When to use const and const reference in function args?

Question

When writing a C++ function which has args that are being passed to it, from my understanding const should always be used if you can guarantuee that the object will not be c

Answer 1

The questions are based on some incorrect assumptions, so not really meaningful.

std::string does not model immutable string values. It models mutable values.

There is no such thing as a "const reference". There are references to const objects. The distinction is subtle but important.

Top-level const for a function argument is only meaningful for a function implementation, not for a pure declaration (where it's disregarded by the compiler). It doesn't tell the caller anything. It's only a restriction on the implementation. E.g. int const is pretty much meaningless as argument type in a pure declaration of a function. However, the const in std::string const& is not top level.

Passing by reference to const avoids inefficient copying of data. In general, for an argument passing data into a function, you pass small items (such as an int) by value, and potentially larger items by reference to const. In the machine code the reference to const may be optimized away or it may be implemented as a pointer. E.g., in 32-bit Windows an int is 4 bytes and a pointer is 4 bytes. So argument type int const& would not reduce data copying but could, with a simple-minded compiler, introduce an extra indirection, which means a slight inefficiency -- hence the small/large distinction.

Cheers & hth.,

Answer 2

The general rule is, use const whenever possible, and only omit it if necessary. const may enable the compiler to optimize and helps your peers understand how your code is intended to be used (and the compiler will catch possible misuse).

As for your example, strings are not immutable in C++. If you hand a non-const reference to a string to a function, the function may modify it. C++ does not have the concept of immutability built into the language, you can only emulate it using encapsulation and const (which will never be bullet-proof though).

After thinking @Eamons comment and reading some stuff, I agree that optimization is not the main reason for using const. The main reason is to have correct code.

Answer 3

The main advantage of const reference over const pointer is following: its clear that the parameter is required and cannot be NULL. Vice versa, if i see a const pointer, i immedeately assume the reason for it not being a reference is that the parameter could be NULL.

Answer 4

Asking whether to add const is the wrong question, unfortunately.

Compare non-const ref to passing a non-const pointer

void modifies(T &param);
void modifies(T *param);

This case is mostly about style: do you want the call to look like call(obj) or call(&obj)? However, there are two points where the difference matters. If you want to be able to pass null, you must use a pointer. And if you're overloading operators, you cannot use a pointer instead.

Compare const ref to by value

void doesnt_modify(T const &param);
void doesnt_modify(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++.)

Compare const pointer to non-modifying plus overload

void optional(T const *param=0);
// vs
void optional();
void optional(T const &param); // or optional(T param)

This is related to the non-modifying case above, except passing the parameter is optional. There's the least difference here between all three situations, so choose whichever makes your life easiest. Of course, the default value for the non-const pointer is up to you.

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 C(int const);
typedef void NC(int);
NC *nc = &f;  // nc is a function pointer
C *c = nc;  // C and NC are identical types