What does the explicit keyword in C++ mean?

Question

Someone posted in a comment to another question about the meaning of the explicit keyword in C++. So, what does it mean?

Answer 1

In C++, the compiler is allowed to make one implict conversion to resolve the parameters to a function. What this means is that the compiler can use single parameter constructors 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.

void main ()
{
  DoBar (42);
}

The parameter is not a Foo object. It is an int though and there exists a constructor for Foo that takes an int so that 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)".

Answer 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 char 'x' will be converted to int and will call String(int) constructor. But this is not what the user might have intended. So to prevent such conditions, we can define the class's constructor as explicit.

class String { 
public: 
explicit String (int n); //allocate n bytes
String(const char *p); // initialize sobject with string p 
}

Answer 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.

Answer 4

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.

Answer 5

Another look