Question

如果我有以下这些方法：

public int add (int a , int b) {


 assert(a>= 0 && a < 256 && b >= 0 && b < 256);

 // the addition is simply an xor operation in GF (256) since we are working on modulo(2) then 1+1=0 ; 0+0=0; 1+0=0+1=1;

    return a ^ b;

}

第二种方法是：

public int FFMulFast(int a, int b){
int t = 0;;

if (a == 0 || b == 0)

return 0;

 // The multiplication is done by using lookup tables. We have used both logarithmic and exponential table for mul
// the idea is firstly look to Logarithmic table then add their powers and find the corresponding of this to exponential table

t = (Log[(a & 0xff)] & 0xff) + (Log[(b & 0xff)] & 0xff);

if (t > 255) t = t - 255;

return Exp[(t & 0xff)];

}

现在我想用这些方法计算多项式f（x）= a0 + a1x + a2 x（pow 2）+ ... a2 x（pow k-1）其中这些系数a0，a1，a2 i有生成如下：

 public void generate (int k) {
byte a [] = new byte [k];

Random rnd = new SecureRandom () ;

a.nextBytes (a); // the element of byte array are also negative as for example -122; -14; etc 

}

现在我想计算我的多项式，但我不确定它是否有效，因为这个负系数。我知道JAVA只支持有符号字节，但我不确定下面的方法是否能正常工作：

private int evaluate(byte x, byte[] a) {

    assert x != 0; // i have this x as argument to another method but x has //only positive value so is not my concern , my concern is second parameter of //method which will have also negative values
    assert a.length > 0;
    int r = 0;
    int xi = 1;
    for (byte b : a) {

        r = add(r, FFMulFast(b, xi));
        xi = FFMulFast(xi, x);
}
    return r;
}

有什么建议吗？此外，如果这个不工作，任何人都可以建议我如何在不使用掩码的情况下将负值转换为正值，因为这样会将数据类型更改为int然后我不能使用getBytes（a）方法

Answer 1

for (byte b : a) {
        r = add(r, FFMulFast(b, xi));
        xi = FFMulFast(xi, x);
}

如果add()和FFMulFast()方法预期为正值，则必须使用：

for (byte b : a) {
        r = add(r, FFMulFast(b & 0xff, xi));
        xi = FFMulFast(xi, x & 0xff);
}

我可以使用十进制中具有负值的字节作为正值

1 个答案: