Does C++ pass objects by value or reference?

Solution 1:

Arguments are passed by value, unless the function signature specifies otherwise:

• in void foo(type arg), arg is passed by value regardless of whether type is a simple type, a pointer type or a class type,
• in void foo(type& arg), arg is passed by reference.

In case of arrays, the value that is passed is a pointer to the first element of the array. If you know the size of the array at compile time, you can pass an array by reference as well: void foo(type (&arg)[10]).

Solution 2:

C++ always gives you the choice: All types T (except arrays, see below) can be passed by value by making the parameter type T, and passed by reference by making the parameter type T &, reference-to-T.

When the parameter type is not explicitly annotated to be a reference (type &myVariable), it is always passed by value regardless of the specific type. For user-defined types too (that's what the copy constructor is for). Also for pointers, even though copying a pointer does not copy what's pointed at.

Arrays are a bit more complicated. Arrays cannot be passed by value, parameter types like int arr[] are really just different syntax for int *arr. It's not the act of passing to a function which produces a pointer from an array, virtually every possible operation (excluding only a few ones like sizeof) does that. One can pass a reference-to-an-array, but this explicitly annotated as reference: int (&myArray)[100] (note the ampersand).

Solution 3:

C++ makes both pass by value and pass by reference paradigms possible.

You can find two example usages below.

http://www.learncpp.com/cpp-tutorial/72-passing-arguments-by-value/

http://www.learncpp.com/cpp-tutorial/73-passing-arguments-by-reference/

Arrays are special constructs, when you pass an array as parameter, a pointer to the address of the first element is passed as value with the type of element in the array.

When you pass a pointer as parameter, you actually implement the pass by reference paradigm yourself, as in C. Because when you modify the data in the specified address, you exactly modify the object in the caller function.