Consider this simple example:
template <class Type>
class smartref {
public: smartref() : data(new Type) { } operator Type&(){ return *data; }
private: Type* data;
};
class person {
public: void think() { std::cout << "I am thinking"; }
};
int main() { smartref<person> p; p.think(); // why does not the compiler try substituting Type&?
}How do conversion operators work in C++? (i.e) when does the compiler try substituting the type defined after the conversion operator?
07 Answers
Some random situations where conversion functions are used and not used follow.
First, note that conversion functions are never used to convert to the same class type or to a base class type.
Conversion during argument passing
Conversion during argument passing will use the rules for copy initialization. These rules just consider any conversion function, disregarding of whether converting to a reference or not.
struct B { };
struct A { operator B() { return B(); }
};
void f(B);
int main() { f(A()); } // called!Argument passing is just one context of copy initialization. Another is the "pure" form using the copy initialization syntax
B b = A(); // called!Conversion to reference
In the conditional operator, conversion to a reference type is possible, if the type converted to is an lvalue.
struct B { };
struct A { operator B&() { static B b; return b; }
};
int main() { B b; 0 ? b : A(); } // called!Another conversion to reference is when you bind a reference, directly
struct B { };
struct A { operator B&() { static B b; return b; }
};
B &b = A(); // called!Conversion to function pointers
You may have a conversion function to a function pointer or reference, and when a call is made, then it might be used.
typedef void (*fPtr)(int);
void foo(int a);
struct test { operator fPtr() { return foo; }
};
int main() { test t; t(10); // called!
}This thing can actually become quite useful sometimes.
Conversion to non class types
The implicit conversions that happen always and everywhere can use user defined conversions too. You may define a conversion function that returns a boolean value
struct test { operator bool() { return true; }
};
int main() { test t; if(t) { ... }
}(The conversion to bool in this case can be made safer by the safe-bool idiom, to forbid conversions to other integer types.) The conversions are triggered anywhere where a built-in operator expects a certain type. Conversions may get into the way, though.
struct test { void operator[](unsigned int) { } operator char *() { static char c; return &c; }
};
int main() { test t; t[0]; // ambiguous
}
// (t).operator[] (unsigned int) : member
// operator[](T *, std::ptrdiff_t) : built-inThe call can be ambiguous, because for the member, the second parameter needs a conversion, and for the built-in operator, the first needs a user defined conversion. The other two parameters match perfectly respectively. The call can be non-ambiguous in some cases (ptrdiff_t needs be different from int then).
Conversion function template
Templates allow some nice things, but better be very cautious about them. The following makes a type convertible to any pointer type (member pointers aren't seen as "pointer types").
struct test { template<typename T> operator T*() { return 0; }
};
void *pv = test();
bool *pb = test();The "." operator is not overloadable in C++. And whenever you say x.y, no conversion will automatically be be performed on x.
0Conversions aren't magic. Just because A has a conversion to B and B has a foo method doesn't mean that a.foo() will call B::foo().
The compiler tries to use a conversion in four situations
- You explicitly cast a variable to another type
- You pass the variable as an argument to a function that expects a different type in that position (operators count as functions here)
- You assign the variable to a variable of a different type
- You use the variable copy-construct or initialize a variable of a different type
There are three types of conversions, other than those involved with inheritance
- Built-in conversions (e.g. int-to-double)
- Implicit construction, where class B defines a constructor taking a single argument of type A, and does not mark it with the "explicit" keyword
- User-defined conversion operators, where class A defines an operator B (as in your example)
How the compiler decides which type of conversion to use and when (especially when there are multiple choices) is pretty involved, and I'd do a bad job of trying to condense it into an answer on SO. Section 12.3 of the C++ standard discusses implicit construction and user-defined conversion operators.
(There may be some conversion situations or methods that I haven't thought of, so please comment or edit them if you see something missing)
Implicit conversion (whether by conversion operators or non-explicit constructors) occurs when passing parameters to functions (including overloaded and default operators for classes). In addition to this, there are some implicit conversions performed on arithmetic types (so adding a char and a long results in the addition of two longs, with a long result).
Implicit conversion does not apply to the object on which a member function call is made: for the purposes of implicit conversion, "this" is not a function parameter.
The compiler will attempt one(!) user-defined cast (implicit ctor or cast operator) if you try to use an object (reference) of type T where U is required.
The . operator, however, will always try to access a member of the object (reference) on its left side. That's just the way it's defined. If you want something more fancy, that's what operator->() can be overloaded for.
You should do
((person)p).think();The compiler doesn't have the information for automatically casting to person, so you need explicit casting.
If you would use something like
person pers = p;Then the compiler has information for implicit casting to person.
You can have "casting" through constructors:
class A
{
public: A( int );
};
A a = 10; // Looks like a cast from int to AThese are some brief examples. Casting (implicit, explicit, etc) needs more to explain. You can find details in serious C++ books (see the questions about C++ books on stack overflow for good titles, like this one).
1//Virtual table Fuction(VFT)
#include <iostream>
using namespace std;
class smartref {
public:
virtual char think() { }//for Late bindig make virtual function if not make virtual function of char think() {} then become early binding and pointer call this class function smartref() : data(new char) { } operator char(){ return *data; }
private: char* data;
};
class person:public smartref
{
public: char think() { std::cout << "I am thinking"; }
};
int main() { smartref *p;//make pointer of class person o1;//make object of class p=&o1;//store object address in pointer p->think(); // Late Binding in class person
return 0;
}
