Does copy occur if variable not defined as a reference?

[seeLTrun]

I'm sure this has been asked before in this forum, but what I need is confirmation on this: If a variable that is assigned a reference to some struct is defined without the &, i.e.

Some_Type_t some_type = some_object.getSomeType();

instead of

Some_Type_t& some_type = some_object.getSomeType();

where method getSomeType() returns a reference to Some_Type_t, will some_type be a reference to the the Some_Type_t object or will it be a copy of the object?

I've looked in many books and places online but can't seem to find the answer. Thanks.

[Danny]

I'm sure this has been asked before in this forum, but what I need is confirmation on this: If a variable that is assigned a reference to some struct is defined without the &, i.e.

Some_Type_t some_type = some_object.getSomeType();

instead of

Some_Type_t& some_type = some_object.getSomeType();

where method getSomeType() returns a reference to Some_Type_t, will some_type be a reference to the the Some_Type_t object or will it be a copy of the object?

It will be a copy, not a reference. This means that any change that you apply to some_type after its initialization will not affect any other object.
If however some_type was declared as a reference variable:

Some_Type_t& some_type = some_object.getSomeType();

some_type would indeed be an alias of another object, so chaging it would affect the actual object to which it is bound.
C++ is rather explicit (unlike Java for instance) in this respect: when you want a reference or a pointer, you must use & and * , respectively in the declaration of a variable. Otherwise, the result is an independent object.