java中二进制字符串的异或
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Xor of binary strings in java
提问by Kailash
i am using the following code to XOR 2 strings
我正在使用以下代码对 2 个字符串进行 XOR
String lseq = "0011111111101111111111111100101101111100110000001011111000010100";
String tseq = "0011111111100000110011001100110011001100110011001100110011001100";
StringBuilder sb = new StringBuilder();
for(int i = 0; i < lseq.length(); i++)
sb.append((lseq.charAt(i) ^ tseq.charAt(i)));
String result = sb.toString();
System.out.println(result);
The above code gives me correct output :
上面的代码给了我正确的输出：
0000000000001111001100110000011110110000000011000111001011011000
I need to XOR one more string
我需要再异或一个字符串
String hseq = "0011111111110010010111110100010111100000101101001110000100011110";
if i try sb.append((lseq.charAt(i) ^ tseq.charAt(i) ^ hseq.charAt(i));
如果我尝试 sb.append((lseq.charAt(i) ^ tseq.charAt(i) ^ hseq.charAt(i));
I am getting result :
我得到结果：
48484848484848484848484948484948494848494848494949484848494848494848494849494848484949494948484848484948494948494949484948484948
which is wrong. I need help in doing xor of 3 binary strings
这是错误的。我需要帮助做 3 个二进制字符串的异或
采纳答案by Elliott Frisch
I would do it like this
我会这样做
private static boolean bitOf(char in) {
return (in == '1');
}
private static char charOf(boolean in) {
return (in) ? '1' : '0';
}
public static void main(String[] args) {
String lseq ="0011111111101111111111111100101101111100110000001011111000010100";
String tseq ="0011111111100000110011001100110011001100110011001100110011001100";
String hseq ="0011111111110010010111110100010111100000101101001110000100011110";
StringBuilder sb = new StringBuilder();
for (int i = 0; i < lseq.length(); i++) {
sb.append(charOf(bitOf(lseq.charAt(i)) ^ bitOf(tseq.charAt(i))
^ bitOf(hseq.charAt(i))));
}
String result = sb.toString();
System.out.println(result);
}
Which outputs
哪些输出
0011111111111101011011000100001001010000101110001001001111000110
回答by Guido
You can use BigInteger, it will make your code simpler. It has a constructor where you can pass a String and the base you want to use (2 in your case).
您可以使用BigInteger，它会使您的代码更简单。它有一个构造函数，您可以在其中传递一个字符串和要使用的基数（在您的情况下为 2）。
Then you can do a XOR calling public BigInteger xor(BigInteger val)(see docsor take a look to openjdk BigInteger's code)
然后您可以进行 XOR 调用public BigInteger xor(BigInteger val)（请参阅文档或查看openjdk BigInteger 的代码）
回答by ian
You could also consider something like this:
你也可以考虑这样的事情：
String lseq = "0011111111101111111111111100101101111100110000001011111000010100";
String tseq = "0011111111100000110011001100110011001100110011001100110011001100";
String hseq = "0011111111110010010111110100010111100000101101001110000100011110";
StringBuilder sb = new StringBuilder();
for(int i = 0; i < lseq.length(); i++)
sb.append((lseq.charAt(i) - '0' ^ tseq.charAt(i) - '0' ^ hseq.charAt(i) - '0'));
String result = sb.toString();
System.out.println(result);
Under the hood, a charis treated like an intin that the numeric value represents a pre-defined character. We can subtract the value of the character '0' from our character (knowing that the value of '1' is only 1 more than '0') and get either 0 or 1, which can be used with the ^operator.
在幕后， achar被视为 an int，因为数值表示预定义的字符。我们可以从我们的字符中减去字符'0'的值（知道'1'的值只比'0'多1）得到0或1，可以和^运算符一起使用。
