CODEWITHSUNDEEP
c++
elle
elle

Reputation: 3245

why is the value being changed even in a const function?

#include<iostream>
using namespace std;

class temp
    {
      int value1; 
      public :
        void fun() const
        {
        ((temp*)this)->value1 = 10;
        }
        void print()
        {
            cout<<value1<<endl;
        }
     };
int main()
{
  temp t;
  t.fun();
  t.print();
}

Upvotes: 5

Views: 204

Answers (3)

Tony Delroy
Tony Delroy

Reputation: 106096

C++ tries to prevent accidental errors, but it doesn't go out of its way to fight a programmer who's determined to have things their own way. If you use cast operators you're telling it "trust me, I know what's there", demanding it ignore it's own knowledge of the program. It's precisely because the C-style casting operator you've used is dangerous and can be easily misused that C++ introduces static_cast, const_cast and reinterpret_cast, which communicate the programmer's intent in such a way that the compiler can still say "hey, hold on there, that'd require more than just the type of leniency you're asking for". reinterpret_cast is the big daddy though... no arguing with that... just as brutal as the C-cast and rarely needed in a high-level application. Precisely because it's rarely needed, verbose and easily seen, it draws scrutiny. Code littered with C-style casts can easily hide bugs.

Upvotes: 3

Chubsdad
Chubsdad

Reputation: 25497

$5.4/5 is about explicit type conversion (which is what is being used here)

The conversions performed by

— a const_cast (5.2.11),

— a static_cast (5.2.9),

— a static_cast followed by a const_cast,

— a reinterpret_cast (5.2.10), or

— a reinterpret_cast followed by a const_cast,

can be performed using the cast notation of explicit type conversion. The same semantic restrictions and behaviors apply. If a conversion can be interpreted in more than one of the ways listed above, the interpretation that appears first in the list is used, even if a cast resulting from that interpretation is ill-formed. If a conversion can be interpreted in more than one way as a static_cast followed by a const_cast, the conversion is ill-formed.

In this case, ((temp*)this) got treated as (const_cast<temp *>(this)) and was well-formed. This removed away the constness, thereby allowing to change the class member value.

Upvotes: 5

GManNickG
GManNickG

Reputation: 503855

Because you're casting away const...

When you cast something, the responsibility is yours for making sure that it doesn't do something dumb.

Note that if temp t; is changed to const temp t;, you get undefined behavior, for modifying a const value.

Coincidentally I literally just touched on this in my blog. (Almost the same function, too.)

Upvotes: 14

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