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c++castingconstants
Maddyfire
Maddyfire

Reputation: 61

Does const_cast return the type specified inside angular bracket

I am little confused with the return type of const_cast? Does the type inside the angular brackets <> is the return type?

const int i = 5;
int b = const_cast<int&>(i);

Is const_cast returning int& (integer reference), if yes then why how we storing it in integer? Or we should modify the code to:

const int i = 5; 
int & b  = const_cast<int&>(i);

Upvotes: 0

Views: 65

Answers (3)

juanchopanza
juanchopanza

Reputation: 227468

Does const_cast return the type specified inside angular bracket

It evaluates to an expression of the type in the angular brackets. What matters here is how you use that expression:

const int i = 5;
int b = const_cast<int&>(i);

In this case, b is just a copy of i. The cast is not required. 

const int i = 5; 
int & b  = const_cast<int&>(i);

Here, b refers to i. Note that using it to modify the value of i is undefined behaviour.

Upvotes: 3

songyuanyao
songyuanyao

Reputation: 172964

Does the type inside the angular brackets <> is the return type?

Yes, the return type of const_cast<int&>(i) is int&. After that you assign it to a int, and the value get copied.

And in int & b = const_cast<int&>(i);, you assign it to a int&, now b is the reference to i. Pay attention to that any modification on b will cause undifined behaviour.

Upvotes: 0

rodrigo
rodrigo

Reputation: 98436

Yes. But instead of returning you should say that the type of the resulting expression is the type in <>.

When that type is a reference, as in const_cast<int&>, it means that it is a l-value. In the first case there is no difference, as it is immediately converted to an r-value anyway. But then, the const in such an r-value is ignored, so:

int b = const_cast<int&>(i); //ok
int b = const_cast<int>(i);  //also ok
int b = i;  //hey! also ok

In the second case there is a difference, because there is no l-value to r-value conversion:

int &b  = const_cast<int&>(i); //ok
int &b  = const_cast<int>(i); //error: cannot bind a reference to an r-value
int &b  = i; //error: cannot bind a `int&` to a `const int&`

Upvotes: 3

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