Use of the & Operator in C++ Function Signatures

Use of the & operator in C++ function signatures

A reference is not a pointer, they're different although they serve similar purpose.
You can think of a reference as an alias to another variable, i.e. the second variable having the same address. It doesn't contain address itself, it just references the same portion of memory as the variable it's initialized from.

string s = "Hello, wordl";
string* p = &s; // Here you get an address of s
string& r = s; // Here, r is a reference to s    

s = "Hello, world"; // corrected
assert( s == *p ); // this should be familiar to you, dereferencing a pointer
assert( s == r ); // this will always be true, they are twins, or the same thing rather

string copy1 = *p; // this is to make a copy using a pointer
string copy = r; // this is what you saw, hope now you understand it better.

Use of * and & together in function signature

I need to use pointer with address of

Here, & does not mean "address of"; it means the type "reference to".

It's clearer if you write it not like this:

vector<int>* &primes

but like this:

vector<int>*& primes

Though the choice of whitespace is artificial, that better documents that this & is "part of the type".

Have some types:

std::vector<T> = A vector of Ts
std::vector<T>& = A reference to a vector of Ts
std::vector<T>* = A pointer to a vector of Ts
std::vector<T>*& = A reference to a pointer to a vector of Ts
std::vector<T>*** = A pointer to a pointer to a pointer to a vector of Ts
std::vector<T>**& = A reference to a pointer to a pointer to a vector of Ts

…and so forth.

As for why you need a vector<int>*& for printPrimes to do its job, we could not tell you without actually being able to see it. I will say that it seems unlikely it needs a pointer at all, and that if it wants to modify that pointer it's going to cause problems with the new and delete in the calling scope.

In fact, all that dynamic allocation is completely pointless and only complicates things.

The following was likely intended instead:

void printPrimes(long long l, long long r, vector<int>& primes) {
 // some code
}
vector<int> sieve() {
 vector<int> prime;
 return prime;

}
int main() {
    vector<int> primes = sieve();
    printPrimes(l, r, primes);
}

Use of '&' operator before a function name in C++

In C++, when the ref-sign (&) is used before the function name in the declaration of a function it is associated with the return value of the function and means that the function will return by reference.

int& foo(); // Function will return an int by reference.

When not used within a declaration context, putting the ref-sign before a function name results in calling the address-of operator returning the address of the function. This can be used to e.g. create a pointer to a function.

// Some function.
int sum(int a, int b) {
    return a + b;
}

int main() {
    // Declare type func_ptr_t as pointer to function of type int(int, int).
    using func_ptr_t = std::add_pointer<int(int, int)>::type;

    func_ptr_t func = ∑ // Declare func as pointer to sum using address-of.
    int sum = func(1, 2);   // Initialize variable sum with return value of func.
}

In C, the only use of & is for the address-of operator. References does not exist in the C language.

What does '&' do in a C++ declaration?

The "&" denotes a reference instead of a pointer to an object (In your case a constant reference).

The advantage of having a function such as

foo(string const& myname)

over

foo(string const* myname)

is that in the former case you are guaranteed that myname is non-null, since C++ does not allow NULL references. Since you are passing by reference, the object is not copied, just like if you were passing a pointer.

Your second example:

const string &GetMethodName() { ... }

Would allow you to return a constant reference to, for example, a member variable. This is useful if you do not wish a copy to be returned, and again be guaranteed that the value returned is non-null. As an example, the following allows you direct, read-only access:

class A
{
  public:
  int bar() const {return someValue;}
  //Big, expensive to copy class
}

class B
{
public:
 A const& getA() { return mA;}
private:
 A mA;
}
void someFunction()
{
 B b = B();
 //Access A, ability to call const functions on A
 //No need to check for null, since reference is guaranteed to be valid.
 int value = b.getA().bar(); 
}

You have to of course be careful to not return invalid references.
Compilers will happily compile the following (depending on your warning level and how you treat warnings)

int const& foo() 
{
 int a;

 //This is very bad, returning reference to something on the stack. This will
 //crash at runtime.
 return a; 
}

Basically, it is your responsibility to ensure that whatever you are returning a reference to is actually valid.

What does the & (ampersand) at the end of member function signature mean?

Ref-qualifiers - introduced in C++11

Ref-qualifiers is not C++17 feature (looking at the tag of the question), but was a feature introduced in C++11.

struct Foo
{
  void bar() const &  { std::cout << "const lvalue Foo\n"; }
  void bar()       &  { std::cout << "lvalue Foo\n"; }
  void bar() const && { std::cout << "const rvalue Foo\n"; }
  void bar()       && { std::cout << "rvalue Foo\n"; }
};

const Foo&& getFoo() { return std::move(Foo()); }

int main()
{
  const Foo c_foo;
  Foo foo;

  c_foo.bar();            // const lvalue Foo
  foo.bar();              // lvalue Foo
  getFoo().bar();         // [prvalue] const rvalue Foo
  Foo().bar();            // [prvalue] rvalue Foo

  // xvalues bind to rvalue references, and overload resolution
  // favours selecting the rvalue ref-qualifier overloads.
  std::move(c_foo).bar(); // [xvalue] const rvalue Foo
  std::move(foo).bar();   // [xvalue] rvalue Foo
}

Note that an rvalue may be used to initialize a const lvalue reference (and in so expanding the lifetime of the object identified by the rvalue), meaning that if we remove the rvalue ref-qualifier overloads from the example above, then the rvalue value categories in the example will all favour the remaining const & overload:

struct Foo
{
  void bar() const & { std::cout << "const lvalue Foo\n"; }
  void bar()       & { std::cout << "lvalue Foo\n"; }
};

const Foo&& getFoo() { return std::move(Foo()); }

int main()
{
  const Foo c_foo;
  Foo foo;

  // For all rvalue value categories overload resolution
  // now selects the 'const &' overload, as an rvalue may
  // be used to initialize a const lvalue reference.
  c_foo.bar();            // const lvalue Foo
  foo.bar();              // lvalue Foo
  getFoo().bar();         // const lvalue Foo
  Foo().bar();            // const lvalue Foo
  std::move(c_foo).bar(); // const lvalue Foo
  std::move(foo).bar();   // const lvalue Foo
}

See e.g. the following blog post for for a brief introduction:

Andrzej's C++ blog - Ref-qualifiers

rvalues cannot invoke non-`const` `&` overloads

To possibly explain the intent of your recollected quote from the CppCon talk,

"... that the only true way of overloading operator= ..."

we visit [over.match.funcs]/1, /4 & /5 [emphasis mine]:

/1 The subclauses of [over.match.funcs] describe the set of candidate functions and the argument list submitted to overload
resolution in each context in which overload resolution is used. ...

/4 For non-static member functions, the type of the implicit object parameter is

(4.1) — “lvalue reference to cv X” for functions declared without a ref-qualifier or with the & ref-qualifier

(4.2) — “rvalue reference to cv X” for functions declared with the && ref-qualifier

where X is the class of which the function is a member and cv is the
cv-qualification on the member function declaration. ...

/5 ... For non-static member functions declared without a ref-qualifier, an additional rule applies:

(5.1) — even if the implicit object parameter is not const-qualified, an rvalue can be bound to the parameter as long as
in all other respects the argument can be converted to the type of the
implicit object parameter. [ Note: The fact that such an argument is
an rvalue does not affect the ranking of implicit conversion
sequences. — end note ]

From /5 above, the following overload (where the explicit & ref-qualifier has been omitted)

struct test
{
    test& operator=(const test&) { return *this }
}

allows assigning values to r-values, e.g.

int main()
{
    test t1;
    t1 = test(); // assign to l-value
    test() = t1; // assign to r-value
}

However, if we explicitly declare the overload with the & ref-qualifier, [over.match.funcs]/5.1 does not apply, and as long we do not supply an overload declared with the && ref-qualifier, r-value assignment will not be allowed.

struct test
{
    test& operator=(const test&) & { return *this; }
};

int main()
{
    test t1;
    t1 = test(); // assign to l-value
    test() = t1; // error [clang]: error: no viable overloaded '='
}

I won't place any opinion as to whether explicitly including the & ref-qualifier when declaring custom assignment operator overloads is "the only true way of overload operator=", but would I dare to speculate, then I would guess that the intent behind such a statement is the exclusion of to-r-value assignment.

As a properly designed assignment operator should arguably never be const (const T& operator=(const T&) const & would not make much sense), and as an rvalue may not be used to initialize a non-const lvalue reference, a set of overloads for operator= for a given type T that contain only T& operator=(const T&) & will never proviade a viable overload that can be invoked from a T object identified to be of an rvalue value category.

What does && mean at the end of a function signature (after the closing parenthesis)?

This is a ref-value qualifier. Here is a basic example:

// t.cpp
#include <iostream>

struct test{
  void f() &{ std::cout << "lvalue object\n"; }
  void f() &&{ std::cout << "rvalue object\n"; }
};

int main(){
  test t;
  t.f(); // lvalue
  test().f(); // rvalue
}

Output:

$ clang++ -std=c++0x -stdlib=libc++ -Wall -pedantic t.cpp
$ ./a.out
lvalue object
rvalue object

Taken from here.

Ternary operator and function signature

Not with the ternary operator. The type of a ternary expression is the common type of its second and third operands; if they have no common type you can't use it. So just use an ordinary if statement:

if (booleanValue)
    MyClassInstance->Foo(24);
else
    MyClassInstance->Foo("a string");

Use of the & Operator in C++ Function Signatures