将C++转换为C#

edv*_*ld0 1 c# c++

在c ++中:

byte des16[16];
....
byte *d = des16+8;
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在c#?

byte des16[16];
    [0] 207 'Ï' unsigned char
    [1] 216 'Ø' unsigned char
    [2] 108 'l' unsigned char
    [3] 93 ']'  unsigned char
    [4] 249 'ù' unsigned char
    [5] 249 'ù' unsigned char
    [6] 100 'd' unsigned char
    [7] 0   unsigned char
    [8] 76 'L'  unsigned char
    [9] 50 '2'  unsigned char
    [10]    104 'h' unsigned char
    [11]    118 'v' unsigned char
    [12]    104 'h' unsigned char
    [13]    191 '¿' unsigned char
    [14]    171 '«' unsigned char
    [15]    199 'Ç' unsigned char

after
 byte *d = des16+8;
d = "L2hvh¿«Ç†¿æ^ òÎL2hvh¿«Ç"
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Man*_*red 9

C#(一般来说)没有指针.也许以下是您的追求:

byte[] des16 = new byte[16];
byte byteAtIndex8 = des16[8];
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此代码为您提供索引8处的元素.

如果我正确地读取了你的代码,你就是试图在索引8处获取元素的地址.这一般来说 - 用C#是不可能的(除非你使用不安全的代码).