Question

Does temporary object creation depends on the compiler?

In the code bellow, I call a function sending a char* but the function requires an Object reference. However, Object has a constructor that uses a char*.. So a temporary object is automatically created and send to the print function.

class Object {
    string text;
  public:
    Object ( const char* value ) { text = value; }
    void print() const { printf( "[%s]\n", text.c_str() ); }
};

void print( const Object& obj ) { obj.print(); }

int main() {
    print( "hello" );
}

Does this behavior depends on the compiler?
You can see the output here: http://codepad.org/AABw5Ulz

Answer 1

When you declare such a constructor

 Object ( const char* value )

it means the compiler can do an implicit conversion.

So no it is not a compiler dependent feature but it is by desing.

The compiler is simply allowed to take your "hello" and pass it to the Object contructor before sending it to the print(const Object &obj)function.