How to pass objects to functions in C++?

Question

I am new to C++ programming, but I have experience in Java. I need guidance on how to pass objects to functions in C++.

Do I need to pass pointers, references, or no

Answer 1

There are several cases to consider.

Parameter modified ("out" and "in/out" parameters)

void modifies(T ¶m);
// vs
void modifies(T *param);

This case is mostly about style: do you want the code to look like call(obj) or call(&obj)? However, there are two points where the difference matters: the optional case, below, and you want to use a reference when overloading operators.

...and optional

void modifies(T *param=0);  // default value optional, too
// vs
void modifies();
void modifies(T ¶m);

Parameter not modified

void uses(T const ¶m);
// vs
void uses(T param);

This is the interesting case. The rule of thumb is "cheap to copy" types are passed by value — these are generally small types (but not always) — while others are passed by const ref. However, if you need to make a copy within your function regardless, you should pass by value. (Yes, this exposes a bit of implementation detail. C'est le C++.)

...and optional

void uses(T const *param=0);  // default value optional, too
// vs
void uses();
void uses(T const ¶m);  // or optional(T param)

There's the least difference here between all situations, so choose whichever makes your life easiest.

Const by value is an implementation detail

void f(T);
void f(T const);

These declarations are actually the exact same function! When passing by value, const is purely an implementation detail. Try it out:

void f(int);
void f(int const) { /* implements above function, not an overload */ }

typedef void NC(int);       // typedefing function types
typedef void C(int const);

NC *nc = &f;  // nc is a function pointer
C *c = nc;    // C and NC are identical types