What does the explicit keyword mean?
C ++中explicit关键字的含义是什么？
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:
MyString(int size) class with a constructor that constructs a string of the given size. You have a function print(const MyString&) , 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. 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
};
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.
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