Intro

It is universally agreed that C++ is a complex language. One reason is that its syntax is highly overloaded, meaning that the same code could mean many different things. Let’s do a simple mental exercise by considering the following code:

f(x)

How many possible meanings can there be?

Macros

First, let get macros out of the box. If any of f and x is a macro, f(x) can expand to basically anything. Hereafter, we presume that f and x are not macros.

Function Call

f(x) can be a call to function named f with argument x. Duh.

void f(int);

int x = 0;
f(x);  // call to f. Simple as that.

There are a few sub cases:

  1. If f is a function template, then template argument deduction takes place, and implicit template instantiation follows. The compiler automatically generates a specialization (f<int>) for you.

  2. If f is an overload set, then overload resolution takes place. The compiler picks the best candidate for you.

Note that 1 and 2 can both happen in a single call.

Indirect Function Call

If f is a pointer to function, then f(x) is the same as (*f)(x) - indirect function call. The indirection comes from “dereferencing” the pointer to function to get the actual function.

This “shortcut” syntax of indirect function call is inherited from C.

void g(int);

auto f = &g;  // f has type void (*)(int)
auto f = g;   // same as above; function decay

int x = 0;
f(x);     // call g(x)
(*f)(x);  // call g(x)

This is a very powerful technique. It is one way to get polymorphism.

Similarly, when f is a reference to function (yes, functions can have references):

void g(int);

auto& f = g;  // f has type void (&)(int)

int x = 0;
f(x);  // call g(x)

Call Operator

If f is an object whose type provides operator(), then f(x) invokes that call operator. Such f is commonly referred to as a function object or simply functor:

struct F {
    void operator()(int);  // (*)
};

F f;
int x = 0;
f(x);  // calls (*)

Lambdas, std::function, and the return type of std::bind/std::bind_front/std::bind_back, are all examples of such.

Just like free functions, this operator() can be overloaded and templated.

  • Generic lambdas have templated operator()

Surrogate Function

On the other hand, a type can define a conversion operator that converts the instance to a function pointer, thus making it “callable”. That is:

void g(int);

struct F {
    using Fun_Pointer = void(*)(int);
    operator Fun_Pointer() { return g; }
};

F f;
int x = 0;
f(x);  // g(x)

This is known as “Surrogate Function”.

  • A capture-less lambda is also a surrogate function, because it can convert to a function pointer

Object Creation

If f is a type (or an alias to a type) and x is an object, then f(x) creates a new object by calling the constructor:

struct f {
    f(int);
};

int x = 0;
f(x);  // (1)

(1) creates a temporary object and immediately destroys it.

Right?

Wrong.

The above code doesn’t compile. Here’s what the compiler says:

<source>:19:11: error: conflicting declaration 'main()::f x'
   19 |         f(x); // (1)
      |           ^
<source>:18:13: note: previous declaration as 'int x'
   18 |         int x = 0;

(1) is actually a definition - it defines a variable x of type f. It is the same as f x;. The parentheses are redundant, but permitted nonetheless.

struct f {};

f(x);  // Ok: declare a variable `x` of type `f`, then default-initialize it

The parentheses are even allowed in function parameter list, which I don’t even want to talk about:

void fun( int (x), int (y) );  // Why would anyone write it this way?

Anyway, to make f(x) do what we thought it’d do, we can wrap it with a pair of parentheses to make it an expression:

struct f {
    f(int);
};

int x = 0;
(f(x));  // Create a temporary object and immediately destroys it

What about just f(x);? Can this create an (temporary) object?

Yes, if f is a class template, and class template argument deduction (CTAD) kicks in:

template <class T>
struct f {
    f(T);
};

int main() {
    int x = 0;
    f(x);  // (?)
}

Update: Since this post was published, r/sagittarius_ack pointed it out that the above code doesn’t compile, at least on GCC, Clang and MSVC, due to conflicting definition. It does compile on EDG though - EDG treats the line (?) as an expression statement that creates a temporary object and immediately destroys it.

Even more interestingly, consider the following code:

#include <iostream>

template <class T = int>
struct f {
    f()  { std::cout << "1"; }
    f(T) { std::cout << "2"; }
    ~f() { std::cout << "3"; }
};

int x = 0;

int main() {
    f(x);  // (?)
}

There is implementation divergence:

I’m not sure what’s going on.

Invoke Static Call Operator?

Since C++23, call operators are allowed to be static:

struct f {
    static void operator()(int);  // Ok since C++23
};

So, does this make f(x) an invocation to that static method, given x is an int?

No.

The correct way to call that static method is the same as how you usually call a static method - use the :: scope resolution:

f::operator()(x);  // Ok since C++23

By the way, C++ has always allowed you to invoke a static method through an instance (even it is not used at all):

f v;
v(x);  // Ok

In short, despite its look, f(x) is never a call to the static call operator of type f.

  • Similarly, despite operator[] is allowed static, you cannot call it like T[arg]. You have to write T::operator[](arg).

Constructor Declaration

If f is a class type and x is a type or type alias, then within the definition of class f, f(x) declares a constructor:

struct x {};
class f {
    f(x);  // single-argument constructor
};

Function Type

If both f and x are types or type aliases, then f(x) is a function type - functions that take an x and return an f.

struct f {};
struct x {};

f(x);  // (1)

The above code compiles, so (1) must have formed a function type.

Right?

Wrong.

C++ does not allow “empty declarations” - declaration does not declare anything. In short, this is illegal:

int;  // error:  declaration does not declare anything

But f(x); compiles. Are function types an exception to the rule?

No. Here, f(x); defines a new variable called x, same as before. Even though x is already a type prior to the line.

Yes, you read it right. For whatever reasons, C++ allows (re)using class names as variable names:

class Bar {};   // Bar is a class
Bar b;          // Ok, define a variable `b`

int Bar = 0;    // Bar is an object now!
Bar b2;         // Error

// If you really want the class Bar back:
class Bar b2;   // Ok, define a variable `b2`

f(x) is interpreted as a function type elsewhere, when a type is expected:

using F = f(x);          // Ok: declare a type alias F that refers to the function type
std::function<f(x)> ff;  // Ok

Just not f(x);.

What we have So Far

We can produce a summary so far:

  1. If f is a function, function template, or pointer to function, then f(x) is a function call

  2. If f is an object, then f(x) either invokes the call operator (static or not), or invokes the conversion operator that returns a pointer to function and calls that instead

  3. If f is a class type, then f(x); declares an object x of type f, regardless of what x is

  4. If f is a class type and x is an object, and f(x) is part of an expression, then f(x) creates a temporary object via calling the constructor

  5. If f is a class type and x is a class type, then f(x) is a function type

  6. If f is a class template and x is an object, then f(x) creates a temporary object, provided that class template argument deduction succeeds; this is even the case for f(x);

Could there be more?

Argument Dependent Lookup (ADL)

This is a special case of calling a regular function. Basically:

namespace my {
    struct Bar {};

    void f(Bar) {}
}

int main() {
    my::Bar x;
    f(x);  // Ok; same as my::f(x)
}

Note that the f is unqualified, yet the compiler finds my::f because the argument x, a my::Bar, brings in everything under the namespace my.

What this means is, f(x) can mean ns1::f(x), ns1::ns2::f(x), and so on.

Conversion Operator

When f is a type (or a type alias), and x has a conversion operator to f, f(x) may call that conversion operator:

struct f;

struct X {
    operator f();  // (2)
};

X x;
auto y = f(x);  // Calls (2); `y` is an `f`

Then again, as we’ve seen more than once already, the following is a declaration:

f(x);  // Declare a variable `x` of type `f`, then default-initialize it

Functional-style Cast

Generally, when f is a type (or a type alias), f(x) is the syntax for function-style cast - cast the expression x to target type f.

As you can see, it gets the name because it looks like a function call, but is not necessarily a function call. For example, the pointer cast that we all like to do:

class T {
    ... 
} object;        // An object ...
T* x = &object;  // and a pointer to it

using f = const unsigned char*;
auto p = f(x);  // No function call happening; it's just a cast
// Now we can poke around the object's internals like a pro

C-style casts, of which function-style cast is one form, despite its perceived simplicity and elegancy, are generally advised against using (in fact, all explicit casts are recommended to keep to a minimum). There is actually a lot of mechanics going on behind the scenes - C-style casts can do a combination of const_cast, static_cast, reinterpret_cast - and is even an active issue and evolving as of this writing.

  • I would only use C-style casts for fundamental types - int, double, and the like. Pointers are excluded.

Take Away

Now comes the summary, take two. Roughly, f(x) can mean these different things:

  1. Some sort of function call to f
  2. A cast to type f (including calls to constructor and conversion operator)
  3. A function type
  4. A declaration of x

4 is really surprising and should be avoided, 3 is related to 1. That leaves us only 2 and 1.

f(x) should mean a function call. Period. The fact that it sometimes means a declaration is unfortunate; 4 should be avoided. 3 is really a “prototype” of 1. Now, if C-style casts are replaced with C++ casts (const_cast, static_cast, reinterpret_cast), then f(x) can really, really mean a function call in the codebase.

Or maybe I missed something?