b ) {
a_ = a; b_ = b;
}
int getValue() const {
return complexCalculation( a_, b_ );
}
private:
int complexCalculation( int a, int b );
int a_,b_;
};
Semantically the second class and the first class are equivalent, but now we have avoided to perform the complex calculation if the value is not needed, so it is an advantage if the value is only requested in some cases. But at the same time it is a disadvantage if the value is requested many times for the same object: each time the complex calculation will be performed even if the inputs have not changed.
The solution is caching the result. For that we can the result to the class. When the result is requested, if we have already calculated it, we only need to retrieve it, while if we do not have the value we must calculate it. When the inputs change we invalidate the cache. This is when the mutable keyword comes in handy. It tells the compiler that the member is not part of the perceivable state and as such it can be modified within a constant method:
class ComplexProcessor3 {
public:
ComplexProcessor3() : cached_(false) {}
void setInputs( int a, int b ) {
a_ = a; b_ = b;
cached_ = false;
}
int getValue() const {
if ( !cached_ ) {
result_ = complexCalculation( a_, b_ );
cached_ = true;
}
return result_;
}
private:
int complexCalculation( int a, int b );
int a_,b_;
// This are not part of the perceivable state:
mutable int result_;
mutable bool cached_;
};
The third implementation is semantically equivalent to the two previous versions, but avoid having to recalculate the value if the result is already known --and cached.
The mutable keyword is needed in other places, like in multithreaded applications the mutex in classes are often marked as mutable. Locking and unlocking a mutex are mutating operations for the mutex: its state is clearly changing. Now, a getter method in an object that is shared among different threads does not modify the perceived state but must acquire and release the lock if the operation has to be thread safe:
template <typename T>
class SharedValue {
public:
void set( T v ) {
scoped_lock lock(mutex_);
value = v;
}
T get() const {
scoped_lock lock(mutex_);
return value;
}
private:
T value;
mutable mutex mutex_;
};
The getter operation is semantically constant, even if it needs to modify the mutex to ensure single threaded access to the value member.
The const keyword is only considered during compile time checks. C++ provides no facilities to protect your class against any memory access, which is what you are doing with your pointer/reference. Neither the compiler nor the runtime can know if your pointer points to an instance that you declared const somewhere.
EDIT:
Short example (might not compile):
// lets say foo has a member const int Foo::datalength() const {...}
// and a read only acces method const char data(int idx) const {...}
for (int i; i < foo.datalength(); ++i)
{
foo.questionable(); // this will most likely mess up foo.datalength !!
std::cout << foo.data(i); // HERE BE DRAGONS
}
In this case, the compiler might decide, ey, foo.datalength is const,
and the code inside the loop promised not to change foo, so I have to evaluate
datalength only once when I enter the loop. Yippie!
And if you try to debug this error, which will most likely only turn up if you compile with optimizations (not in the debug builds) you will drive yourself crazy.
Keep the promises! Or use mutable with your braincells on high alert!
You have reached circular dependencies. See FAQ 39.11 And yes, modifying const data is UB even if you have circumvented the compiler. Also, you are severely impairing the compiler's capacity to optimize if you don't keep your promises (read: violate const).
Why is Questionable const if you know that you will modify it via a call to its owner? Why does the owned object need to know about the owner? If you really really need to do that then mutable is the way to go. That is what it is there for -- logical constness (as opposed to strict bit level constness).
From my copy of the draft n3090:
9.3.2 The this pointer [class.this]
1 In the body of a non-static (9.3) member function, the keyword this is an rvalue a prvalue expression whose
value is the address of the object for which the function is called. The type of this in a member function
of a class X is X*. If the member function is declared const, the type of this is const X*, if the member
function is declared volatile, the type of this is volatile X*, and if the member function is declared
const volatile, the type of this is const volatile X*.
2 In a const member function, the object for which the function is called is accessed through a const access
path; therefore, a const member function shall not modify the object and its non-static data members.
[Note emphasis mine].
On UB:
7.1.6.1 The cv-qualifiers
3 A pointer or reference to a cv-qualified type need not actually
point or refer to a cv-qualified
object, but it is treated as if it
does; a const-qualified access path
cannot be used to modify an object
even if the object referenced is a
non-const object and can be modified
through some other access path. [
Note: cv-qualifiers are supported by
the type system so that they cannot be
subverted without casting (5.2.11).
—end note ]
4 Except that any class
member declared mutable (7.1.1) can be
modified, any attempt to modify a
const object during its lifetime (3.8)
results in undefined behavior.
Please consider the following code.
struct foo
{
};
template<typename T>
class test
{
public:
test() {}
const T& value() const
{
return f;
}
private:
T f;
};
int main()
{
const test<foo*> t;
foo* f = t.value();
return 0;
}
t is a const variable and value() is a constant member-function which returns const T&. AFAIK, a const type is not assignable to a non-const type. But how foo* f = t.value(); compiles well. How this is happening and how can I ensure value() can be only assigned to const foo*?
Edit
I found that, this is happening on when templates are used. Following code works as expected.
class test
{
public:
test() {}
const foo* value() const { return f; }
private:
foo* f;
};
int main()
{
const test t;
foo* f = t.value(); // error here
return 0;
}
Why the problem is happening when templates are used?
Because you have two levels of indirection - in your main function, that call to value returns a reference to a const pointer to a non-const foo.
This can safely be copied into non-const pointer to a non-const foo.
If you'd instantiated test with const foo *, it would be a different story.
const test<const foo*> t;
foo* f = t.value(); // error
const foo* f = t.value(); // fine
return 0;
Update
From the comment:
value() returns const T& which can
only be assigned to another const
type. But in this case, compiler is
safely allowing the conversion.
Const data can only be read. It cannot be written ("mutated"). But copying some data is a way of reading it, so it's okay. For example:
const int c = 5;
int n = c;
Here, I had some const data in c, and I copied the data into a non-const variable n. That's fine, it's just reading the data. The value in c has not been modified.
Now, suppose your foo had some data in it:
struct foo { int n; };
If I have a non-const pointer to one of those, I can modify the n value through the pointer. You asked your test template to store a pointer to a non-const foo, and then made a const instance of test. Only the pointer address is constant, therefore. No one can change the address stored in the pointer inside test, so it cannot be made to point to another object. However, the object it points to can have its contents modified.
Update 2:
When you made your non-template version of the example, you made a mistake. To get it right, you need to substitute foo * into each place where there's a T.
const T& value() const
Notice that you have a reference to a const T there. So the return value will be a reference to something const: a foo *. It's only the pointer address that can't be modified. The object it points to can have its contents modified.
In your second example, you got rid of the reference part, which changes the meaning and makes the const modifier apply to the object that the pointer points to, instead of applying to the pointer itself.
Use the following template specialization:
template<typename T>
class test<T*>
{
public:
test() {}
const T* value() const
{
return f;
}
private:
T* f;
};
After including this, g++ says:
d.cpp: In function ‘int main()’:
d.cpp:41: error: invalid conversion from ‘const foo*’ to ‘foo*’
There's nothing wrong in your code, having a const reference to a pointer only means that you can't modify the pointer, but the pointed-to object remains perfectly mutable. If inside your main function you try to change the address pointed to by the f member of t you'll see that you can't: encapsulation is perfectly preserved.
This is the same principle that makes the following code valid:
void foo(std::vector<int *> const & v)
{
*v[0] = 0; // op. [] returns const & to int *
}
People new to C++ are usually surprised by this behavior, because for them a const vector should not allow the modification of its elements. And in fact it doesn't, because the pointer stored in the vector does not change (it keeps pointing to the same address). It's the pointed-to object which is modified, but the vector does not care about that.
The only solution is to do as Amit says and provide a specialization of your class for T*.