What does the explicit
keyword mean in C++?
explicit
can be applied to more than just constructors. It's now valid when applied to conversion operators as well. Say you have a class BigInt
with a conversion operator to int
and an explicit conversion operator to std::string
for whatever reason. You'll be able to say int i = myBigInt;
, but you'll have to cast explicitly (using static_cast
, preferably) in order to say std::string s = myBigInt;
. - anyone int x(5);
) - anyone bool
is special in this regard. Those answers and searching "explicit conversion operators" will lead you to more writeups about this feature and be better suited than a comment chain. - anyone The compiler is allowed to make one implicit conversion to resolve the parameters to a function. 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 with converting constructors that shows how it works:
struct Foo {
// Single parameter constructor, can be used as an implicit conversion.
// Such a constructor is called "converting constructor".
Foo(int x) {}
};
struct Faz {
// Also a converting constructor.
Faz(Foo foo) {}
};
// The parameter is of type Foo, not of type int, so it looks like
// we have to pass a Foo.
void bar(Foo foo);
int main() {
// However, the converting constructor allows us to pass an int.
bar(42);
// Also allowed thanks to the converting constructor.
Foo foo = 42;
// Error! This would require two conversions (int -> Foo -> Faz).
Faz faz = 42;
}
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 call bar(42)
. It is now necessary to call for conversion explicitly with bar(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 function print(const MyString&)
(as well as an overload print (char *string)
), and you call print(3)
(when you actually intended to call print("3")
). You expect it to print "3", but it prints an empty string of length 3 instead.Answered 2023-09-20 20:15:42
explicit
-- you declare a constructor explicit
. But yes: your parameters of type Foo
have to be constructed explicite
ly, they won't be silently constructed by just plugging their constructor's parameters into the function. - anyone Suppose, you have a class String
:
class String {
public:
String(int n); // allocate n bytes to the String object
String(const char *p); // initializes object with char *p
};
Now, if you try:
String mystring = 'x';
The character 'x'
will be implicitly converted to int
and then the String(int)
constructor will be called. But, this is not what the user might have intended. So, to prevent such conditions, we shall define the constructor as explicit
:
class String {
public:
explicit String (int n); //allocate n bytes
String(const char *p); // initialize sobject with string p
};
Answered 2023-09-20 20:15:42
String s = {0};
ill-formed, rather than trying to call the other constructor with a null pointer, as String s = 0;
would do. - anyone String mystring('x')
when you meant String mystring("x")
wouldn't you? Also, from the comment above I see the improved behavior of String s = {0}
over String s = 0
thanks to making the int form of the ctor 'explicit'. But, other than knowing the precedence of the ctors how do you know the intent (i.e. how to spot the bug) of this String s{0}
? - anyone 'x'
is being treated as an integer because the char
data type is just a 1-byte integer. - anyone =
) but not with direct initialization (without using =
): the compiler will still call the String(int)
constructor without generating an error if you write String mystring('x');
, as @Arbalest pointed out. The explicit
keyword is meant for preventing implicit conversions that happen in direct initialization and function resolution. A better solution to your example would be a simple overload of the constructor: String(char c);
. - anyone In C++, a constructor with only one required parameter is considered an implicit conversion function. It converts the parameter type to the class type. Whether this is a good thing or not depends on the semantics of the constructor.
For example, if you have a string class with constructor String(const char* s)
, that's probably exactly what you want. You can pass a const char*
to a function expecting a String
, and the compiler will automatically construct a temporary String
object for you.
On the other hand, if you have a buffer class whose constructor Buffer(int size)
takes the size of the buffer in bytes, you probably don't want the compiler to quietly turn int
s into Buffer
s. To prevent that, you declare the constructor with the explicit
keyword:
class Buffer { explicit Buffer(int size); ... }
That way,
void useBuffer(Buffer& buf);
useBuffer(4);
becomes a compile-time error. If you want to pass a temporary Buffer
object, you have to do so explicitly:
useBuffer(Buffer(4));
In summary, if your single-parameter constructor converts the parameter into an object of your class, you probably don't want to use the explicit
keyword. But if you have a constructor that simply happens to take a single parameter, you should declare it as explicit
to prevent the compiler from surprising you with unexpected conversions.
Answered 2023-09-20 20:15:42
useBuffer
expects an lvalue for his argument, useBuffer(Buffer(4))
will not work either because of it. Changing it to take a const Buffer&
or Buffer&&
or just Buffer
would make it work. - anyone explicit
accompanies eitherC++ [class.conv.ctor]
1) A constructor declared without the function-specifier explicit specifies a conversion from the types of its parameters to the type of its class. Such a constructor is called a converting constructor.
2) An explicit constructor constructs objects just like non-explicit constructors, but does so only where the direct-initialization syntax (8.5) or where casts (5.2.9, 5.4) are explicitly used. A default constructor may be an explicit constructor; such a constructor will be used to perform default-initialization or valueinitialization (8.5).
C++ [class.conv.fct]
2) A conversion function may be explicit (7.1.2), in which case it is only considered as a user-defined conversion for direct-initialization (8.5). Otherwise, user-defined conversions are not restricted to use in assignments and initializations.
Explicit conversion functions and constructors can only be used for explicit conversions (direct initialization or explicit cast operation) while non-explicit constructors and conversion functions can be used for implicit as well as explicit conversions.
/*
explicit conversion implicit conversion
explicit constructor yes no
constructor yes yes
explicit conversion function yes no
conversion function yes yes
*/
X, Y, Z
and functions foo, bar, baz
:Let's look at a small setup of structures and functions to see the difference between explicit
and non-explicit
conversions.
struct Z { };
struct X {
explicit X(int a); // X can be constructed from int explicitly
explicit operator Z (); // X can be converted to Z explicitly
};
struct Y{
Y(int a); // int can be implicitly converted to Y
operator Z (); // Y can be implicitly converted to Z
};
void foo(X x) { }
void bar(Y y) { }
void baz(Z z) { }
Conversion of a function argument:
foo(2); // error: no implicit conversion int to X possible
foo(X(2)); // OK: direct initialization: explicit conversion
foo(static_cast<X>(2)); // OK: explicit conversion
bar(2); // OK: implicit conversion via Y(int)
bar(Y(2)); // OK: direct initialization
bar(static_cast<Y>(2)); // OK: explicit conversion
Object initialization:
X x2 = 2; // error: no implicit conversion int to X possible
X x3(2); // OK: direct initialization
X x4 = X(2); // OK: direct initialization
X x5 = static_cast<X>(2); // OK: explicit conversion
Y y2 = 2; // OK: implicit conversion via Y(int)
Y y3(2); // OK: direct initialization
Y y4 = Y(2); // OK: direct initialization
Y y5 = static_cast<Y>(2); // OK: explicit conversion
X x1{ 0 };
Y y1{ 0 };
Conversion of a function argument:
baz(x1); // error: X not implicitly convertible to Z
baz(Z(x1)); // OK: explicit initialization
baz(static_cast<Z>(x1)); // OK: explicit conversion
baz(y1); // OK: implicit conversion via Y::operator Z()
baz(Z(y1)); // OK: direct initialization
baz(static_cast<Z>(y1)); // OK: explicit conversion
Object initialization:
Z z1 = x1; // error: X not implicitly convertible to Z
Z z2(x1); // OK: explicit initialization
Z z3 = Z(x1); // OK: explicit initialization
Z z4 = static_cast<Z>(x1); // OK: explicit conversion
Z z1 = y1; // OK: implicit conversion via Y::operator Z()
Z z2(y1); // OK: direct initialization
Z z3 = Z(y1); // OK: direct initialization
Z z4 = static_cast<Z>(y1); // OK: explicit conversion
explicit
conversion functions or constructors?Conversion constructors and non-explicit conversion functions may introduce ambiguity.
Consider a structure V
, convertible to int
, a structure U
implicitly constructible from V
and a function f
overloaded for U
and bool
respectively.
struct V {
operator bool() const { return true; }
};
struct U { U(V) { } };
void f(U) { }
void f(bool) { }
A call to f
is ambiguous if passing an object of type V
.
V x;
f(x); // error: call of overloaded 'f(V&)' is ambiguous
The compiler does not know wether to use the constructor of U
or the conversion function to convert the V
object into a type for passing to f
.
If either the constructor of U
or the conversion function of V
would be explicit
, there would be no ambiguity since only the non-explicit conversion would be considered. If both are explicit the call to f
using an object of type V
would have to be done using an explicit conversion or cast operation.
Conversion constructors and non-explicit conversion functions may lead to unexpected behaviour.
Consider a function printing some vector:
void print_intvector(std::vector<int> const &v) { for (int x : v) std::cout << x << '\n'; }
If the size-constructor of the vector would not be explicit it would be possible to call the function like this:
print_intvector(3);
What would one expect from such a call? One line containing 3
or three lines containing 0
? (Where the second one is what happens.)
As Bjarne Stroustrup puts it (in "The C++ Programming Language", 4th Ed., 35.2.1, pp. 1011) on the question why std::duration
cannot be implicitly constructed from a plain number:
If you know what you mean, be explicit about it.
Answered 2023-09-20 20:15:42
This answer is about object creation with/without an explicit constructor since it is not covered in the other answers.
Consider the following class without an explicit constructor:
class Foo
{
public:
Foo(int x) : m_x(x)
{
}
private:
int m_x;
};
Objects of class Foo can be created in 2 ways:
Foo bar1(10);
Foo bar2 = 20;
Depending upon the implementation, the second manner of instantiating class Foo may be confusing, or not what the programmer intended. Prefixing the explicit
keyword to the constructor would generate a compiler error at Foo bar2 = 20;
.
It is usually good practice to declare single-argument constructors as explicit
, unless your implementation specifically prohibits it.
Note also that constructors with
can both be used as single-argument constructors. So you may want to make these also explicit
.
An example when you would deliberately not want to make your single-argument constructor explicit is if you're creating a functor (look at the 'add_x' struct declared in this answer). In such a case, creating an object as add_x add30 = 30;
would probably make sense.
Here is a good write-up on explicit constructors.
Answered 2023-09-20 20:15:42
The explicit
keyword makes a conversion constructor to non-conversion constructor. As a result, the code is less error prone.
Answered 2023-09-20 20:15:42
The explicit
-keyword can be used to enforce a constructor to be called explicitly.
class C {
public:
explicit C() =default;
};
int main() {
C c;
return 0;
}
the explicit
-keyword in front of the constructor C()
tells the compiler that only explicit call to this constructor is allowed.
The explicit
-keyword can also be used in user-defined type cast operators:
class C{
public:
explicit inline operator bool() const {
return true;
}
};
int main() {
C c;
bool b = static_cast<bool>(c);
return 0;
}
Here, explicit
-keyword enforces only explicit casts to be valid, so bool b = c;
would be an invalid cast in this case. In situations like these explicit
-keyword can help programmer to avoid implicit, unintended casts. This usage has been standardized in C++11.
Answered 2023-09-20 20:15:42
explicit operator bool()
is also the C++11 version of safe bool, and can be used implicitly in condition checks (and only in condition checks, as far as I'm aware). In your second example, this line would also be valid in main()
: if (c) { std::cout << "'c' is valid." << std:: endl; }
. Apart from this, though, it can't be used without explicit casting. - anyone C c();
and also played around in Compiler Explorer and it appears that making a default constructor explicit
doesn't actually do anything. This answer is simply wrong, even with the code corrections already made. - anyone Cpp Reference is always helpful!!! Details about explicit specifier can be found here. You may need to look at implicit conversions and copy-initialization too.
Quick look
The explicit specifier specifies that a constructor or conversion function (since C++11) doesn't allow implicit conversions or copy-initialization.
Example as follows:
struct A
{
A(int) { } // converting constructor
A(int, int) { } // converting constructor (C++11)
operator bool() const { return true; }
};
struct B
{
explicit B(int) { }
explicit B(int, int) { }
explicit operator bool() const { return true; }
};
int main()
{
A a1 = 1; // OK: copy-initialization selects A::A(int)
A a2(2); // OK: direct-initialization selects A::A(int)
A a3 {4, 5}; // OK: direct-list-initialization selects A::A(int, int)
A a4 = {4, 5}; // OK: copy-list-initialization selects A::A(int, int)
A a5 = (A)1; // OK: explicit cast performs static_cast
if (a1) cout << "true" << endl; // OK: A::operator bool()
bool na1 = a1; // OK: copy-initialization selects A::operator bool()
bool na2 = static_cast<bool>(a1); // OK: static_cast performs direct-initialization
// B b1 = 1; // error: copy-initialization does not consider B::B(int)
B b2(2); // OK: direct-initialization selects B::B(int)
B b3 {4, 5}; // OK: direct-list-initialization selects B::B(int, int)
// B b4 = {4, 5}; // error: copy-list-initialization does not consider B::B(int,int)
B b5 = (B)1; // OK: explicit cast performs static_cast
if (b5) cout << "true" << endl; // OK: B::operator bool()
// bool nb1 = b2; // error: copy-initialization does not consider B::operator bool()
bool nb2 = static_cast<bool>(b2); // OK: static_cast performs direct-initialization
}
Answered 2023-09-20 20:15:42
explicit operator bool()
vs. if
is a special case. There is no way to reproduce it with user defined Bool
, explicit operator Bool()
and a function called If
. - anyone It is always a good coding practice to make your one argument constructors (including those with default values for arg2
,arg3
,...) as already stated.
Like always with C++: if you don't - you'll wish you did...
Another good practice for classes is to make copy construction and assignment private (a.k.a. disable it) unless you really need to implement it. This avoids having eventual copies of pointers when using the methods that C++ will create for you by default. An other way to do this is derive from boost::noncopyable
.
Answered 2023-09-20 20:15:42
= delete
. - anyone Constructors append implicit conversion. To suppress this implicit conversion it is required to declare a constructor with a parameter explicit.
In C++11 you can also specify an "operator type()" with such keyword http://en.cppreference.com/w/cpp/language/explicit With such specification you can use operator in terms of explicit conversions, and direct initialization of object.
P.S. When using transformations defined BY USER (via constructors and type conversion operator) it is allowed only one level of implicit conversions used. But you can combine this conversions with other language conversions
Answered 2023-09-20 20:15:42
Other answers are missing one important factor which I am going to mention here.
Along with "delete" keyword, "explicit" allows you to control the way compiler is going to generate special member functions - default constructor, copy constructor, copy-assignment operator, destructor, move constructor and move-assignment.
Refer https://learn.microsoft.com/en-us/cpp/cpp/explicitly-defaulted-and-deleted-functions
Answered 2023-09-20 20:15:42
delete
keyword does NOT allow you to control the way that the compiler is going to generate special member functions. - anyone