CODEWITHSUNDEEP
c++reference
Moeb
Moeb

Reputation: 10871

Why does this conversion from int to double& not work?

Error!

$ cat double.cc
#include<iostream>

int main() {
    int i = 42;
    double &r = i;
}
$ g++ double.cc
double.cc: In function ‘int main()’:
double.cc:5: error: invalid initialization of reference of type ‘double&’
from expression of type ‘int’
$

Works!

$ cat double.cc
#include<iostream>

int main() {
    int i = 42;
    const double &r = i;
}
$ g++ double.cc
$

Why do we get the error in the first case? I know this is discussed here, but I am not sure I understand why this not allowed, while int to const double& is allowed?

Upvotes: 0

Views: 1046

Answers (3)

Praetorian
Praetorian

Reputation: 109269

When you write

int i = 42;
double const &r = i;

i is implicitly converted to a double, which results in a temporary value, or rvalue. A non-const reference is not allowed to bind to an rvalue, so r needs to be a double const& for the above to work.

Upvotes: 4

Mark B
Mark B

Reputation: 96301

In C++ you can almost think of references as alias to another variable. They must match in non-cv qualified type, and in this case the types are different.

The reason it works with const double& is becaused the compiler creates a temporary double with a copy of i's value and binds the const reference to that temporary. This is allowed because binding to a temporary extends the life of the temporary to the lifetime of the reference.

Upvotes: 4

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361712

int i = 42;
double &r = i;

Here r is a reference to double, but i is not a double, then how can r refer to it? It cannot. So it creates a temporary object of type double out of the expression i. But again, r which is non-const reference, cannot bind to a temporary object.

However, if you make it const reference, then it would work:

double const & r = i;

Note that const reference can bind to a temporary object which is created out of i.

Upvotes: 6

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