Pass by Reference / Value in C++

Solution 1:

I think much confusion is generated by not communicating what is meant by passed by reference. When some people say pass by reference they usually mean not the argument itself, but rather the object being referenced. Some other say that pass by reference means that the object can't be changed in the callee. Example:

struct Object {
    int i;
};

void sample(Object* o) { // 1
    o->i++;
}

void sample(Object const& o) { // 2
    // nothing useful here :)
}

void sample(Object & o) { // 3
    o.i++;
}

void sample1(Object o) { // 4
    o.i++;
}

int main() {
    Object obj = { 10 };
    Object const obj_c = { 10 };

    sample(&obj); // calls 1
    sample(obj) // calls 3
    sample(obj_c); // calls 2
    sample1(obj); // calls 4
}

Some people would claim that 1 and 3 are pass by reference, while 2 would be pass by value. Another group of people say all but the last is pass by reference, because the object itself is not copied.

I would like to draw a definition of that here what i claim to be pass by reference. A general overview over it can be found here: Difference between pass by reference and pass by value. The first and last are pass by value, and the middle two are pass by reference:

    sample(&obj);
       // yields a `Object*`. Passes a *pointer* to the object by value. 
       // The caller can change the pointer (the parameter), but that 
       // won't change the temporary pointer created on the call side (the argument). 

    sample(obj)
       // passes the object by *reference*. It denotes the object itself. The callee
       // has got a reference parameter.

    sample(obj_c);
       // also passes *by reference*. the reference parameter references the
       // same object like the argument expression. 

    sample1(obj);
       // pass by value. The parameter object denotes a different object than the 
       // one passed in.

I vote for the following definition:

An argument (1.3.1) is passed by reference if and only if the corresponding parameter of the function that's called has reference type and the reference parameter binds directly to the argument expression (8.5.3/4). In all other cases, we have to do with pass by value.

That means that the following is pass by value:

void f1(Object const& o);
f1(Object()); // 1

void f2(int const& i);
f2(42); // 2

void f3(Object o);
f3(Object());     // 3
Object o1; f3(o1); // 4

void f4(Object *o);
Object o1; f4(&o1); // 5

1 is pass by value, because it's not directly bound. The implementation may copy the temporary and then bind that temporary to the reference. 2 is pass by value, because the implementation initializes a temporary of the literal and then binds to the reference. 3 is pass by value, because the parameter has not reference type. 4 is pass by value for the same reason. 5 is pass by value because the parameter has not got reference type. The following cases are pass by reference (by the rules of 8.5.3/4 and others):

void f1(Object *& op);
Object a; Object *op1 = &a; f1(op1); // 1

void f2(Object const& op);
Object b; f2(b); // 2

struct A { };
struct B { operator A&() { static A a; return a; } };
void f3(A &);
B b; f3(b); // passes the static a by reference

Solution 2:

When passing by value:

void func(Object o);

and then calling

func(a);

you will construct an Object on the stack, and within the implementation of func it will be referenced by o. This might still be a shallow copy (the internals of a and o might point to the same data), so a might be changed. However if o is a deep copy of a, then a will not change.

When passing by reference:

void func2(Object& o);

and then calling

func2(a);

you will only be giving a new way to reference a. "a" and "o" are two names for the same object. Changing o inside func2 will make those changes visible to the caller, who knows the object by the name "a".

Pass by Reference / Value in C++

Solution 1:

Solution 2:

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