Cast char* to my string implicitly

Question

I have my own string class ( DinString ), nothing special. I was wondering is possible to do something like this

DinString a= "Helo World";

bool Func(DinString string);

Func("test");

Answer 1

Yes, it is. Simply implement an appropriate (non- explicit !) constructor:

class DinString {
public:
    DinString( const char *s ) {
        // ...
    }
};

Please note that this won't work if you 'chain' more than one constructor like this. For instance, the following does not work :

class DinString {
public:
    DinString( const char *s ) {
        // ...
    }
};

class FooString {
public:
    FooString( const DinString & ) { }
};

void f( const FooString &) { }

f( "hello" ); // doesn't call FooString(DinString("hello!")); !

Answer 2

You can define a constructor to do this conversion for you:

class DinString {
  DinString(const char *string) { 
    // do something useful here
  }
};

This works because DinString a= "Helo World"; is not an assignment (despite appearances) and is actually equivalent to writing DinString a("Hello World");

Answer 3

是的，为您的类提供一个转换构造函数，将const char *作为输入。

DinString::DinString(const char *);