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constant references with typedef and templates in c++

I heard the temporary objects can only be assigned to constant references.

But this code gives error

#include <iostream.h>    
template<class t>
t const& check(){
  return t(); //return a temporary object
}    
int main(int argc, char** argv){

const int &resCheck = check<int>(); /* fine */
typedef int& ref;
const ref error = check<int>(); / *error */
return 0;
}

The error that is get is invalid initialization of reference of type 'int&' from expression of type 'const int'

like image 325
Melvin Michael Avatar asked Sep 27 '10 07:09

Melvin Michael


3 Answers

This:

typedef int& ref;
const ref error;

Doesn't do what you think it does. Consider instead:

typedef int* pointer;
typedef const pointer const_pointer;

The type of const_pointer is int* const, not const int *. That is, when you say const T you're saying "make a type where T is immutable"; so in the previous example, the pointer (not the pointee) is made immutable.

References cannot be made const or volatile. This:

int& const x;

is meaningless, so adding cv-qualifiers to references has no effect.

Therefore, error has the type int&. You cannot assign a const int& to it.


There are other problems in your code. For example, this is certainly wrong:

template<class t>
t const& check()
{
    return t(); //return a temporary object
}

What you're doing here is returning a reference to a temporary object which ends its lifetime when the function returns. That is, you get undefined behavior if you use it because there is no object at the referand. This is no better than:

template<class t>
t const& check()
{
    T x = T();
    return x; // return a local...bang you're dead
}    

A better test would be:

template<class T>
T check()
{
    return T();
}

The return value of the function is a temporary, so you can still test that you can indeed bind temporaries to constant references.

like image 60
GManNickG Avatar answered Nov 09 '22 07:11

GManNickG


Your code gives error because the const qualifier in const ref error is just ignored because 8.3.2/1 says

Cv-qualified references are ill-formed except when the cv-qualifiers are introduced through the use of a typedef (7.1.3) or of a template type argument(14.3), in which case the cv-qualifiers are ignored.`

So error has type int& not const int& .

like image 7
Prasoon Saurav Avatar answered Nov 09 '22 08:11

Prasoon Saurav


It's a very common mistake for English speaking people, because of the way the English grammar works.

I consider it extremely unfortunate that the C++ syntax would allow both:

const int // immutable int
int const // immutable int

to have the same meaning.

It doesn't make it easier, really, and isn't composable since:

const int* // mutable pointer to immutable int
int* const // immutable pointer to mutable int

certainly do NOT have the same meaning.

And this, unfortunately for you, what kicks in here, as @GMan explains.

If you wish to avoid this kind of error in the future, take the habit of qualifying your types (const and volatile) on their right, then you'll be able to treat a typedef as simple text replacement.

like image 7
Matthieu M. Avatar answered Nov 09 '22 09:11

Matthieu M.