references question

[nvangogh]

According to c++ primer "once initialized, a reference remains bound to
it's initial object. There is no way to rebind a reference to refer to a
different object." moreover, "Because references are not objects, we may
not define a reference to a reference".

I am confused because this code seems to at least define a reference to
a reference:

#include <iostream>

int main()
{
int i = 0, &r1 = i;
double d = 5.0, &r2 = d;

r2 = r1;
std::cout << r2 << std::endl;

return 0;
}

When run this outputs r2 as 0. Why is this?

 

[Wouter van Ooijen]

nvangogh schreef op 17-Mar-14 2:47 PM:

According to c++ primer "once initialized, a reference remains bound to
it's initial object. There is no way to rebind a reference to refer to a
different object." moreover, "Because references are not objects, we may
not define a reference to a reference".

I am confused because this code seems to at least define a reference to
a reference:

#include <iostream>

int main()
{
int i = 0, &r1 = i;
double d = 5.0, &r2 = d;

r2 = r1;
std::cout << r2 << std::endl;

return 0;
}

When run this outputs r2 as 0. Why is this?

I don't see any reference to a reference in your code.

It prints 0 because the 'r2 = r1' statement is in effect (after
following the references) 'd = i;'.

Wouter van Ooijen

 

[Öö Tiib]

According to c++ primer "once initialized, a reference remains bound to
it's initial object. There is no way to rebind a reference to refer to a
different object." moreover, "Because references are not objects, we may
not define a reference to a reference".

I am confused because this code seems to at least define a reference to
a reference:

Next time please describe what you think that you see. It is impossible
for us to understand what you think that code does if you do not say it
out. For me it seems fully normal code and nothing mystical is going on.
Let me try to describe what I see:

#include <iostream>

int main()
{
int i = 0, &r1 = i;

'i' is int variable object with value 0, 'r1' is reference (alias) of that object.

double d = 5.0, &r2 = d;

'd' is double variable object with value 5.0, 'r2' is reference (alias) of that object.

r2 = r1;

Value of 'i' (0) is assigned to 'd'; int is converted to double and 'd' becomes 0.0.

std::cout << r2 << std::endl;

Value of 'd' (0.0) is output to 'std::cout'.

return 0;
}

When run this outputs r2 as 0. Why is this?

I did describe above. Possible cause for your confusion is that you do not
understand something of what happens. Please describe what it was/is.

Value of of 'r2' is always value of object what it refers. There are no
"reference values" unlike with pointers where you have pointer's value and
value of pointed at object. Therefore you can never take reference to
reference itself only to object that it refers.