What does the explicit
keyword mean in C++?
C++ – What does the explicit keyword mean
cc++-faqconstructorexplicitexplicit-constructor
Related Solutions
A pointer can be re-assigned:
int x = 5; int y = 6; int *p; p = &x; p = &y; *p = 10; assert(x == 5); assert(y == 10);
A reference cannot be re-bound, and must be bound at initialization:
int x = 5; int y = 6; int &q; // error int &r = x;
A pointer variable has its own identity: a distinct, visible memory address that can be taken with the unary
&
operator and a certain amount of space that can be measured with thesizeof
operator. Using those operators on a reference returns a value corresponding to whatever the reference is bound to; the reference’s own address and size are invisible. Since the reference assumes the identity of the original variable in this way, it is convenient to think of a reference as another name for the same variable.int x = 0; int &r = x; int *p = &x; int *p2 = &r; assert(p == p2); // &x == &r assert(&p != &p2);
You can have arbitrarily nested pointers to pointers offering extra levels of indirection. References only offer one level of indirection.
int x = 0; int y = 0; int *p = &x; int *q = &y; int **pp = &p; **pp = 2; pp = &q; // *pp is now q **pp = 4; assert(y == 4); assert(x == 2);
A pointer can be assigned
nullptr
, whereas a reference must be bound to an existing object. If you try hard enough, you can bind a reference tonullptr
, but this is undefined and will not behave consistently./* the code below is undefined; your compiler may optimise it * differently, emit warnings, or outright refuse to compile it */ int &r = *static_cast<int *>(nullptr); // prints "null" under GCC 10 std::cout << (&r != nullptr ? "not null" : "null") << std::endl; bool f(int &r) { return &r != nullptr; } // prints "not null" under GCC 10 std::cout << (f(*static_cast<int *>(nullptr)) ? "not null" : "null") << std::endl;
You can, however, have a reference to a pointer whose value is
nullptr
.Pointers can iterate over an array; you can use
++
to go to the next item that a pointer is pointing to, and+ 4
to go to the 5th element. This is no matter what size the object is that the pointer points to.A pointer needs to be dereferenced with
*
to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses->
to access its members whereas a reference uses a.
.References cannot be put into an array, whereas pointers can be (Mentioned by user @litb)
Const references can be bound to temporaries. Pointers cannot (not without some indirection):
const int &x = int(12); // legal C++ int *y = &int(12); // illegal to take the address of a temporary.
This makes
const &
more convenient to use in argument lists and so forth.
static_cast
is the first cast you should attempt to use. It does things like implicit conversions between types (such as int
to float
, or pointer to void*
), and it can also call explicit conversion functions (or implicit ones). In many cases, explicitly stating static_cast
isn't necessary, but it's important to note that the T(something)
syntax is equivalent to (T)something
and should be avoided (more on that later). A T(something, something_else)
is safe, however, and guaranteed to call the constructor.
static_cast
can also cast through inheritance hierarchies. It is unnecessary when casting upwards (towards a base class), but when casting downwards it can be used as long as it doesn't cast through virtual
inheritance. It does not do checking, however, and it is undefined behavior to static_cast
down a hierarchy to a type that isn't actually the type of the object.
const_cast
can be used to remove or add const
to a variable; no other C++ cast is capable of removing it (not even reinterpret_cast
). It is important to note that modifying a formerly const
value is only undefined if the original variable is const
; if you use it to take the const
off a reference to something that wasn't declared with const
, it is safe. This can be useful when overloading member functions based on const
, for instance. It can also be used to add const
to an object, such as to call a member function overload.
const_cast
also works similarly on volatile
, though that's less common.
dynamic_cast
is exclusively used for handling polymorphism. You can cast a pointer or reference to any polymorphic type to any other class type (a polymorphic type has at least one virtual function, declared or inherited). You can use it for more than just casting downwards – you can cast sideways or even up another chain. The dynamic_cast
will seek out the desired object and return it if possible. If it can't, it will return nullptr
in the case of a pointer, or throw std::bad_cast
in the case of a reference.
dynamic_cast
has some limitations, though. It doesn't work if there are multiple objects of the same type in the inheritance hierarchy (the so-called 'dreaded diamond') and you aren't using virtual
inheritance. It also can only go through public inheritance - it will always fail to travel through protected
or private
inheritance. This is rarely an issue, however, as such forms of inheritance are rare.
reinterpret_cast
is the most dangerous cast, and should be used very sparingly. It turns one type directly into another — such as casting the value from one pointer to another, or storing a pointer in an int
, or all sorts of other nasty things. Largely, the only guarantee you get with reinterpret_cast
is that normally if you cast the result back to the original type, you will get the exact same value (but not if the intermediate type is smaller than the original type). There are a number of conversions that reinterpret_cast
cannot do, too. It's used primarily for particularly weird conversions and bit manipulations, like turning a raw data stream into actual data, or storing data in the low bits of a pointer to aligned data.
C-style cast and function-style cast are casts using (type)object
or type(object)
, respectively, and are functionally equivalent. They are defined as the first of the following which succeeds:
const_cast
static_cast
(though ignoring access restrictions)static_cast
(see above), thenconst_cast
reinterpret_cast
reinterpret_cast
, thenconst_cast
It can therefore be used as a replacement for other casts in some instances, but can be extremely dangerous because of the ability to devolve into a reinterpret_cast
, and the latter should be preferred when explicit casting is needed, unless you are sure static_cast
will succeed or reinterpret_cast
will fail. Even then, consider the longer, more explicit option.
C-style casts also ignore access control when performing a static_cast
, which means that they have the ability to perform an operation that no other cast can. This is mostly a kludge, though, and in my mind is just another reason to avoid C-style casts.
Related Topic
- C++ – The Definitive C++ Book Guide and List
- C++ – Why is “using namespace std;” considered bad practice
- C++ – the “–>” operator in C/C++
- C++ – the copy-and-swap idiom
- C++ – The Rule of Three
- C++ – What are the basic rules and idioms for operator overloading
- C++11 introduced a standardized memory model. What does it mean? And how is it going to affect C++ programming
- Java – Why is processing a sorted array faster than processing an unsorted array
Best Answer
The compiler is allowed to make one implicit conversion to resolve the parameters to a function. What this means is that the compiler can use constructors callable with a single parameter to convert from one type to another in order to get the right type for a parameter.
Here's an example class with a constructor that can be used for implicit conversions:
Here's a simple function that takes a
Foo
object:and here's where the
DoBar
function is called:The argument is not a
Foo
object, but anint
. However, there exists a constructor forFoo
that takes anint
so this constructor can be used to convert the parameter to the correct type.The compiler is allowed to do this once for each parameter.
Prefixing the
explicit
keyword to the constructor prevents the compiler from using that constructor for implicit conversions. Adding it to the above class will create a compiler error at the function callDoBar (42)
. It is now necessary to call for conversion explicitly withDoBar (Foo (42))
The reason you might want to do this is to avoid accidental construction that can hide bugs.
Contrived example:
MyString
class with a constructor that constructs a string of the given size. You have a functionprint(const MyString&)
(as well as an overloadprint (char *string)
), and you callprint(3)
(when you actually intended to callprint("3")
). You expect it to print "3", but it prints an empty string of length 3 instead.