问题
What does the explicit keyword mean in C++?
回答1:
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:
class Foo
{
public:
// single parameter constructor, can be used as an implicit conversion
Foo (int foo) : m_foo (foo)
{
}
int GetFoo () { return m_foo; }
private:
int m_foo;
};
Here's a simple function that takes a Foo object:
void DoBar (Foo foo)
{
int i = foo.GetFoo ();
}
and here's where the DoBar function is called.
int main ()
{
DoBar (42);
}
The argument is not a Foo object, but an int. However, there exists a constructor for Foo that takes an int 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 call DoBar (42). It is now necessary to call for conversion explicitly with DoBar (Foo (42))
The reason you might want to do this is to avoid accidental construction that can hide bugs. Contrived example:
- You have a
MyString(int size)class with a constructor that constructs a string of the given size. You have a functionprint(const MyString&), 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.
回答2:
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
};
回答3:
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 ints into Buffers. 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.
回答4:
Explicit conversion constructors (C++ only)
The explicit function specifier controls unwanted implicit type conversions. It can only be used in declarations of constructors within a class declaration. For example, except for the default constructor, the constructors in the following class are conversion constructors.
class A
{
public:
A();
A(int);
A(const char*, int = 0);
};
The following declarations are legal:
A c = 1;
A d = "Venditti";
The first declaration is equivalent to A c = A( 1 );.
If you declare the constructor of the class as explicit, the previous declarations would be illegal.
For example, if you declare the class as:
class A
{
public:
explicit A();
explicit A(int);
explicit A(const char*, int = 0);
};
You can only assign values that match the values of the class type.
For example, the following statements are legal:
A a1;
A a2 = A(1);
A a3(1);
A a4 = A("Venditti");
A* p = new A(1);
A a5 = (A)1;
A a6 = static_cast<A>(1);
回答5:
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
- default arguments for all parameters, or
- default arguments for the second parameter onwards
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.
回答6:
The keyword explicit accompanies either
- a constructor of class X that cannot be used to implicitly convert the first (any only) parameter to type X
C++ [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).
- or a conversion function that is only considered for direct initialization and explicit conversion.
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.
Overview
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
*/
Example using structures 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) { }
Examples regarding constructor:
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
Examples regarding conversion functions:
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
Why use 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.)
Using the explicit keyword in a class interface enforces the user of the interface to be explicit about a desired conversion.
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.
回答7:
The explicit keyword makes a conversion constructor to non-conversion constructor. As a result, the code is less error prone.
回答8:
The explicit-keyword can be used to enforce a constructor to be called explicitly.
class C{
public:
explicit C(void) = default;
};
int main(void){
C c();
return 0;
}
the explicit-keyword in front of the constructor C(void) 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(void) const{
return true;
}
};
int main(void){
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.
回答9:
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
}
回答10:
This has already been discussed (what is explicit constructor). But I must say, that it lacks the detailed descriptions found here.
Besides, 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.
回答11:
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
- up integral ranks (char to int, float to double);
- standart conversions (int to double);
- convert pointers of objects to base class and to void*;
来源：https://stackoverflow.com/questions/121162/what-does-the-explicit-keyword-mean