What is the Method Signature in the following
int DoSomething(int a, int b);
Return type is a part of signature or not???
Return type is not part of the method signature in C#. Only the method name and its parameters types (but not the parameter names) are part of the signature. You cannot, for example, have these two methods:
int DoSomething(int a, int b);
string DoSomething(int a, int b);
To be clear: Methods cannot be overloaded based on their return type. They must have a unique name, unique parameter types, or pass their arguments differently (e.g. using out or ref).
Edit: To answer your original question, the method signature for your method is:
DoSomething(int, int)
Note that this all applies to normal methods. If you're talking about delegates, then you should see keyboardP's answer. (Short version: return type IS part of the signature for a delegate).
Is the return type is a part of signature or not?
It depends on why you are asking the question. Why do you care?
There are two definitions of method signature. The C# language definition does not include the return type, and uses the signature of the method to determine whether two overloads are allowed. Two methods with the same signature are not allowed in a type. Since C# does not consider the return type to be a part of the signature, C# does not allow two methods that differ only in return type to be declared in the same type.
The CLR, however, does include the return type in the signature. The CLR allows for two methods to be in the same type that differ only in return type.
To be more specific: in C# the signature consists of the methods:
name
number of type parameters
number of formal parameters
type of each formal parameter
out/ref/value-ness of each formal parameter
with the following additional notes:
generic type parameter constraints are not part of the signature
return type is not part of the signature
type parameter and formal parameter names are not part of the signature
two methods may not differ only in out/ref
In the CLR the signature consists of:
name
number of type parameters
number of formal parameters
type of each formal parameter including modopts and modreqs
return type including modopts and modreqs
ref/value-ness of each formal parameter
Note that the CLR does not distinguish between "ref int" and "out int" at all when considering signatures. Note that the CLR does distinguish between modopt/modreq types. (The way that the C# compiler deals with modopt/modreq types is too complex to summarize here.)
From MSDN:
A return type of a method is not part of the signature of the method
for the purposes of method overloading. However, it is part of the
signature of the method when determining the compatibility between a
delegate and the method that it points to.
To clarify, in your example the return type is not part of the signature. However, when you're matching the signature of a delegate, it is considered part of the signature. From MSDN:
Any method that matches the delegate's signature, which consists of
the return type and parameters, can be assigned to the delegate. This
makes is possible to programmatically change method calls, and also
plug new code into existing classes. As long as you know the
delegate's signature, you can assign your own delegated method.
So I believe it's based on context. Most of the time, and as shown in your code, the return type is not part of it. However, in the context of delegation, it is considered part of it.
From MSDN:
The signature of a method consists of the name of the method and the type and kind (value, reference, or output) of each of its formal parameters, considered in the order left to right. The signature of a method specifically does not include the return type
Edit: That is from old documentation. It seems the definition of 'signature' has changed since then. Now a method has two different signatures, one for the purpose of overloading and one for the purposes of determining delegate compatibility. See keyboardP's answer below for more details.
DoSomething(int a, int b);
is the method signature,
int is the return type.
take a look at this :Signatures and overloading
The signature does not contain the return type. Neither the parameter names. In your case it would be DoSomething(int, int)
In the case when method have generic arguments understanding of signature becomes more confusing.
Generic types declaring on class level are considered as normal types.
But generic types declaring on method level are considered as index in method's generic arguments.
For example these all methods have different signatures.
class MyClass<TValue>
{
public void F(TValue v) { } // generics: 0, arg: TValue
public void F<X>(TValue v) { } // generics: 1, arg: TValue
public void F<X, Y>(TValue v) { } // generics: 2, arg: TValue
public void F<X>(X v) { } // generics: 1, arg: 0
public void F<X, Y>(X v) { } // generics: 2, arg: 0
public void F<X, Y>(Y v) { } // generics: 2, arg: 1
}
Related
On MSDN (https://learn.microsoft.com/en-us/dotnet/standard/generics/), it says:
Generic methods can appear on generic or nongeneric types. It is important to note that a method is not generic just because it belongs to a generic type, or even because it has formal parameters whose types are the generic parameters of the enclosing type. A method is generic only if it has its own list of type parameters. In the following code, only method G is generic.
Isn't M for all intents and purposes generic? If you call method M on some instance of type T, M can only have input and output parameters of type T.
Would there be any issues with using such a nongeneric method M?
class A
{
T G<T>(T arg)
{
T temp = arg;
//...
return temp;
}
}
class Generic<T>
{
T M(T arg)
{
T temp = arg;
//...
return temp;
}
}
Isn't M for all intents and purposes generic?
If you use reflection to inspect those methods, you'll find that IsGenericMethod and IsGenericMethodDefinition are true for A.G, and false for Generic<>.M. This is true even though Generic<>.M's return type is T, whose IsGenericParameter property is true.
You're asking a question about semantics (i.e. the linguistic meaning of "generic method"), and Microsoft has defined the meaning of these concepts as they pertain to C#. So when Microsoft says that a generic method is one that takes generic type arguments, that's the definition that's going to be accepted, both by developers and by the framework itself.
If you call method M on some instance of type T, M can only have input and output parameters of type T.
Yes, the difference is in the scope of the type T. For example, if you have an instance of Generic<string>, you cannot pass its M an int. With the generic method, the generic arguments can be changed at the method level, whereas a non-generic method on a generic class uses types that match the generic types of the class that it belongs to.
Would there be any issues with using such a nongeneric method M?
There are no "issues" with using a non-generic method like M inside of a generic class. That's how generic classes were meant to work. Just be aware of the differences in behavior I mentioned above.
Given this code:
class C
{
C()
{
Test<string>(A); // fine
Test((string a) => {}); // fine
Test((Action<string>)A); // fine
Test(A); // type arguments cannot be inferred from usage!
}
static void Test<T>(Action<T> a) { }
void A(string _) { }
}
The compiler complains that Test(A) can't figure out T to be string.
This seems like a pretty easy case to me, and I swear I've relied far more complicated inference in other generic utility and extension functions I've written. What am I missing here?
Update 1: This is in the C# 4.0 compiler. I discovered the issue in VS2010 and the above sample is from a simplest-case repro I made in LINQPad 4.
Update 2: Added some more examples to the list of what works.
Test(A);
This fails because the only applicable method (Test<T>(Action<T>)) requires type inference, and the type inference algorithm requires that each each argument be of some type or be an anonymous function. (This fact is inferred from the specification of the type inference algorithm (§7.5.2)) The method group A is not of any type (even though it is convertable to an appropriate delegate type), and it is not an anonymous function.
Test<string>(A);
This succeeds, the difference being that type inference is not necessary to bind Test, and method group A is convertable to the required delegate parameter type void Action<string>(string).
Test((string a) => {});
This succeeds, the difference being that the type inference algorithm makes provision for anonymous functions in the first phase (§7.5.2.1). The parameter and return types of the anonymous function are known, so an explicit parameter type inference can be made, and a correspondense is thereby made between the types in the anonymous function (void ?(string)) and the type parameter in the delegate type of the Test method’s parameter (void Action<T>(T)). No algorithm is specified for method groups that would correspond to this algorithm for anonymous functions.
Test((Action<string>)A);
This succeeds, the difference being that the untyped method group parameter A is cast to a type, thereby allowing the type inference of Test to proceed normally with an expression of a particular type as the only argument to the method.
I can think of no reason in theory why overload resolution could not be attempted on the method group A. Then—if a single best binding is found—the method group could be given the same treatment as an anonymous function. This is especially true in cases like this where the method group contains exactly one candidate and it has no type parameters. But the reason it does not work in C#4 appears to be the fact that this feature was not designed and implemented. Given the complexity of this feature, the narowness of its application, and the existance of three easy work-arounds, I am not going to be holding my breath for it!
I think it's because it's a two-step inference:
It has to infer that you want to convert A to a generic delegate
It has to infer what the type of the delegate parameter should be
I'm not sure if this is the reason, but my hunch is that a two-step inference isn't necessarily easy for the compiler.
Edit:
Just a hunch, but something is telling me the first step is the problem. The compiler has to figure out to convert to a delegate with a different number of generic parameters, and so it can't infer the types of the parameters.
This looks like a vicious circle to me.
Test method expects a parameter of delegate type constructed from generic type Action<T>. You pass in a method group instead: Test(A). This means compiler has to convert your parameter to a delegate type (method group conversion).
But which delegate type? To know the delegate type we need to know T. We didn't specify it explicitly, so compiler has to infer it to figure out the delegate type.
To infer the type parameters of the method we need to know the types of the method arguments, in this case the delegate type. Compiler doesn't know the argument type and thus fails.
In all other cases either type of argument is apparent:
// delegate is created out of anonymous method,
// no method group conversion needed - compiler knows it's Action<string>
Test((string a) => {});
// type of argument is set explicitly
Test((Action<string>)A);
or type parameter is specified explicitly:
Test<string>(A); // compiler knows what type of delegate to convert A to
P.S. more on type inference
You're passing the name of the Method A. The .Net framework CAN convert it to an Action, but it's implicit and it will not take responsibility for it.
But still, a method name is NOT an explicit Action<> Object. And therefor it won't infer the type as an Action type.
I could be wrong, but I imagine the real reason C# cannot infer the type is due to method overloading and the ambiguity that arises. For example, suppose I have the following methods: void foo (int) and void foo (float). Now if I write var f = foo. Which foo should the compiler pick? Likewise, the same problem happens with your example using Test(foo).
Why I cant overload Method with dynamic and object Parameters.
void Method(dynamic arg) // Member with same signature is already declared
{
Console.WriteLine("Dynamic");
}
void Method(object arg) // Member with same signature is already declared
{
Console.WriteLine("Not Dynamic");
}
But I can overload Method with dynamic and another type except object or dynamic it self.
void Method(dynamic arg)
{
Console.WriteLine("Dynamic");
}
void Method(string arg)
{
Console.WriteLine("Not Dynamic");
}
As you can see in the first example compiler can't discriminate two overloads.
I think the second example should not work too because string can be send to both overloads. However Compiler Prefers to call the overload with known type.
I've done some test to see how actually it works.
Method(""); // Prints: Not Dynamic
Method((dynamic)""); // Prints: Not Dynamic
Method(null); // Prints: Not Dynamic
Method(???); // Dynamic for any other type except string and null.
Consider first example works and compiler would decide like this:
Method(new object());// Prints: Not Dynamic
Method((object)5); // Prints: Not Dynamic
Method(???); // Prints: Dynamic for any other type except object and null.
So where is the confusion to the compiler? Why I cant have such overloads?
From the C# 5.0 specification, 4.7 The dynamic type:
The type dynamic has special meaning in C#. Its purpose is to allow dynamic binding, which is described in detail in §7.2.2.
dynamic is considered identical to object except in the following respects:
• Operations on expressions of type dynamic can be dynamically bound (§7.2.2).
• Type inference (§7.5.2) will prefer dynamic over object if both are candidates.
Because of this equivalence, the following holds:
• There is an implicit identity conversion between object and dynamic, and between constructed types that are the same when replacing dynamic with object
• Implicit and explicit conversions to and from object also apply to and from dynamic.
• Method signatures that are the same when replacing dynamic with object are considered the same signature
In fact, once your code is compiled, a method having dynamic as a parameter is effectively the same as if that same method were declared using object. It's simply that inside the method, the dynamic parameter can be used dynamically (i.e. with run-time binding).
Because of this, your first example pair of methods is essentially the same as if you'd declared both methods with the parameter type of object.
In your second example, just as you can have overloads where one has a parameter of object and the other a parameter of string, so too can you pair a method using dynamic with a method of the same name using string.
Per the overload resolution rules for C# (also in the specification), as long as the method parameter lists are different in ways that allow one method overload to be chosen as the "best" overload, no compile-time error will occur. In your specific example, passing anything other than a string to the method automatically excludes the method with string as a parameter. In the case where you do pass string, the overload resolution rules select the string overload as the "best" overload, so no compile error occurs.
I have two overloaded generic methods:
T Foo<T>(T t) { Console.WriteLine("T"); return t; }
T Foo<T>(int i) { Console.WriteLine("int"); return default(T); }
When I try to call Foo as follows on my computer:
Foo(5);
I get no compiler errors or warnings, and the first method with the generic argument is called (i.e. the output is T). Will this be the case in all C# incarnations and on all platforms? In that case, why?
On the other hand, if I explicitly specify the type in the generic call:
Foo<int>(5);
the second method with the int argument is called, i.e. the output is now int. Why?
I am using different argument names in my two method overloads, so the output from the following calls are as expected:
Foo<int>(t: 5); // output 'T'
Foo<int>(i: 5); // output 'int'
If I am calling the first method, I can even leave out the type specification:
Foo(t: 5); // output 'T'
But if I try to compile this:
Foo(i: 5);
I get an error The type arguments for method 'Foo(int)' cannot be inferred from the usage. Try specifying the type arguments explicitly. Why cannot the compiler deal with this call?
Note These tests have been performed with LinqPad on a Windows 8 x64 system (in case that is relevant to the results...)
Last question
Since you specified (by parameter name) that it should call the overload that takes an int parameter, the compiler has no idea what to pass for T.
First question
Because of this, Foo(5) only matches one overload (Foo<T>()).
Therefore, it must only call Foo<T>().
Second question
When you explicitly specify a type argument (<int>), both overloads are applicable.
In that case, Foo(int) is better, since its parameter is not of generic type.
As per the C# spec §7.5.3.2:
Otherwise, if MP has more specific parameter types than MQ, then MP is better than MQ. Let {R1, R2, …, RN} and {S1, S2, …, SN} represent the uninstantiated and unexpanded parameter types of MP and MQ. MP’s parameter types are more specific than MQ’s if, for each parameter, RX is not less specific than SX, and, for at least one parameter, RX is more specific than SX:
A type parameter is less specific than a non-type parameter.
(emphasis added)
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Generic methods and method overloading
Ok, I hit this one by accident... Giving this situation:
class Program {
static void Main( string[ ] args ) {
var obj = new gen<int>( );
Console.Write( obj[ 1 ] );
Console.ReadKey( );
}
}
class gen<T> {
public int this[ T i ] { get { return 2; } }
public int this[ int i ] { get { return 1; } }
}
It will always print 1. I would have expected for the compiler to complain, or the runtime to crash and burn and melt the CPU, but no, it is happy to print '1'
Of course I can make a choice to return either if I use any other type for the generic parameter. For giggles, I try using UInt as the generic type parmater, and I can differentiate between the calls, so the questions are:
Why C# does not freak out? Shouldn't Anders Hejlsberg feel a disturbance in the force?
How can I restrict a generic parameter from certain types? As in this T can be anything but ints (but long are OK)
I believe this is specified in section 7.5.3.2 of the C# 4 spec (Better Function Member).
Otherwise, if MP has more specific parameter types than MQ, then MP is better than MQ [...]
A type parameter is less specific than a nontype parameter
[...]
So here, the member with the T parameter is less specific than the member with the int parameter.
Try to design your way out of this one simply by not overloading like this. It's hard for indexers of course, but you could always provide methods instead (or possibly as well, as a fall-back).
EDIT: Note that if you have overloaded methods where both are type parameters, the compiler will complain:
public class Foo<T1, T2>
{
public void Bar(T1 t1) {}
public void Bar(T2 t2) {}
}
...
Foo<int, int> foo = new Foo<int, int>();
foo.Bar(10); // Error
Here neither method is more specific.
How can I restrict a generic parameter from certain types? As in this T can be anything but ints (but long are OK)
This is really an entirely separate question, but basically you can't. You can constrain type parameters in various ways, but not by explicitly including and excluding types. See the MSDN page on constraints for more information.
As Eric Lippert says:
The C# specification says that when you have a choice between calling ReallyDoIt(string) and ReallyDoIt(string) – that is,
when the choice is between two methods that have identical signatures,
but one gets that signature via generic substitution – then we pick
the “natural” signature over the “substituted” signature.
Also this process described in C# spec 7.5.3.2 (Better function member):
In case the parameter type sequences {P1, P2, …, PN} and {Q1, Q2, …, QN} are equivalent (i.e. each Pi has an identity conversion to the corresponding Qi), the following tie-breaking rules are applied, in order, to determine the better function member.
If MP is a non-generic method and MQ is a generic method, then MP is better than MQ (as John pointed, this is true when you have generic method, not generic type)
...
Otherwise, if MP has more specific parameter types than MQ, then MP is better than MQ. Let {R1, R2, …, RN} and {S1, S2, …, SN} represent the uninstantiated and unexpanded parameter types of MP and MQ. MP’s parameter types are more specific than MQ’s if, for each parameter, RX is not less specific than SX, and, for at least one parameter, RX is more specific than SX:
A type parameter is less specific than a non-type parameter (this is your case - thus methods are not generic, and inferred parameter type equals to non-generic parameter type)
The C# compiler always chooses the more-specific versus the more-generic method if the call could fit in both =). That's why it doesn't freak out, he just follows his rules.
C# compiler doesn't freak out because both method are valid, and both can be called.
Here's an example that return "2":
Gen<Form> gen = new Gen<Form>();
textBox1.Text = gen[this].ToString();
Where "this" is a form. Of course, using an index accessor as an object instead of a number... Well, whatever, it works.
But like everybody else said, the compiler will prefer the explicit over the implicit.