在Java中将具有偶数奇偶校验字节的UTF-8字符转换为ASCII 7位

Question

A device I am trying to communicate with takes ASCII 7-bit characters with even parity. When trying to convert a UTF-8 character I cast it to an integer then to a binary string. check the string and then set the parity bit if needed.

However when converting it back using Byte.parseByte I get a NumberFormatError if the signed bit is set. How can I get round this?

public byte addParity(byte b){
    int a = (int)b;
    int c = 0;
    String s = Integer.toBinaryString(a);
    for(int i=0; i!=(8-s.length());)
    {
        s = "0" +s;
    }

    for(int i=0; i<s.length(); i++){

        if(s.substring(i, i+1).equals("1"))c++;
    }
    if(c%2==0)return b;
    else return Byte.parseByte(("1"+s.substring(1)),2);         

}

Answer 1

You are getting the error because Byte.parseByte refuses to parse values out of the range of byte (-128..127). So it refuses to parse something like "10001011" which is 139 in decimal. A quick fix could be using Integer.parseInt instead and casting the result to byte:

else return (byte) Integer.parseInt(("1"+s.substring(1)),2);

I'd however step back and redo the whole thing with bitwise arithmetic.

Answer 2

要设置一点，它应该足以执行以下操作：

return (byte) (b | 0x80);