I know this has been asked a lot, but the only answers I could find was when the const-ness was actually casted away using (int*) or similar. Why isn't the const qualifier working on pointer type member variables on const objects when no cast is involved?
#include <iostream>
class bar {
public:
void doit() { std::cout << " bar::doit() non-const\n"; }
void doit() const { std::cout << " bar::doit() const\n"; }
};
class foo {
bar* mybar1;
bar mybar2;
public:
foo() : mybar1(new bar) {}
void doit() const {
std::cout << "foo::doit() const\n";
std::cout << " calling mybar1->doit()\n";
mybar1->doit(); // This calls bar::doit() instead of bar::doit() const
std::cout << " calling mybar2.doit()\n";
mybar2.doit(); // This calls bar::doit() const correctly
}
// ... (proper copying elided for brevity)
};
int main(void)
{
const foo foobar; // NOTE: foobar is const
foobar.doit();
}
The code above yields the following output (tested in gcc 4.5.2 and vc100):
foo::doit() const calling mybar1->doit() bar::doit() non-const <-- Why ? calling mybar2.doit() bar::doit() const
Once you have a const object, it cannot be assigned to a non-const reference or use functions that are known to be capable of changing the state of the object. This is necessary to enforce the const-ness of the object, but it means you need a way to state that a function should not make changes to an object.
In a function declaration, the keyword const may appear inside the square brackets that are used to declare an array type of a function parameter. It qualifies the pointer type to which the array type is transformed.
The const qualifier explicitly declares a data object as something that cannot be changed. Its value is set at initialization. You cannot use const data objects in expressions requiring a modifiable lvalue. For example, a const data object cannot appear on the lefthand side of an assignment statement. C only.
Because the data type being pointed to is const, the value being pointed to can't be changed. We can also make a pointer itself constant. A const pointer is a pointer whose address can not be changed after initialization.
In constant pointers, the pointer points to a fixed memory location, and the value at that location can be changed because it is a variable, but the pointer will always point to the same location because it is made constant here.
The const keyword specifies that a variable's value is constant and tells the compiler to prevent the programmer from modifying it.
When a foo instance is const, its data members are const too, but this applies differently for pointers than you might at first think:
struct A {
int *p;
};
A const obj;
The type of obj.p is int * const, not int const *; that is, a constant pointer to int, not a pointer to constant int.
For another way to look at it, let's start with a function:
template<class T>
T const& const_(T const &x) {
return x;
}
Now imagine we have an A instance, and we make it const. You can imagine that as applying const_ on each data member.
A nc;
// nc.p has type int*.
typedef int *T; // T is the type of nc.p.
T const &p_when_nc_is_const = const_(nc.p);
// "T const" is "int * const".
const T &be_wary_of_where_you_place_const = const_(nc.p);
// "const T" is "int * const".
// "const T" is *not* "const int *".
The variable be_wary_of_where_you_place_const shows that "adding const" is not the same as prepending "const" to the literal text of a type.
I am going to answer my own question in this case. Fred Nurk's answer is correct but does not really explain the "why". mybar1
and *mybar1
are different. The first refer to the actual pointer and the latter the object. The pointer is const (as mandated by the const-ness on foo; you can't do mybar1 = 0
), but not the pointed to object, as that would require me to declare it const bar* mybar1
. The declaration bar* mybar1
is equivalent to bar* const mybar1
when the foo object is const (i.e. pointer is const, not pointed to object).
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