removing constness in templated function

Question

I have a little issue with removing the constness of using Templated function.

#include <iostream>
using namespace std;

template< typename T>
void fct(T&  param)
{
  const_cast<T>(param) = 40;    
}


int _tmain(int argc, _TCHAR* argv[])
{
  int x = 30;
  const int  cx = x;
  const int& rx = x;
  fct(cx);
  return 0;
}

when I run this I get :

error C2440: 'const_cast' : cannot convert from 'int' to 'int'

How could I use const_cast into my function.

No matter the error, what you're trying to do is a ***bad idea***. — Some programmer dude, Jan 16 '18 at 11:18
@KillzoneKid How can `T` be `int`? He's passing a `const int&`, which is matched against a `T&` so `T` is deduced to be `const int`. — oisyn, Jan 16 '18 at 11:45

Sandro · Answer 1 · 2019-02-12T12:05:17.477

6

you can try to do something like this:

template< typename T>
void fct(const T&  param)
{
  const_cast<T&>(param) = 40;    
}

template type T must be a reference or your cast does not make any sense

edited Feb 12 '19 at 12:05

answered Jan 16 '18 at 11:18

Sandro

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Modifying a const object through a non-const access path is UB – AMA Jan 16 '18 at 11:22
Yes @AMA, but from curiousity I wanna know how to be able to remove the const ;D – Blood-HaZaRd Jan 16 '18 at 11:25
2

Aside from the fact that the indended goal is bad, I like this answer better for its simplicity. No need for type traits, let te compiler take out the `const` of the deduced type by explicitely specifying it. Of course, the different versions don't do exactly the same: this one binds to rvalues, the other one doesn't. Which can either be a pro or a con, depending on your requirements. – oisyn Jan 16 '18 at 13:49

songyuanyao · Accepted Answer · 2018-01-16T13:45:29.743

4

const_cast<T>(param) = 40; doesn't do what you want, for both fct(cx); and fct(rx);, T is deduced as const int.

If you want to remove the constness, i.e. get a reference to non-const, you can use std::remove_const:

const_cast<typename std::remove_const<T>::type &>(param) = 40;

For T is deduced as const int, typename std::remove_const<T>::type results in int, then the above code is same as const_cast<int &>(param)....

Note that cx is a constant, trying to modify it via reference got from const_cast leads to UB. For rx it's fine, it refers to a non-constant in fact.

edited Jan 16 '18 at 13:45

answered Jan 16 '18 at 11:26

songyuanyao

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Sorry, I am using C++11 :/ not 14 * – Blood-HaZaRd Jan 16 '18 at 11:32
@Blood-HaZaRd Revised. – songyuanyao Jan 16 '18 at 11:34
2

@Blood-HaZaRd [cppreference.com](http://en.cppreference.com/w/) is fine. – songyuanyao Jan 16 '18 at 11:38
yes I did read that but The thing that I missed is the member type `type`. – Blood-HaZaRd Jan 16 '18 at 11:41
@Blood-HaZaRd That's why C++14 makes it simpler; `typename std::remove_const::type` => `std::remove_const_t`. – songyuanyao Jan 16 '18 at 11:42
Doesn't this fail if you pass a non const to it? – Yakk - Adam Nevraumont Jan 16 '18 at 12:00
@Yakk if I believe the wording and example from [cppreference](http://en.cppreference.com/w/cpp/types/remove_cv), it just leaves unqualified types untouched. Which makes sense. – Quentin Jan 16 '18 at 12:06
@quen no: the const cast that does nothing – Yakk - Adam Nevraumont Jan 16 '18 at 13:35
@Yakk [it looks fine as well](http://eel.is/c++draft/expr.const.cast#2) :) – Quentin Jan 16 '18 at 13:45
@Quentin Ok; I was unsure. – Yakk - Adam Nevraumont Jan 16 '18 at 15:10

removing constness in templated function

2 Answers2