I have a custom class named Optional< T >. This class overrides the implicit operator, something like this:
public static implicit operator Optional<T>(T value)
{
return new Optional<T>(value);
}
public static implicit operator T(Optional<T> value)
{
return value.value;
}
Ok, till now all fine, the direct operations like these work:
Optional<int> val = 33; //ok
int iVal = val; //ok
But, this doesn't work:
Optional<Optional<int>> val = 33; //error, cannot convert implicitly
So I'm wondering how can be supported the previous case.
Cheers.
You cannot chain user-defined implicit operators. If you define an implicit conversion from A to B, and from B to C, there is no implicit conversion from A to C that calls both. You'll either need to create a separate user-defined conversion from A to C, or have an explicit conversion to do half of the conversion for you.
I have a struct type in C#. I want to be able to convert null implicitly to this type. For instance, null could be represented by a special value of the struct type and the cast operator should return a struct with this value.
In C++, I could use an implicit cast operator overload of type std::nullptr_t. Is there a comparable type in C#?
I have had the idea to use a special NullType class which has no instances. This works but looks somehow ugly. Is there a better way?
Example:
class NullType
{
private NullType(){} // ensures that no instance will ever be created
}
struct X
{
private static readonly int nullValue = -1;
private int val;
public X(int val){ this.val= val; }
public static implicit operator X(NullType t)
{
return new X(nullValue);
}
}
class MainClass
{
public static void Main(string[] args)
{
X x = null; // Works now!
}
}
No, conversion operators only allow conversion from types, not from a specific value of a type, aka null, so there is no null conversion operator.
The best fitting operator without introducing own types would be this:
public static implicit operator X(object t)
But obviously you don't want to use this. It isn't very safe to use (t can be any value, and the only way to handle non-null cases in an exception).
That said, I think the way you've created it now, using a class that can't be initialized is the best way to do this. In fact the only question is: why do you want to use null, instead of 'just' an default value instance on your struct (X.Null).
I have a generic struct and two synonyms with defined generic type parameter. I want to define type conversions between these two types like this:
using X = A<int>;
using Y = A<long>;
struct A<T>
{
private T data;
public A(T data)
{
this.data = data;
}
public static explicit operator X(Y value)
{
return new X((int)value.data);
}
public static implicit operator Y(X value)
{
return new Y(value.data);
}
}
and here is an example how I want to use the struct:
class Program
{
static void Main(string[] args)
{
X x = new X(1);
Y y = new Y(2);
x = (X)y; // explicit cast with precision loss
y = x; // implicit cast without precision loss
}
}
Unfortunately each "using" creates new specific definition of the struct and C# compiler treats synonym structures not as a subset of a generic structure, but as an individual structure. That's why compiler reports errors:
"ambigious between X.static implicit operator Y(X) and Y.static implicit operator Y(X)".
"User-defined conversion must convert to or from the enclosing type".
Does anybody knows a way to realize this type conversions without changing type to class?
UPDATE: I'm using .NET 4.
The error
User-defined conversion must convert to or from the enclosing type.
means that your conversion operators need to convert to/from A<T>. Yours are converting to/from A<int/string>. That's not the same thing (much less general at least).
So this cannot work. You have to find some other way to do the conversions. Maybe runtime casting can help here (define the operators as acting on A<T> and do casting inside of them).
I think this problem is unrelated to the type synonyms. In fact they made the question harder to understand.
I've distilled this question down to the simplest code sample I can think of. Why is the explicit operator not invoked from the generic method?
class Program
{
static void Main(string[] args)
{
A a = new A();
B b = (B)a; // works as expected
b = Cast<B>(a);
}
static TResult Cast<TResult>(object o)
{
return (TResult)o; // throws Invalid Cast Exception
}
}
class A
{
}
class B
{
public static explicit operator B(A a)
{
return new B();
}
}
Because a generic method has one set of IL, based around TResult. It doesn't do different IL per-caller. And the generic TResult does not have any operators.
Also: the operator here would need to be between object and TResult, and you can't define operators involving object.
For example: Cast<int> and Cast<string> and Cast<Guid> all have the exact same IL.
You can cheat with dynamic:
return (TResult)(dynamic)o;
What it comes down to is that the implicit and explicit operators aren't true conversion operators; they're entirely compile time syntactic sugar. Once the code is compiled nothing about the conversion operators remains.
When the compiler sees:
B b = (B) new A();
It says, "is there any native conversion (implicit or explicit) from an A to a B?" (This would be the case if A extended B, for example), or for one of the few special cased language conversions such as double to int (implicit) or int to double (explicit).)
If not, it then looks for user defined conversion operators (it only looks in the definition of A and B, it doesn't look in the definition of C for an implicit conversion from A and to B, for example). If it finds one, then it injects that operator as a static method call, so the code ends up looking like:
B b = B.SomeAutogneratedName(new A());
That way by the time you get to runtime it's just executing another method, something the runtime knows how to do. The only actual runtime conversion operators that are allowed are the handful baked into the language (i.e. from any base type to a parent type, and between certain primitive types).
I wonder if there is a possibility in c++ to achieve the same cast overloading like in this C# example:
class A {
public static implicit operator A(string s) {
return new A();
}
public static implicit operator A(double d) {
return new A();
}
static void Main(string[] args) {
A a = "hello";
A b = 5.0;
}
}
In C++ it should be something like this:
#include <string>
using namespace std;
class A{
/*SOME CAST OVERLOADING GOES HERE*/
};
void main(){
A a = "hello";
A b = 5.0;
}
Can you help me how I can make this cast overloading?
This is typically achieved with constructors:
class A
{
public:
A(const std::string & s) { /*...*/ }
A(double d) { /*...*/ }
//...
};
Usage: A a("hello"), b(4.2), c = 3.5, d = std::string("world");
(If you declare the constructor explicit, then only the first form ("direct initialization", with the parentheses) is allowed. Otherwise the two forms are entirely identical.)
A one-parameter constructor is also called a "conversion constructor" for this very reason: You can use it to construct an object from some other object, i.e. "convert" a double, say, to an A.
Implicit conversion is widely used. For example, it'll apply in the following situation:
void f(A, const A&) { /* ... */ }
int main() { f(1.5, std::string("hello")); }
This constructs temporaries A(1.5) and A("hello") and passes them to f. (Temporaries also bind to constant references, as you can see with the second argument.)
Update: (Credits to #cHao for looking this up.) According to the standard (12.3.4), "At most one user-defined conversion (constructor or conversion function) is implicitly applied to a single value." (This refers to implicit conversion; direct-initialization a la A a("hello"); doesn't fall under this*. See this related question.) So unfortunately you cannot say f("hello", "world");. (Of course you can add a const char * constructor, which in the new C++11 you can even forward to the string constructor with almost no effort.)
*) I think this is actually a bit more subtle. Only user-defined conversions are affected by the rule. So first off, you get a fundamental conversion from char (&)[6] to const char * for free. Then you get one implicit conversion from const char * to std::string. Finally, you have an explicit conversion-construction from std::string to A.
A conversion operator will provide an implicit conversion from your type to some other type. For example:
class A
{
public:
operator std::string () const { return "foo"; }
};
However, implicit conversions can be dangerous because it can be employed when you don't expect it to be employed and would prefer a compiler error. Therefore, it's often better to implement a convert constructor on the destination object, rather than a conversion operator on the source object.
class A
{
public:
A(const std::string& rhs) { /* ... */ }
};
There is still some implicit conversion possible with this, however. Consider:
string f = "foo";
A foo = f;
f is implicitly converted to an A because the convert constructor is available. It might be best for this to result in a compiler error, however. By marking the convert constructor explicit, you make it so that you can easily convert from one type to another, but only when you really intended to.
#include <string>
using namespace std;
class A
{
public:
A() {};
explicit A(const std::string& ) {};
};
int main()
{
string f = "foof";
A a1 = f; // WONT COMPILE
A a(f); // WILL COMPILE
}
There's no need for casting ... this can easily be done by creating constructors for you class that take the correct argument-type, and only use a single argument so they can be called implicitly by the compiler.
If your constructors take more than one non-default argument, then they can't be used in conversion operations.
Furthermore, if you ever want to avoid ambiguity in the selection of constructors based on inferred type conversion, you can always use the explicit keyword with your constructors. For instance, imagine a scenario like:
struct test
{
int a;
test (signed int b): a(b) {}
test (unsigned int b): a(b) {}
};
int main()
{
test A = 5.0;
return 0;
}
This will result in a compiler error due to ambiguity in the conversion of the double type to a scaler-type ... the double type could be implicitly converted to both types. This can be fixed through the use of the explicit keyword to-do the following:
struct test
{
int a;
test (signed int b): a(b) {}
explicit test (unsigned int b): a(b) {}
};
Now the unsigned int version of the constructor for test will only be called if it's passed an unsigned int. A conversion operation from a double type will use the default int version, removing the ambiguity problem.
What it means in C# ?
In C#, the semantics is not about overloading the operator = (which you can't), but provide a cast operator from any type to the type A, or from the type A to any type.
This means the following code (completed with the A-to-type conversions):
class A {
public static implicit operator A(string s) {
return new A();
}
public static implicit operator A(double d) {
return new A();
}
public static implicit operator string(A a) {
return string.Empty;
}
public static implicit operator double(A a) {
return 0.0;
}
static void Main(string[] args) {
A a = "hello"; // line A
A b = 5.0; // line B
a = "World"; // line C
a = 3.14; // line D
double d = a ; // line E
string s = a ; // line F
}
}
works for assignment, either in simple assignments (like lines C and D) or assignment in declarations (like lines A and B). The line E and F demonstrate how we can convert from the user type to other types.
How to do it in C++ ?
In C++, the lines A and B are object constructions, which will call the relevant constructors.
The lines C and D are assignment, which will call the relevant assignment operator.
So the C++ code for the C# code you provided must provide both constructors and assignments for string and doubles, as follows:
class A
{
public:
A(const std::string & s) { /*...*/ }
A(double d) { /*...*/ }
A & operator = (const std::string & s) { /*...*/ ; return *this ; }
A & operator = (double d) { /*...*/ ; return *this ; }
// etc.
} ;
This way, you can have
void main()
{
A a0 = "Hello" ; // constructor
A a1("Hello") ; // constructor
A a2 = 5.0 ; // constructor
A a3(5.0) ; // constructor
a0 = "Hello World" ; // assignment
a0 = 3.14 ; // assignment
}
What about cast operators in C++ ?
The cast operators in C++ work like the following C# converter operators:
class A
{
static public operator string(A a) { /*... ; return a string */ }
static public operator double(A a) { /*... ; return a double */ }
// etc.
}
In C++, the cast operators are written as such:
class A
{
public:
operator std::string() { /*... ; return a string */ }
operator double() { /*... ; return a double */ }
// etc.
} ;
void main()
{
A a ;
std::string s ;
double d ;
// etc.
s = a ; // cast operator
d = a ; // cast operator
}
Why is it so complicated in C++ ?
In C#, a cast/conversion operator can be written from the class type to any type, or from any type to the class type.
In C++, for conversion, you must choose between one or more ways, that is: constructor, assignment operator or cast operator, because in C++, you have a fine grained control on types and operation, and thus, you must use that fine grained API.
Constructions means the object does not exist, so there is no point in constructing it to some default value and then assigning it another value (this can cost in speed). In C#, the reference/value type is null/zeroed before the assignment, so this is not an issue.
Assignment means the object does already exist, so it is possible one can reuse the same internals to accommodate the new value. In C#, the original reference object is discarded, and another is created (if the creation costs, that you'll pay that price)
As for the cast, it is used in the current case to convert an existing class into another for which you have no control: You have no right to extend the std::string to accommodate a constructor for your class, and there is no way to add a constructor to a built-in type like double.
You can provide a constructor with one argument of your desired type:
class Foo {
int bar;
public:
Foo(const int bar) : bar(bar) {};
}
Foo foo = 5;