CODEWITHSUNDEEP
c++castinglvaluervalue
T.J.
T.J.

Reputation: 1506

behaviour of const_cast

I was reading about const_cast operator in c++

1.First weird thing thing i can't understand is

const_cast operator syntax i.e.

-const_cast--<--Type-->--(--expression--)--------------------><

what i have understand about this syntax is that it helps to cast away constness of anexpressionof type Type .But consider this code

class  ConstTest {   

private:
    int year;
public:
    ConstTest() : year(2007) {}
    void  printYear() const;
};

int main() {
    ConstTest c;
    c.printYear();
    return  0;
}

void ConstTest::printYear() const {
    ConstTest  *c  = const_cast<ConstTest*>(this);
    c->year  = 42;
    std::cout  <<  "This  is the  year "  << year  << std::endl;
}

Here in line ConstTest *c = const_cast<ConstTest*>(this), I think that the const of this pointer should be cast away, but the output shows that it is the object which this refers to that loses its const-ness.

I feel that the code should have been ConstTest *c = const_cast<ConstTest>(*this), but this produces an error. I know i am wrong at many interpretations. Please correct them all.

2.my second problem is the statement given below

The result of a const_cast expression is an rvalue unless Type is a reference type. In this case, the result is an lvalue.

Why is this so, and why it is not true in case of pointers?

Upvotes: 1

Views: 699

Answers (1)

Steve Jessop
Steve Jessop

Reputation: 279245

it helps to cast away constness of an expression of type Type

No, Type is the type of the result, not the type of the operand.

What i think is const of this pointer should be casted away

this has type const ConstTest*. const_cast<ConstTest*>(this) has type ConstTest*. That's what "casting away const" from a pointer-to-const means.

I feel code should have been ConstTest *c = const_cast<ConstTest>(*this)

The result of const_cast<T> has type T, that's how it's defined. Maybe you would have defined it differently, but tough luck, you don't get a ConstTest* by writing const_cast<ConstTest>, you get it by writing const_cast<ConstTest*>. Your preferred syntax is not available.

You can either do ConstTest &c = const_cast<ConstTest&>(*this) or ConstTest *c = const_cast<ConstTest*>(this), so pick your favorite.

The result of a const_cast expression is an rvalue unless Type is a reference type. In this case, the result is an lvalue.

why so and why it is not true in case of pointers?

It is true of pointers. ConstTest* is not a reference type, and the result of const_cast<ConstTest*>(this) is an rvalue. You then assign that value to the variable c.

Upvotes: 4

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