爲什麼C＃中的8位和16位整數沒有算術運算符（+， - ，*，/，％）？

Question

我震驚知道有沒有算術運算符+，-，*，/，並%在C＃中8個和16位整數。我正在閱讀第23頁的「C＃5.0 Pocket Reference」，如下所示。

下面的代碼無法編譯。

class Program 
{ 
    static void With16Bit() 
    { 
     short a = 1; 
     short b = 2; 
     short c = a + b; 
     Console.WriteLine(c); 
    } 

    static void With8Bit() 
    { 
     byte a = 1; 
     byte b = 2; 
     byte c = a + b; 
     Console.WriteLine(c); 
    } 

    static void Main(string[] args) 
    { 
     With8Bit(); 
     With16Bit(); 
    } 
} 

爲什麼C＃設計人員會這樣做？他們對此有何考慮？

Answer 1

有算術與INT8，的Int16;但結果是INT32，所以你必須投：

class Program 
{ 
    static void With16Bit() 
    { 
     short a = 1; 
     short b = 2; 
     short c = (short) (a + b); // <- cast, since short + short = int 
     Console.WriteLine(c); 
    } 

    static void With8Bit() 
    { 
     byte a = 1; 
     byte b = 2; 
     byte c = (byte) (a + b); // <- cast, since byte + byte = int 
     Console.WriteLine(c); 
    } 

    static void Main(string[] args) 
    { 
     With8Bit(); 
     With16Bit(); 
    } 
} 

Answer 2

普萊舍記住，當你在short和byte進行加法運算的默認結果將是整數。

所以：

字節+字節= INT
短+短= INT

所以，如果你想找回你需要轉換回實際值。

試試這個：

 short a = 1; 
     short b = 2; 
     short c =(short) (a + b); 
     Console.WriteLine(c); 

     byte a = 1; 
     byte b = 2; 
     byte c =(byte) (a + b); 
     Console.WriteLine(c); 

從Source：

此行爲是因爲設計者沒有考慮到 byte和short爲實際數字，但他們認爲他們作爲 只序列號bits。所以對它們執行算術運算 沒有任何意義，所以如果是這樣的話int和long將爲 服務。

Answer 3

從kek444 answer

All operations with integral numbers smaller than Int32 are widened to 32 bits 
before calculation by default. The reason why the result is Int32 is 
simply to leave it as it is after calculation. If you check the 
MSIL arithmetic opcodes, the only integral numeric type they operate 
with are Int32 and Int64. It's "by design".