TITLE: Non-const reference arguments and temporary generation


PROBLEM: jliberty@zdi.ziff.com (Jesse Liberty)

Is it legal to pass an int  to a function which expects an unsigned int & ?


RESPONSE: clamage@Eng.Sun.COM (Steve Clamage), 12 Aug 94

No. When you pass an object to a non-const reference, the types must be
compatible. The reason is that the function might modify the actual parameter
(since the reference is to a non-const). If the actual parameter is not
the same type as the reference expects, the compiler must create a
temporary object of the correct type initialized with the value of the
actual argument and pass a reference to the temp. If the function then
attempts to modify the actual argument, it modifies the temp instead; the
temp disappears when the function exits and the actual argument is unchanged.
This would be quite a surprise to the programmer who called the function.

What if the compiler didn't create a temp, but just passed a reference to
the actual argument, whatever it happened to be? Then the function would
have the wrong idea about the actual argument, possibly overwriting areas
outside the actual argument or storing data in the wrong format (long vs
float, for example, which might be the same size).

If the formal parameter of the function is a reference to a const, then
the compiler can safely create a temp and pass a reference to it. The
function promises not to modify the actual argument, so a copy is as good
as the original. Once again, though, the value of the actual argument
must be converted to the type expected by the function -- hence the temp.

In your particular case, 'int' and 'unsigned int' are not compatible.
Either may be implicitly converted to the other, which is why you get
no complaint when passing one to a const reference to the other.