Here’s a slogan that needs more currency:

const is a contract.

Let’s start off with the one place this is not true: variable declarations.

int main() { int x = 42; const int y = 43; // ... }

Here y is an object of const type. This means that it’s constant, immutable; nobody can change its value, and its value cannot change. This is what newcomers to the language usually expect const to mean. This is the easy part.

“But what if I const_cast away the const and modify y anyway?” Then you have undefined behavior.

const_cast<int&>(y) += 1; // Undefined behavior!

An object defined const is really and truly constant, immutable, unchanging.

In function prototypes

So what about in a function prototype?

void f(const int& p) { ... }

Here p is a reference to a const-qualified int . Is that int ’s value constant, immutable, unchanging? No!

static int g = 0; void f(const int& p) { printf("%d ", p); g = 1; printf("%d

", p); } int main() { f(g); }

The above program is required to print 0 1 . The compiler is not allowed to assume that the value referenced by p is immutable!

So what good is const ?

const is a contract.

This keyword const is a contract — a promise — between the function f and its caller. By putting const in its signature, f is saying, “Hello caller, I promise never to modify the referent of p .” (At least, not directly via p .)

The caller can look at that and know that f is promising not to modify the referent of p ; therefore, it is safe to pass something to it that the caller doesn’t have permission to modify.

The compiler enforces contract law

Alice’s aunt (who lives in Maine) owns a very rare and delicate Ming vase. Alice promised her aunt that she wouldn’t break the vase, or modify it in any way; and so Alice’s aunt has grudgingly permitted Alice to touch the vase.

void alice(const Ming& vase) { // ... } int main() { Ming vase; alice(vase); }

Notice that Alice’s aunt in Maine actually can modify the vase; it’s not literally immutable. But Alice herself has made a promise — a contract — that says she, Alice, will not attempt to modify the vase.

Now Bob says, “Let me touch the vase too!”

void bob(Ming& vase);

Bob isn’t making any promises like the one Alice gave her aunt. Alice strongly suspects that Bob is going to try to modify the vase; after all, he didn’t promise that he wouldn’t! What should Alice do? Is it okay for Alice to let Bob touch her aunt’s vase?

No, of course it’s not okay! Alice doesn’t let Bob touch the vase.

void bob(Ming& vase); void alice(const Ming& vase) { bob(vase); // ERROR }

Now Charlie asks to touch the vase; and he adds, significantly, “I promise not to modify the thing I touch.” Only because Charlie promises not to modify the vase, Alice will let Charlie touch it:

void charlie(const Ming&); // "I promise not to modify the thing I touch." void alice(const Ming& vase) { charlie(vase); // "OK, Charlie, since you promised." }

The compiler does not permit Alice to ignore the promise she’s made to her aunt.

Naughty on purpose with const_cast

C++ does have a thing called const_cast that Alice can use to be naughty on purpose:

void bob(Ming& vase) { vase.smash(); // OK; and honestly this is exactly what I expected Bob to do } void alice(const Ming& vase) { vase.smash(); // ERROR; the compiler prevents this bob(vase); // ERROR; the compiler prevents this const_cast<Ming&>(vase).smash(); // OK, but naughty bob(const_cast<Ming&>(vase)); // OK, but naughty } int main() { Ming vase; alice(vase); }

In this example, Alice is not ignoring the promise she’s made to her aunt; she is actively and explicitly breaking that promise. Normally, you don’t want to do this in your programs. But C++ allows it…

…as long as the vase object is not literally immutable!

int main() { const Ming vase; const_cast<Ming&>(vase).smash(); // Undefined behavior! }

Pointers behave just like references in this respect

When we take a function parameter of type const int * , we’re saying “Please give me the address of an int , and I promise not to modify that int .”

And the const can apply at each different level, independently:

When we take a function parameter of type const int ** , we’re saying “Please give me a pointer to an int* , and I promise not to modify the int it points to (but I might modify the int* itself).”

When we take a function parameter of type int *const * , we’re saying “Please give me a pointer to a pointer to an int , and I promise not to modify that int* (but I might modify the int it points to).”

When we take a function parameter of type const int *const * , we’re saying “Please give me a pointer to an int* , and I promise not to modify either the int* or the int it points to.”

(Very minor handwaving here: “pointer to int* ” and “pointer to const int* ” are actually different types, and behave according to covariance and contravariance, which is fun times. Intrepid readers can find more details in the standard, until I find a better explanation to link to.)

const on a member function affects the this pointer

Here are two non-member (“free”) functions. One promises not to modify the Ming it receives via parameter p ; the other makes no such promise.

void handle_with_care(const Ming *p); void juggle(Ming *p);

Here are two member functions. One promises not to modify the Ming it receives via the hidden this parameter; the other makes no such promise.

struct Ming { void polish() const; void smash(); };

Guideline:

After you write a member function, look at your implementation. Does it mutate the this object? If it does not, then you should mark the member function const .

Recall that const is a contract. Suppose Alice has been permitted to touch her aunt’s vase, on condition that she not modify it in any way. Should the compiler allow Alice to polish the vase? Should the compiler allow Alice to smash the vase?

Before you write a member function, think about what it does. If someone has a const Foo& that they’ve promised not to modify, should they be allowed to call your member function? If so, then you must mark the member function const !

const on a by-value parameter is not a meaningful contract

We’ve discussed const Foo& and const Foo * parameters. What about const Foo by itself?

void replicate(Ming vase); void counterfeit(const Ming vase);

Here we have two functions that accept a Ming vase by copy. If we read the function signatures as contracts, we see that replicate says, “Give me a copy of your vase; and I make no promises about what I’ll do with the copy.” Whereas counterfeit says, “Give me a copy of your vase; and I promise not to modify my copy.”

That’s not much of a promise! Why should your aunt care whether you modify a copy of her vase? She knows her original won’t be modified. It’s none of her business what you do with your copy.

Therefore, C++ treats these two function signatures as exactly equivalent. There is no difference, from the compiler’s point of view, between taking Ming by value and taking const Ming by value. (There is a difference inside the body of the function, but Alice’s aunt doesn’t care what you do in there.)

Therefore, most C++ programmers (including me!) recommend that you never take a parameter “by const value” like this:

void counterfeit(const Ming vase); // BAD STYLE!

There’s a benefit to adopting this guideline. Especially for students coming to C++ from Java, a very very common failure mode is to accidentally leave off the ampersand on an otherwise idiomatic pass-by-const-reference:

bool operator==(const Ming& a, const Ming b) { // ... }

Because passing by-const-value is definitely bad, we know there’s something wrong with const Ming b . Having been alerted to a problem, it doesn’t take us long to identify the missing ampersand. (There’s a tool called clang-tidy that can help with this… at least, I think it can.)

const on a by-value parameter can be a pessimization

I looked for this kind of typo in my current employer’s codebase — code written by professional programmers with years of C++ experience. I found an absolutely astounding number of them! The most common and egregious variation in our codebase looks like this:

class Widget { std::string name_; public: void set_name(const std::string name) { name_ = name; } };

This function will pass all the unit tests you throw at it, but it makes an additional heap-allocation compared to either

void set_name(const std::string& name) { name_ = name; }

or

void set_name(std::string name) { name_ = std::move(name); }

We’ll see more about pessimizations later in this blog post.

const on a by-reference return type

Sometimes we want our caller to promise something to us. Let’s flip around our Ming vase example. Now instead of Alice directly permitting Bob to touch the vase, Alice will expose a public API method that allows any caller to gain access to the vase. Here’s our initial code, without any const anywhere.

struct Alice { Ming *vase_; explicit Alice(Ming *vase) : vase_(vase) {} Ming& getVase() { return *vase_; } }; struct Bob { void ask_for_access(Alice& a) { Ming& vase = a.getVase(); vase.smash(); // You probably saw this coming. } }; int main() { Ming vase; Alice alice(&vase); Bob bob; bob.ask_for_access(alice); }

Alice’s aunt wouldn’t approve of this version! Let’s make Alice promise not to modify the vase. When Alice asks her aunt for a pointer to the vase, she must additionally promise not to modify that vase:

struct Alice { const Ming *vase_; explicit Alice(const Ming *vase) : vase_(vase) {} Ming& getVase() { return *vase_; } // ERROR! };

Now the compiler gives us an error in getVase() . See, Bob — or anybody — can call getVase to get a reference to the vase, and then they can touch it; and then they can smash it; and that would break the promise that Alice made to her aunt. The compiler will not let Alice ignore her promise!

So when Alice returns the reference from getVase , Alice must extract a promise from the caller. The caller must promise not to modify the vase! We indicate that by putting const in the appropriate place on the return type.

const Ming& getVase() { return *vase_; } // OK! // "I'll let you touch this vase, but you must promise not to modify it."

Now the compiler is happy with Alice… but Bob is unhappy!

struct Bob { void ask_for_access(Alice& a) { Ming& vase = a.getVase(); // ERROR! Binding `Ming&` to `const Ming&` discards qualifiers. vase.smash(); } };

The only way for Bob to touch Alice’s aunt’s vase is for Bob to promise that he won’t smash it.

struct Bob { void ask_for_access(Alice& a) { const Ming& vase = a.getVase(); // OK, binding `const Ming&` to `const Ming&` vase.smash(); // NOT ALLOWED } };

Of course, if Bob makes his own copy of the vase, he can do whatever he wants with the copy.

struct Bob { void ask_for_access(Alice& a) { Ming vase = a.getVase(); // OK, making a copy via the copy constructor vase.smash(); // OK, smashing the copy } };

Before we leave this example, we should make it fully const-correct. That means applying our earlier guideline:

Before you write a member function, think about what it does. If someone has a const Foo& that they’ve promised not to modify, should they be allowed to call your member function? If so, then you must mark the member function const !

Here’s the fully const-correct code, with the one remaining ERROR exactly where we want it. (Bob has promised Alice not to modify the vase, so the compiler won’t let him smash it.)

struct Alice { const Ming *vase_; // Alice promises not to smash this vase. explicit Alice(const Ming *vase) : vase_(vase) {} // Alice promises not to smash whatever vase she's given. const Ming& getVase() const { return *vase_; } // Alice extracts a similar promise from her callers. // Also, getVase() does not modify this Alice object! }; struct Bob { void ask_for_access(const Alice& a) const { // This function doesn't modify the Alice object, and doesn't modify this Bob, either. const Ming& vase = a.getVase(); // OK, binding `const Ming&` to `const Ming&` vase.smash(); // ERROR: cannot smash a const Ming vase } }; int main() { Ming vase; Alice alice(&vase); Bob bob; bob.ask_for_access(alice); }

const on a return type is not a meaningful contract (and is sometimes a pessimization)

We’ve discussed const Foo& and const Foo * return types. What about const Foo by itself as a return type?

const Ming getImmutableVase();

Const-by-value return types are wacky. Here’s a Wandbox showing their behavior, but before you look at it, think about what a const-by-value return type should mean. It means the function is giving you a new copy of its return object; but that copy is marked const .

const is a contract.

The function is extracting from you a promise that you will not modify the return object — even though you’re getting your own copy of it! The function is saying: “I’ll give you a copy of what I have; but you have to promise not to modify your copy.” This doesn’t make much sense! In the example above, when Bob made a copy of the vase, we didn’t care if Bob smashed his copy. And we shouldn’t care in this case either.

Suppose Alice has

const Ming getVase() const { return *vase_; }

And suppose Bob does

Ming vase = alice->getVase(); // make a copy vase.smash(); // smash the copy

This is totally fine! And it remains fine even if you ask the compiler to fill in type Ming for you:

auto vase = alice->getVase(); // `auto` is deduced as `Ming` vase.smash(); // smash the copy

(And yes, copy elision happens in both cases.) But here’s the weird part:

alice->getVase().smash(); // ERROR: cannot smash a const Ming vase

The compiler remembers the const-qualification and enforces your promise not to modify the return value, even though that promise is effectively meaningless!

Just to add a dash of confusion: the situation for primitive types such as int is different. Primitive return types behave like parameter types: const-qualifying them is rightly seen as pointless and so the compiler just ignores the const in those cases.

const int getInt(); const Ming getMing(); static_assert(is_same_v< decltype(getInt()), int >); // the const is quietly dropped! static_assert(is_same_v< decltype(getMing()), const Ming >);

Okay, now you can go look at the Wandbox example and have your mind blown a little bit by the inconsistency of it all.

This inconsistency is amusing, but it never really gets in our way, because the inconsistency happens only when we const-qualify a by-value return — and that’s meaningless, so we should never do that! Guidelines:

const on a by-value parameter is never right. (It often indicates a missing ampersand.) const on a by-value return type is never right. (It often indicates a missing ampersand.)

The missing ampersand typo

Many times in student code — and even in production code written by professional programmers — I’ve seen APIs like this:

class Person { std::string firstname_; std::string lastname_; int age_; public: // ... int age() const { return age_; } const std::string firstname() const { return firstname_; } const std::string lastname() const { return lastname_; } };

Did you spot the red flag? Those last two methods were intended to be either

const std::string& firstname() const { return firstname_; } // return by-reference, as long as the caller promises not to modify the name

or

std::string firstname() const { return firstname_; } // return by-copy

We don’t know for sure which version was intended, but we know that what’s there now is a typo.

The misplaced const typo

Furthermore — more in student code than in professional code — I often see this mistake:

const int age() { return age_; } const std::string firstname() { return firstname_; } const std::string lastname() { return lastname_; }

This usually means that the student is confused about where the const goes on a method. In the above position, const is saying that age() returns a non-modifiable int ; but age() doesn’t promise anything about whether it modifies *this itself. Compare to:

int age() const { return age_; } std::string firstname() const { return firstname_; } std::string lastname() const { return lastname_; }

In this trailing position, const means that age() promises not to modify *this — which means now the compiler will allow us to call age() on a const-qualified Person object. This is what we want! If I have a handle to a Person whom I’ve promised not to modify, I should still be able to call .age() on it, right? Right.

Returning move-semantic types by const value is a pessimization

Consider this code:

std::string foo(); const std::string bar(); int main() { std::string x; x = foo(); // OK: move-assignment x = bar(); // OK, but: copy-assignment! }

Do you see what’s happening here? The assignment x = foo() ends up calling string::operator=(string&&) because the right-hand side is a prvalue of type string . But x = bar() cannot call string::operator=(string&&) because binding the parameter of type string&& to a prvalue of type const string would discard qualifiers. So instead it calls string::operator=(const string&) , which does a whole new heap-allocation in order to make a copy of the returned string.

So in the code from the previous section —

const std::string firstname() const { return firstname_; }

— not only did the typo make it difficult to understand our intent, but, even if we had meant to return by value, we were accidentally pessimizing a lot of places in our code!

Be aware that Scott Meyers’ Effective C++, Third Edition (Item 3, page 18) once “implied” that returning by const value could be a good thing.

const Rational operator*(const Rational& lhs, const Rational& rhs);

This advice was removed via an erratum in 2009, and in 2014 Meyers’ Effective Modern C++ (Item 25, page 172) gave a similar C++11 example with no return-by-const-value in sight:

Matrix operator+(Matrix&& lhs, const Matrix& rhs);

Guidelines for reliably const-correct code

const is a contract. When const appears in a function signature, it means that someone is making a promise to someone else not to modify something. Taking parameters by const reference is frequently useful, and makes a meaningful promise to your caller. Taking parameters by const value is never meaningful to the caller. The function signatures in your header files should never contain “const-by-value” parameters. Returning by const reference is sometimes useful, and extracts a meaningful promise from your caller. Returning by const value is never meaningful for scalar types, and is actually harmful for move-semantic class types. Your functions should never return “by const value.” Violations of these guidelines frequently indicate typos, pessimizations, or misunderstandings of the role of const .

Grep your codebase today!

# Look for pass-by-const-value typos git grep '[(,] *const [A-Za-z0-9_:]* [A-Za-z0-9_:]* *[,)=]' # Look for return-by-const-value typos git grep 'const [A-Za-z0-9_:]* [A-Za-z0-9_:]*([^")]*) const' git grep 'const [A-Za-z0-9_:]* [A-Za-z0-9_:]*([^")]*\( \|$\)'

The latter regex will inevitably have some false positives, because it cannot distinguish between

const bool get_value(int); // a (BAD) function declaration const bool my_value(true); // a (FINE) variable definition

Of course this ambiguity is eliminated if you use idiomatic variable definitions —

const bool my_value = true;

— but that’s a subject for a different post. :)