Top-level const doesn't influence a function signature

Solution 1:

allow these two functions simultaneously as different function since they are really different as to "whether parameter can be written or not". Intuitively, it should be!

Overloading of functions is based on the parameters the caller provides. Here, it's true that the caller may provide a const or non-const value but logically it should make no difference to the functionality that the called function provides. Consider:

f(3);
int x = 1 + 2;
f(x);

If f() does different thing in each of these situations, it would be very confusing! The programmer of this code calling f() can have a reasonable expectation of identical behaviour, freely adding or removing variables that pass parameters without it invalidating the program. This safe, sane behaviour is the point of departure that you'd want to justify exceptions to, and indeed there is one - behaviours can be varied when the function's overloaded ala:

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

So, I guess this is what you find non-intuitive: that C++ provides more "safety" (enforced consistent behaviour through supporting only a single implementation) for non-references than references.

The reasons I can think of are:

• So when a programmer knows a non-const& parameter will have a longer lifetime, they can select an optimal implementation. For example, in the code below it may be faster to return a reference to a T member within F, but if F is a temporary (which it might be if the compiler matches const F&) then a by-value return is needed. This is still pretty dangerous as the caller has to be aware that the returned reference is only valid as long as the parameter's around.

    T f(const F&);
    T& f(F&);    // return type could be by const& if more appropriate

• propagation of qualifiers like const-ness through function calls as in:

    const T& f(const F&);
    T& f(F&);

Here, some (presumably F member-) variable of type T is being exposed as const or non-const based on the const-ness of the parameter when f() is called. This type of interface might be chosen when wishing to extend a class with non-member functions (to keep the class minimalist, or when writing templates/algos usable on many classes), but the idea is similar to const member functions like vector::operator[](), where you want v[0] = 3 allowed on a non-const vector but not a const one.

When values are accepted by value they go out of scope as the function returns, so there's no valid scenario involving returning a reference to part of the parameter and wanting to propagate its qualifiers.

Hacking the behaviour you want

Given the rules for references, you can use them to get the kind of behaviour you want - you just need to be careful not to modify the by-non-const-reference parameter accidentally, so might want to adopt a practice like the following for the non-const parameters:

T f(F& x_ref)
{
    F x = x_ref;  // or const F is you won't modify it
    ...use x for safety...
}

Recompilation implications

Quite apart from the question of why the language forbids overloading based on the const-ness of a by-value parameter, there's the question of why it doesn't insist on consistency of const-ness in the declaration and definition.

For f(const int) / f(int)... if you are declaring a function in a header file, then it's best NOT to include the const qualifier even if the later definition in an implementation file will have it. This is because during maintenance the programmer may wish to remove the qualifier... removing it from the header may trigger a pointless recompilation of client code, so it's better not to insist they be kept in sync - and indeed that's why the compiler doesn't produce an error if they differ. If you just add or remove const in the function definition, then it's close to the implementation where the reader of the code might care about the constness when analysing the function behaviour. If you have it const in both header and implementation file, then the programmer wishes to make it non-const and forgets or decides not to update the header in order to avoid client recompilation, then it's more dangerous than the other way around as it's possible the programmer will have the const version from the header in mind when trying to analyse the current implementation code leading to wrong reasoning about the function behaviour. This is all a very subtle maintainence issue - only really relevant to commercial programming - but that's the basis of the guideline not to use const in the interface. Further, it's more concise to omit it from the interface, which is nicer for client programmers reading over your API.

Solution 2:

Since there is no difference to the caller, and no clear way to distinguish between a call to a function with a top level const parameter and one without, the language rules ignore top level consts. This means that these two

void foo(const int);
void foo(int);

are treated as the same declaration. If you were to provide two implementations, you would get a multiple definition error.

There is a difference in a function definition with top level const. In one, you can modify your copy of the parameter. In the other, you can't. You can see it as an implementation detail. To the caller, there is no difference.

// declarations
void foo(int);
void bar(int);

// definitions
void foo(int n)
{
  n++;
  std::cout << n << std::endl;
}

void bar(const int n)
{
  n++; // ERROR!
  std::cout << n << std::endl;
}

This is analogous to the following:

void foo()
{
  int = 42;
  n++;
  std::cout << n << std::endl;
}

void bar()
{
  const int n = 42;
  n++; // ERROR!
  std::cout << n << std::endl;
}

Solution 3:

In "The C++ Programming Language", fourth edition, Bjarne Stroustrup writes (§12.1.3):

Unfortunately, to preserve C compatibility, a const is ignored at the highest level of an argument type. For example, this is two declarations of the same function:

void f(int);
void f(const int);

So, it seems that, contrarily to some of the other answers, this rule of C++ was not chosen because of the indistinguishability of the two functions, or other similar rationales, but instead as a less-than-optimal solution, for the sake of compatibility.

Indeed, in the D programming language, it is possible to have those two overloads. Yet, contrarily to what other answers to this question might suggest, the non-const overload is preferred if the function is called with a literal:

void f(int);
void f(const int);

f(42); // calls void f(int);

Of course, you should provide equivalent semantics for your overloads, but that is not specific to this overloading scenario, with nearly indistinguishable overloading functions.

Solution 4:

As the comments say, inside the first function the parameter could be changed, if it had been named. It is a copy of the callee's int. Inside the second function, any changes to the parameter, which is still a copy of the callee's int, will result in a compile error. Const is a promise you won't change the variable.