[翻譯] C# 的常見錯誤

出處

http://www.dotnetcurry.com/csharp/1417/csharp-common-mistakes

線上工具

https://dotnetfiddle.net

引言

C#是個好棒棒的言語,但是它仍會有超乎你想像的行為,
而且就算你是有經驗的開發者,你也要看一看這篇文章.
這篇文章不講幹話,還會給你代碼喔

C# Quiz

Null Value

Null 很危險啦, 你別在 Null 身上調用方法
(譯注:在公司的維運人員應該還蠻常見這個錯誤的 一 ω 一)

We are all aware that null values can be dangerous, if not handled properly.
Dereferencing a null-valued variable (i.e. calling a method on it or accessing one of its properties)
will result in a NullReferenceException, as demonstrated with the following sample code:

1
2
object nullValue = null;
bool areNullValuesEqual = nullValue.Equals(null);

就安全的角度,好像我們要不停的檢查 reference type 是不是 null ,
雖然這件事常常發生,好像也很難說成是非預期的行為了…
(譯注:又有種中槍的感覺)

To be on the safer side, we should always make sure that reference type values are not null before dereferencing them.
Failing to do so could result in an unhandled exception in a specific edge case.
Although such a mistake occasionally happens to everyone, we could hardly call it unexpected behavior.

看看這個代碼, null 值在 runtime 的時候不會有 type 的

1
2
string nullString = (string)null;
bool isStringType = nullString is string;

No, null 值在 runtime 的時候不會有 type 的
No, null 值在 runtime 的時候不會有 type 的
No, null 值在 runtime 的時候不會有 type 的
很重要所以說三次,
當然你也別想呼叫 GetType() 方法

The correct answer is No.

A null value has no type at runtime.

In a way, this also affects reflection.
Of course, you can’t call GetType() on a null value because a NullReferenceException would get thrown:

1
2
object nullValue = null;
Type nullType = nullValue.GetType();

純量呢？

1
2
3
int intValue = 5;
Nullable<int> nullableIntValue = 5;
bool areTypesEqual = intValue.GetType() == nullableIntValue.GetType();

那我們可不可能用反射(reflection)區分 nullable 跟 non-nullable 的值？
答案是不可能, 看看後面的代碼

Is it possible to distinguish between a nullable and a non-nullable value type using reflection?

The answer is No.

The same type will be returned for both variables in the above code: System.Int32.
This does not mean that reflection has no representation for Nullable, though.

1
2
3
Type intType = typeof(int);
Type nullableIntType = typeof(Nullable<int>);
bool areTypesEqual = intType == nullableIntType;

上面兩段程式在 runtime 拿到的 type 很不一樣喔,
一個是System.Int32一個是 System.Nullable'1\[System.Int32\]

當 null 遇上多載方法 (Handling Null values in Overloaded methods)

1
2
3
4
5
6
7
8
9
private string OverloadedMethod(object arg)
{
    return "object parameter";
}

private string OverloadedMethod(string arg)
{
    return "string parameter";
}

上面有兩個OverloadedMethod
猜猜看,傳入 null 時會呼叫哪一個方法？

1
var result = OverloadedMethod(null);

有人會猜編譯失敗嗎？
MAGIC ! 竟然可以編譯成功, 而回傳的值是 “string parameter” ,
一般來說,在編譯時期會作型別檢查,相同簽章的方法參數可以被轉型成另一個型別時,是可以編譯成功的喔.
而有明確型別的方法將被優先調用(譯注:求這段.Net Framework 的原碼來看一下,知道的人請告訴我)

如果要指定 null 參數呼叫的多載方法就要對 null 轉型唷,可以參考下面的方法.

1
var result = OverloadedMethod((object)null);

算術運算 (Arithmetic Operations)

好像很少用位移運算吼？
回憶一下 左移移 右移移

1
var shifted = 0b1 << 1; // = 0b10
1
var shifted = 0b1 >> 1; // = 0b0

bits 跑到底並不會重頭開始喔,一直移位到爆掉就變 0 了.
(這裡會用 32 是因為 int 是 32bit 的數值,你可以試試放超過 32 的數值到 for loop 裡會發生什麼事)

The bits don’t wrap around when they reach the end.
That’s why the result of the second expression is 0.
The same would happen if we shifted the bit far enough to the left (32 bits because integer is a 32-bit number):

1
2
3
4
5
var shifted = 0b1;
for (int i = 0; i < 32; i++)
{
shifted = shifted << 1;
}

The result would again be 0.

那我們是不是可以一次移 32 bit,讓它一次變成 0 呢？
靠北啊 竟然不行捏, 你只會拿到 1,
這跟運算子(operator)基本運算有關,在作位元運算的時候,
會拿第一個運算數除以第二個運算數後取餘數,
這導致我們只會拿 32 % 32 的結果 , 也就是 1 啦
(譯注:這段其實我不是很確定,如果錯誤請糾正)

However, the bit shifting operators have a second operand.
Instead of shifting to the left by 1 bit 32 times, we can shift left by 32 bits and get the same result.

1
var shifted = 0b1 << 32;

Right? Wrong.

The result of this expression will be 1. Why?

Because that’s how the operator is defined. Before applying the operation,
the second operand will be normalized to the bit length of the first operand with the modulo operation,
i.e. by calculating the remainder of dividing the second operand by the bit length of the first operand.

The first operand in the example we just saw was a 32-bit number, hence: 32 % 32 = 0.
Our number will be shifted left by 0 bits. That’s not the same as shifting it left by 1 bit 32 times.

好棒棒 你竟然可以看到這裡,
那我們繼續討論 & (and) 跟 | (or) 運算子吧,
這兩個運算子跟一般的運算子有點不一樣

  • 通常只要看運算子的第一個運算數就能得知結果
  • 在有掛 [Flag] attribute 的列舉它們好好用(看一下範例)
1
2
3
4
5
6
7
8
[Flags]
private enum Colors
{
    None = 0b0,
    Red = 0b1,
    Green = 0b10,
    Blue = 0b100
}
1
2
Colors color = Colors.Red | Colors.Green;
bool isRed = (color & Colors.Red) == Colors.Red;

上面這個刮號可不能省略喔, 因為(&)運算符的優先順序低於(==)運算符,
不過這段程式沒有刮號的話連編譯都不會過,真是好加在
另外在 .NET framework 4.0 之後的版本提供更棒的方法去檢查 flags

1
bool isRed = color.HasFlag(Colors.Red);

Math.Round()

猜一下這個值會是多少？

1
var rounded = Math.Round(1.5);

猜 2 的就答對了, 下一題
猜一下這個值會是多少？

1
var rounded = Math.Round(2.5);

還是 2 ,
因為預設會取最接近的偶數

No. The result will be 2 again. By default,
the midpoint value will be rounded to the nearest even value.
You could provide the second argument to the method to request such behavior explicitly:

1
var rounded = Math.Round(2.5, MidpointRounding.ToEven);

這個行為可以透過MidpointRounding參數改變

1
var rounded = Math.Round(2.5, MidpointRounding.AwayFromZero);

另外要小心浮點數的精度問題,
以下的例子結果會是 1,( 因為 float 的 0.1 實際上小於 0.1 一 ω 一 )
這提醒我們在處理精確數值時,應轉換成整數處理.
(譯注:使用 dotnet fiddle 時並不會有這個問題, 在 windows 環境下測試的確會有問題)

1
2
3
var value = 1.4f;

var rounded = Math.Round(value + 0.1f);

類別初始化

最佳實踐建我我們應該避免在建構子初始化類別,
特別是靜態建構子.
在初始化一個類別的順序如下

  1. 靜態欄位
  2. 靜態建構子
  3. 實體欄位
  4. 實體建構子

看看這個例子

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
public static class Config
{
    public static bool ThrowException { get; set; } = true;
}

public class FailingClass
{
    static FailingClass()
    {
        if (Config.ThrowException)
        {
            throw new InvalidOperationException();
        }
    }
}

當我們嘗試實例化 FailingClass 時,你會得到 Exception;
值得注意的事,你拿到的會是TypeInitializationException
而並不是InvalidOperationException,

那麼我們是不是可以試著透過 try catch 補捉錯誤,
並修改靜態屬性,重新實體化 class 呢？
答案是不行

一個靜態建構值,如果它拋出一個異常,
那麼無論何時你想創建一個實例或以任何其他方式訪問這個類,
這個異常都會被重新拋出.

1
2
3
4
5
6
7
try
{
    var failedInstance = new FailingClass();
}
catch (TypeInitializationException) { }
Config.ThrowException = false;
var instance = new FailingClass();

這個類別在程序重啟前是不能再被使用了(會拋出錯誤),
這在 C# 是個非常糟糕的實踐,
千萬別這樣設計你的類別.

The static constructor for a class is only called once.
If it throws an exception, then this exception will be rethrown
whenever you want to create an instance or access the class in any other way.

The class becomes effectively unusable until the process (or the application domain) is restarted.
Yes, having even a minuscule chance that the
static constructor will throw an exception, is a very bad idea.

繼承與類別初始化

繼承的類別初始化執行順序更加複雜,看看下面的例子

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
public class BaseClass
{
    public BaseClass()
    {
        VirtualMethod(1);
    }

    public virtual int VirtualMethod(int dividend)
    {
        return dividend / 1;
    }
}

public class DerivedClass : BaseClass
{
    int divisor;
    public DerivedClass()
    {
        divisor = 1;
    }

    public override int VirtualMethod(int dividend)
    {
        return base.VirtualMethod(dividend / divisor);
    }
}

當我們初始化 DerivedClass

1
var instance = new DerivedClass();

你會得到一個除 0 的錯誤 DivideByZeroException
這與執行順序有關

  1. 呼叫 BaseClass 建構子
  2. 執行 DerivedClass VirtualMethod (overrid BaseClass)
  3. divisor 未賦值拋出 DivideByZeroException

多形 Polymorphism

這個例子只是要說明多形的概念與應用,
你可以透過轉形呼叫基底類別的方法.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
var instance = new DerivedClass();
var result = instance.Method(); // -> Method in DerivedClass
result = ((BaseClass)instance).Method(); // -> Method in BaseClass
// The correct answer is: by using the new modifier.

public class BaseClass
{
    public virtual string Method()
    {
        return "Method in BaseClass ";
    }
}

public class DerivedClass : BaseClass
{
    public new string Method()
    {
        return "Method in DerivedClass";
    }
}

It’s typically used to hide the interface methods from the consumers of the class implementing it,
unless they cast the instance to that interface.
But it works just as well if we want to have two different implementations of a method inside a single class.
It’s difficult to think of a good reason for doing it, though.

另外一個例子是明確實作介面方法,
如果你的類別已經有同名的方法的話.
雖然沒有什麼好理由建議你這樣作.
(譯注:實務上我有在遇到歷史共業這樣作過…)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
var instance = new DerivedClass();
var result = instance.Method(); // -> Method in DerivedClass
result = ((IInterface)instance).Method(); // -> Method belonging to IInterface
It’s explicit interface implementation.

public interface IInterface
{
    string Method();
}

public class DerivedClass : IInterface
{
    public string Method()
    {
        return "Method in DerivedClass";
    }

    string IInterface.Method()
    {
        return "Method belonging to IInterface";
    }
}

迭代器 Iterators

小心 Iterators 的陷阱
看看以下代碼:

1
2
3
4
5
6
7
8
private IEnumerable<int> GetEnumerable(StringBuilder log)
{
    using (var context = new Context(log))
    {
        return Enumerable.Range(1, 5);
    }
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
public class Context : IDisposable
{
    private readonly StringBuilder log;

    public Context(StringBuilder log)
    {
        this.log = log;
        this.log.AppendLine("Context created");
    }

    public void Dispose()
    {
        this.log.AppendLine("Context disposed");
    }
}

假設我們 foreach 呼叫 GetEnumerable 方法,
你預期 Context 類別會有什麼樣的行為？
我們會印出以下的 output 嗎？

Context created
1
2
3
4
5
Context disposed

1
2
3
4
5
var log = new StringBuilder();
foreach (var number in GetEnumerable(log))
{
    log.AppendLine($"{number}");
}

不是的,
實際上印出的是

Context created
Context disposed
1
2
3
4
5

這點很重要,
因為實務上你很有可能 using dbconnetion 之類的物件,
那麼你在取得真正的資料之前,
你的連線就已經中斷了

This means that in our real world database example, the code would fail –
the connection would be closed before the values could be read from the database.

看看以下的修正

1
2
3
4
5
6
7
8
9
10
private IEnumerable<int> GetEnumerable(StringBuilder log)
{
    using (var context = new Context(log))
    {
        foreach (var i in Enumerable.Range(1, 5))
        {
            yield return i;
        }
    }
}

譯注:看到這裡對 yield return 的使用情境才比較有感啊…

如果你不太熟yield return,其實它只是個語法糖,允許增量執行,
參考以下範例,或許能更容易理解

1
2
3
4
5
6
7
8
9
10
11
12
13
14
private IEnumerable<int> GetCustomEnumerable(StringBuilder log)
{
    log.AppendLine("before 1");
    yield return 1;
    log.AppendLine("before 2");
    yield return 2;
    log.AppendLine("before 3");
    yield return 3;
    log.AppendLine("before 4");
    yield return 4;
    log.AppendLine("before 5");
    yield return 5;
    log.AppendLine("before end");
}
1
2
3
4
5
6
7
var log = new StringBuilder();
log.AppendLine("before enumeration");
foreach (var number in GetCustomEnumerable(log))
{
    log.AppendLine($"{number}");
}
log.AppendLine("after enumeration");

before enumeration
before 1
1
before 2
2
before 3
3
before 4
4
before 5
5
before end
after enumeration

值得注意的事, 如果你在 loop 當中重複執行以上的代碼,
那麼 Iterators 也會重複執行

1
2
3
4
5
6
7
8
9
10
var log = new StringBuilder();
var enumerable = GetCustomEnumerable(log);
for (int i = 1; i <= 2; i++)
{
    log.AppendLine($"enumeration #{i}");
    foreach (var number in enumerable)
    {
        log.AppendLine($"{number}");
    }
}

輸出如下,可以明顯看到 GetCustomEnumerable 方法,
實際上被隱含的執行了兩次,
這在 Code Review 的階段也是難以被察覺的.

enumeration #1
before 1
1
before 2
2
before 3
3
before 4
4
before 5
5
before end
enumeration #2
before 1
1
before 2
2
before 3
3
before 4
4
before 5
5
before end

比較好的作法是將 IEnumerable ToList(),
如果你真的需要對 IEnumerable 的結果作 loop 的操作

1
2
3
4
5
6
7
8
9
10
var log = new StringBuilder();
var enumerable = GetCustomEnumerable(log).ToList();
for (int i = 1; i <= 2; i++)
{
    log.AppendLine($"enumeration #{i}");
    foreach (var number in enumerable)
    {
        log.AppendLine($"{number}");
    }
}

輸出結果

before 1
before 2
before 3
before 4
before 5
before end
enumeration #1
1
2
3
4
5
enumeration #2
1
2
3
4
5

譯者小結

如果真的能夠預期所有的行為的開發人員,
真的是好棒棒,
對我來說 static class constructor 的行為是超乎預期的,
然後對 yield return 的使用場景更有感覺了.
本來預計農曆年就可以完成的翻譯,
竟然也拖了這麼久,看來我英文還是不行啊.

希望對大家有幫助,也請多多看原文 :)

(fin)

Please enable JavaScript to view the Gitalk. :D
Please enable JavaScript to view the LikeCoin. :P
Please enable JavaScript to view the LikeCoin. :P