C#:无法从ulong转换为byte

奇怪的是我如何在C++中完成它,而不是在C#中.

为了清楚起见,我将两个函数粘贴在C++中,然后粘贴到C#中,并使用注释"// error"标记C#代码中有问题的行.这两个函数的作用是对参数进行编码,然后将其添加到名为byte1seeds的全局变量中.

这些是C++中的函数

//Global var:

unsigned char byte1seeds[3];

unsigned long GenerateValue( unsigned long * Ptr )
{
unsigned long val = *Ptr;
for( int i = 0; i < 32; i++ )
    val = (((((((((((val >> 2)^val) >> 2)^val) >> 1)^val) >> 1)^val) >> 1)^val)&1)|((((val&1) << 31)|(val >> 1))&0xFFFFFFFE);
return ( *Ptr = val );
}

void SetupCountByte( unsigned long seed )
{
if( seed == 0 ) seed = 0x9ABFB3B6;
unsigned long mut = seed;
unsigned long mut1 = GenerateValue( &mut );
unsigned long mut2 = GenerateValue( &mut );
unsigned long mut3 = GenerateValue( &mut );
GenerateValue( &mut );
unsigned char byte1 = (mut&0xFF)^(mut3&0xFF);
unsigned char byte2 = (mut1&0xFF)^(mut2&0xFF);
if( !byte1 ) byte1 = 1;
if( !byte2 ) byte2 = 1;
byte1seeds[0] = byte1^byte2;
byte1seeds[1] = byte2;
byte1seeds[2] = byte1;
}

现在C#代码:

我已经改变了函数GenerateValue.它没有指针作为参数,它有一个ulong参数.

要调用它并更改两个值,我使用:

ulong mut1 = GenerateValue(mut);

mut = mut1;

以下是已翻译的函数(有问题的行标有"// error");

//Global var:
public static byte[] byte1seeds = new byte[3];

public static ulong GenerateValue(ulong val)
{
    for( int i = 0; i < 32; i++ )
        val = (((((((((((val >> 2)^val) >> 2)^val) >> 1)^val) >> 1)^val) >> 1)^val)&1)|((((val&1) << 31)|(val >> 1))&0xFFFFFFFE);
    return val ;
}

public static void SetupCountByte( uint seed )
{
    if( seed == 0 ) seed = 0x9ABFB3B6;
    ulong mut = seed;
    ulong mut1 = GenerateValue(mut);
    mut = mut1;
    ulong mut2 = GenerateValue(mut);
    mut = mut2;
    ulong mut3 = GenerateValue(mut);
    mut = mut3;
    mut = GenerateValue(mut);
    byte byte1 = (mut & 0xFF) ^ (mut3 & 0xFF); //error
    byte byte2 = (mut1 & 0xFF) ^ (mut2 & 0xFF); //error
    if( byte1 != 0 )
        byte1 = 1;
    if( byte2 != 0 )
        byte2 = 1;
    byte1seeds[0] = byte1^byte2; //error
    byte1seeds[1] = byte2;
    byte1seeds[2] = byte1;
}

错误是:

无法将类型'ulong'隐式转换为'byte'.存在显式转换(您是否错过了演员？)

编辑:有问题的第3行的错误是:

无法将类型'int'隐式转换为'byte'.存在显式转换(您是否错过了演员？)

问题出在这里:如何解决这些错误？

提前致谢!

添加一个(byte)来投射它.由于您可能会失去精度,您必须告诉编译器该值将适合一个字节,即

byte byte1 = (byte)((mut & 0xFF) ^ (mut3 & 0xFF));
byte byte2 = (byte)((mut1 & 0xFF) ^ (mut2 & 0xFF));