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;
Related
The following operator overload is defined inside my Term class:
public static Term operator *(int c, Term t) {...}
This class also defines an implicit conversion from a Variable to Term:
public static implicit operator Term(Variable var) {...}
I would like to understand why the following does not compile:
static void Main(string[] args)
{
Variable var = ...; // the details don't matter
Console.WriteLine(2 * var); // var isn't implicitly converted to Term...
Console.ReadKey();
}
The compiler says:
Operator '*' cannot be applied to operands of type 'int' and 'OOSnake.Variable'
Why isn't my overload of operator * found?
EDIT: As per the suggestion in the comments, here is a small complete example that re-produces the error:
namespace Temp
{
class A {
}
class B
{
public static implicit operator B(A a) { return new B(); }
public static B operator *(int c, B b) { return new B(); }
}
class Program
{
static void Main(string[] args)
{
Console.WriteLine(2 * new A());
}
}
}
Basically, operator overload resolution doesn't include implicit user-defined conversions in order to find the operators that could be applicable.
From section 7.3.4 of the C# 5 specification:
An operation of the form x op y, where op is an overloadable binary operator, x is an expression of type X, and y is an expression of type Y, is processed as follows:
The set of candidate user-defined operators provided by X and Y for the operation operator op(x, y) is determined. The set consists of the union of the candidate operators provided by X and the candidate operators provided by Y, each determined using the rules of ยง7.3.5. If X and Y are the same type, or if X and Y are derived from a common base type, then shared candidate operators only occur in the combined set once.
And 7.3.5 doesn't include implicit user-defined conversions in its search for a set of operators.
Note that this also wouldn't work if the implicit conversion to Term was declared in the Variable class - although that would be more reasonable to specify and implement, as the compiler could look at the set of conversions from the operand type to other types, and use those for overload resolution.
However, this is only a matter of looking for the operators to start with. The compiler is happy to perform implicit conversions when it considers whether or not an overload is applicable. For example, in your case, if you add:
class A
{
public static B operator *(A a, B b) { return new B(); }
}
Then this is valid:
A a = new A();
Console.WriteLine(a * a);
Console.WriteLine(2 * var) doesn't contain any hint that you would want to convert var to type Term. The compiler sees the int and multiplication operator and a variable of type 'Variable'.
Edit: To clarify, in order for your example to work the compiler would have to go through all the types in scope and see if one just happens to have an implicit conversion from type 'A'.
And if there also happens do be a class C like:
class C
{
public static implicit operator C(A a) { return new A(); }
public static B operator *(int i, C c) { return new C(); }
}
there's no telling what would happen.
Which is why the compiler doesn't do that :)
If you want to make your example to work, you must move the implicit operator that converts A to B to the class A, like this:
namespace Temp
{
class A
{
public static implicit operator B(A a) { return new B(); }
}
class B
{
public static B operator *(int c, B b) { return new B(); }
}
class Program
{
static void Main(string[] args)
{
Console.WriteLine(2 * ((B)new A()));
}
}
}
But you cant use this 2 * new A() because(Jon Skeet answer):
Basically, operator overload resolution doesn't include implicit user-defined conversions in order to find the operators that could be applicable.
https://stackoverflow.com/a/34161798/815590
If I were to, say, create a thin wrapper around the float type for whatever reason, like so:
public class WrappedFloat
{
private float value;
public WrappedFloat(float value)
{
this.value = value;
}
public static implicit operator float(WrappedFloat wrapped)
{
return wrapped.value;
}
public static implicit operator WrappedFloat(float value)
{
return new WrappedFloat(value);
}
}
This code is apparently perfectly valid:
new WrappedFloat(4.0F) + new WrappedFloat(3.0F)
Since WrappedFloat doesn't define any arithmetic operators, there must be something about it's implicit conversion to float allowing this. But operator overloading is just syntactic sugar for methods, right? It's not like I can call methods (i.e. CompareTo) on WrappedFloat just because float has them. So what's so special about operators here? What are the rules for allowing this?
Operators are, for the purposes of this discussion, just like static methods. Imagine you had an actual static method for each operator overload:
public static int Plus(int a, int b) { return a + b; }
public static float Plus(float a, float b) { return a + b; }
public static long Plus(long a, long b) { return a + b; }
public static string Plus(string a, string b) { return a + b; }
public static double Plus(double a, double b) { return a + b; }
//...
Now imagine you have:
Plus(new WrappedFloat(4.0F), new WrappedFloat(3.0F));
What would you expect to happen here? Overload resolution would run, it would see that there is an overload for which both of the parameters have an implicit conversion to, and that overload (accepting float values) becomes the unique best match.
The exact same thing happens with operators. It pulls together all of the overloads, sees which ones have argument lists for which the provided arguments can be implicitly converted to, and then chooses the best match among them if there are multiple.
But operator overloading is just syntactic sugar for methods, right?
It is not syntactic sugar for "methods", it is syntactic sugar for one kind of method implemented in a specific way. The typical method call on an object with . syntax is not applicable here. MS chose a design for programmer convenience. When I call
wrappedFloat.Method()
I dont expect it to look for every possibility on wrappedFloat, but when I do
wrappedFloat1 + wrappedFloat2
I expect it to. I agree with MS here. One difference here, is that in first case it is a call on the instance, the second one is a static method call where the formal parameter is passed as an argument.
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 have (for example) an object of type A that I want to be able to cast to type B (similar to how you can cast an int to a float)
Data types A and B are my own.
Is it possible to define the rules by which this casting occurs?
Example
int a = 1;
float b = (float)a;
int c = (int)b;
Yes, this is possible using C# operator overloading. There are two versions explicit and implicit.
Here is a full example:
class Program
{
static void Main(string[] args)
{
A a1 = new A(1);
B b1 = a1;
B b2 = new B(1.1);
A a2 = (A)b2;
}
}
class A
{
public int Foo;
public A(int foo)
{
this.Foo = foo;
}
public static implicit operator B(A a)
{
return new B(a.Foo);
}
}
class B
{
public double Bar;
public B(double bar)
{
this.Bar = bar;
}
public static explicit operator A(B b)
{
return new A((int)b.Bar);
}
}
Type A can be cast implicitly to type B but type B must be cast explicitly to type A.
Assuming you want that to be an explcit operation you'll need to write an explicit cast operator like so:
public static explicit operator MyTypeOne(MyTypeTwo i)
{
// code to convert from MyTypeTwo to MyTypeOne
}
You can then use it like so:
MyTypeOne a = new MyTypeOne();
MyTypeTwo b = (MyTypeTwo)a;
I'd question whether you want to actually cast one type to another, or whether you actually want to convert instead. I'd say you should avoid writing cast operators for conversions, if you are just aiming to take advantage of a nice syntax :)
Also, in general it is advised not to use implicit casts, as they allow for unintended type converstions. From MSDN documentation on implicit:
However, because implicit conversions
can occur without the programmer's
specifying them, care must be taken to
prevent unpleasant surprises. In
general, implicit conversion operators
should never throw exceptions and
never lose information so that they
can be used safely without the
programmer's awareness.
You cant overload the cast operator in c# but you can use explicit and implicit conversion operators instead:
"Using Conversion Operators (C# Programming Guide)"
This question already has answers here:
Closed 13 years ago.
Possible Duplicate:
C# okay with comparing value types to null
I was working on a windows app in a multithreaded environment and would sometimes get the exception "Invoke or BeginInvoke cannot be called on a control until the window handle has been created." So I figured that I'd just add this line of code:
if(this.Handle != null)
{
//BeginInvokeCode
}
But that didn't solve the problem. So I dug a little further, and realized that IntPtr (the type that Form.Handle is) is a struct which can't be nullable. This was the fix that worked:
if(this.Handle != IntPtr.Zero)
{
//BeginInvokeCode
}
So then it hit me, why did it even compile when I was checking it for null? So I decided to try it myself:
public struct Foo { }
and then:
static void Main(string[] args)
{
Foo f = new Foo();
if (f == null) { }
}
and sure enough it didn't compile saying that "Error 1 Operator '==' cannot be applied to operands of type 'ConsoleApplication1.Foo' and ''". Ok, so then I started looking at the metadata for IntPtr and started adding everything to my Foo struct that was there in the IntPtr struct (ISerializable, ComVisible) but nothing helped. Finally, when I added the operator overloading of == and !=, it worked:
[Serializable]
[ComVisible(true)]
public struct Foo : ISerializable
{
#region ISerializable Members
public void GetObjectData(SerializationInfo info, StreamingContext context)
{
throw new NotImplementedException();
}
#endregion
public override bool Equals(object obj)
{
return base.Equals(obj);
}
public override int GetHashCode()
{
return base.GetHashCode();
}
public static bool operator ==(Foo f1, Foo f2) { return false; }
public static bool operator !=(Foo f1, Foo f2) { return false; }
}
This finally compiled:
static void Main(string[] args)
{
Foo f = new Foo();
if (f == null) { }
}
My question is why? Why if you override == and != are you allowed to compare to null? The parameters to == and != are still of type Foo which aren't nullable, so why's this allowed all of a sudden?
It looks like the issue is that when MS introduced nullable types, they made it so that every struct is implicitly convertable to its nullable type (foo?), so the code
if( f == null)
is equivalent to
if ( (Nullable<foo>)f == (Nullable<foo>)null)
Since MSDN states that "any user-defined operators that exist for value types may also be used by nullable types", when you override operator==, you allow that implicit cast to compile, as you now have a user-defined == -- giving you the nullable overload for free.
An aside:
Seems like in your example, there is some compiler optimization
The only thing that is emitted by the compiler that even hints there was a test is this IL:
ldc.i4.0
ldc.i4.0
ceq
stloc.1 //where there is an unused boolean local
Note that if you change main to
Foo f = new Foo();
object b = null;
if (f == b) { Console.WriteLine("?"); }
It no longer compiles. But if you box the struct:
Foo f = new Foo();
object b = null;
if ((object)f == b) { Console.WriteLine("?"); }
if compiles, emits IL, and runs as expected (the struct is never null);
This has nothing to do with serialization or COM - so it's worth removing that from the equation. For instance, here's a short but complete program which demonstrates the problem:
using System;
public struct Foo
{
// These change the calling code's correctness
public static bool operator ==(Foo f1, Foo f2) { return false; }
public static bool operator !=(Foo f1, Foo f2) { return false; }
// These aren't relevant, but the compiler will issue an
// unrelated warning if they're missing
public override bool Equals(object x) { return false; }
public override int GetHashCode() { return 0; }
}
public class Test
{
static void Main()
{
Foo f = new Foo();
Console.WriteLine(f == null);
}
}
I believe this compiles because there's an implicit conversion from the null literal to Nullable<Foo> and you can do this legally:
Foo f = new Foo();
Foo? g = null;
Console.WriteLine(f == g);
It's interesting that this only happens when == is overloaded - Marc Gravell has spotted this before. I don't know whether it's actually a compiler bug, or just something very subtle in the way that conversions, overloads etc are resolved.
In some cases (e.g. int, decimal) the compiler will warn you about the implicit conversion - but in others (e.g. Guid) it doesn't.
All I can think is that your overloading of the == operator gives the compiler a choice between:
public static bool operator ==(object o1, object o2)
and
public static bool operator ==(Foo f1, Foo f2)
and that with both to choose from it is able to cast the left to object and use the former. Certainly if you try to run something based on your code, it doesn't step into your operator overload. With no choice between operators, the compiler is clearly carrying out some further checking.
struct doesn't define the overloads "==" or "!=" which is why you got the original error. Once the overloads were added to your struct the comparision was legal (from a compiler prospective). As the creator of the operator overload is it your responsibility to handle this logic (obviously Microsoft missed this in this case).
Depending on your implementation of your struct (and what it represents) a comparison to null may be perfectly valid which is why this is possible.
I believe when you overload an operator you are explicitly subscribing to the notion that you will handle all of the logic necessary with the specific operator. Hence it is your responsibility to handle null in the operator overload method, if it ever gets hit. In this case as I am sure you've probably noticed the overloaded methods never get hit if you compare to null.
Whats really interesting is that following Henks answer here, i checked out the following code in reflector.
Foo f1 = new Foo();
if(f1 == null)
{
Console.WriteLine("impossible");
}
Console.ReadKey();
This is what reflector showed.
Foo f1 = new Foo();
Console.ReadKey();
Compiler cleans it up and hence the overloaded operator methods never even get called.
I recomend you to take a look to those pages:
http://www.albahari.com/valuevsreftypes.aspx
http://msdn.microsoft.com/en-us/library/s1ax56ch.aspx
http://msdn.microsoft.com/en-us/library/490f96s2.aspx