function overloading: direct-match vs trivial-conversion

[subramanian100in]

Suppose T is a type.

Consider the two functions:

void fn(T& first)
{
....
}

void fn(T const & second)
{
....
}

Suppose I have the declaration
T obj;

If I call fn(obj), then the function, void fn(T& first) will be
called. This is because:
When both
1) 'T' to 'T&' (this is a direct match)
2) 'T' to 'T const &' (this involves trivial conversion)
are present, direct match takes higher precedence. If direct match is
not found, only then trivial conversion is considered. That is, if the
first function void fn(T& first) is not present, only then the second
function void fn(T const & second) will be called.

Is my above understanding correct ?

Kindly clarify.

Thanks
V.Subramanian

[Looney]

On Apr 13, 3:26 pm, "[e-mail], India"
<[e-mail]:

Suppose T is a type.

Consider the two functions:

void fn(T& first)
{
...

}

void fn(T const & second)
{
...

}

Suppose I have the declaration
T obj;

If I call fn(obj), then the function, void fn(T& first) will be
called. This is because:
When both
1) 'T' to 'T&' (this is a direct match)
2) 'T' to 'T const &' (this involves trivial conversion)
are present, direct match takes higher precedence. If direct match is
not found, only then trivial conversion is considered. That is, if the
first function void fn(T& first) is not present, only then the second
function void fn(T const & second) will be called.

Is my above understanding correct ?

Kindly clarify.

Thanks
V.Subramanian

yes that is correct .
check out http://accu.org/index.php/journals/268
for a details