开发者

Guid == null should not be allowed by the compiler

开发者 https://www.devze.com 2022-12-19 06:54 出处:网络
The behaviour described below is specific to .net-3.5 only I just ran across the most astonishing behavior in the C# compiler;

The behaviour described below is specific to .net-3.5 only

I just ran across the most astonishing behavior in the C# compiler;

I have开发者_如何学Python the following code:

Guid g1 = Guid.Empty;
bool b1= (g1 == null);

Well, Guid is not nullable therefore it can never be equal to null. The comparison i'm making in line 2 always returns false.

If you make the same thing for an integer, the compiler issues an warning saying the result will always be false:

int x=0;
bool b2= (x==null);

My question is: Why does the compiler lets you compare a Guid to null?

According to my knowledge, it already knows the result is always false.

Is the built-in conversion done in such a way that the compiler does assume null is a possible value?

Am I missing anything here?


Mark is correct. Value types that define their own equality operators automatically get lifted-to-nullable versions defined as well, for free. The nullable equality operator that takes two nullable guids is applicable in this situation, will be called, and will always return false.

In C# 2, this produced a warning, but for some reason, this stopped producing a warning for guid-to-null but continues to produce a warning for int-to-null. I don't know why; I haven't had time to investigate yet.

I apologize for the error; I probably screwed up one of the warning-detection code paths when rewriting the nullable logic in C# 3. The addition of expression trees to the language majorly changed the order in which nullable arithmetic operations are realized; I made numerous mistakes moving that code around. It's some complicated code.


The comparison is valid because the compiler converts the Guid to a Nullable<Guid> and then it makes sense.

There is a bug report on the warning not being issued here.

See here here for a fuller explanation from Eric Lippert.


Actually there is a case when Guild == null will return true.

However it is kinda hard to explain.

In ORM mapping frameworks (openAccess for example) when you have a Guid field which will have a default value of Guid.Empty of course it is possible to have the fallowing scenario :

  • You add a new Guid field + a Property
  • You upgrade the old database schema.. in this case all values will be NULL in the database.
  • If you populate an object having this null column of type Guild of course the Object WILL get an Guid.Empty value HOWEVER if you use an LINQ query ... in the LINQ query it looks the Guid is not yet populated so you need to use == null. Maybe it is a bug but this is the way it is.

In short (using OpenAccess but probably not only) :

var item = GetItems().Where(i => i.SomeGuidField == null); will work and u will get items with null guid this is after an schema update. item.First().SomeGuidField will return Empty Guid

var item = GetItems().Where(i => i.SomeGuidField == Guid.Empty); will not work even if after the population of the item it will be Guid.Empty and will return empty result.


Of course this is not only an issue for Guid. The same behavior is seen with any struct type which is not a pre-defined type of C# provided that the struct overloads operator == in the usual way. Other examples in the framework include DateTime and TimeSpan.

This deserves a compile-time warning since, while technically legal because of the lifted operator, this is not a useful comparison since it always gives false. As such, it is an indication of a programmer mistake.

As Eric Lippert said in his answer, the compile-time warning existed with the Visual C# 2.0 compiler. In versions 3.0 through 5.0, the warning was accidentally omitted (for these "user-defined" struct types, but not for pre-defined value types like int, and not for enum types).

Since C# 6.0 (based on Roslyn), the compiler detects this code problem once again. However, because of backwards compatibility(?!), the warning is not issued unless you compile your code with the so-called strict feature.

To enable strict when you use a .csproj file (most usual case), unload the project from Visual Studio, edit the file, insert the XML element:

<Features>strict</Features>

into each <PropertyGroup> (there will usually be more than one) of the .csproj file. You then get the warning (can be "promoted" to an error if you use Treat warnings as errors).

If you cannot edit the .csproj and if you call msbuild.exe from the command line for compiling, use the switch:

/p:Features=strict

to msbuild.exe.

If you do not use .csproj files because you compile directly with csc.exe (the C# compiler), use the switch:

/features:strict

to csc.exe on the command line.

0

精彩评论

暂无评论...
验证码 换一张
取 消