按位不对一串位操作

Question

I have a string with the value 0111000000 . How can I perform a bitwise not operation on this string?

If I convert it to an integer, use the ~ operator and convert it back to a binary string, the resulting string has extra bits. I want the output to be exactly 1000111111 .

The following code works fine, but it's not a good method. Is there another better way of doing this?

   String bstr="";
   while(m!=str.length())
   {             
        char a=str.charAt(m);
        if(a=='1')
        {
             a='0';
             bstr=bstr+a;
             m++;   
        }
        else
        {
             a='1'; 
             bstr=bstr+a;
             m++;
         }
}

Answer 1

try this

    char[] a = s.toCharArray();
    for(int i = 0; i < a.length; i++) {
        a[i] = a[i]=='0' ? '1' : '0';
    }
    s = new String(a);

this also works fine

    int i = ~Integer.parseInt(s, 2);
    String tmp = Integer.toBinaryString(i);
    s = tmp.substring(tmp.length()- s.length());

Answer 2

Keep track of how many bits there are in your bit-string. After converting to an integer and using a ~value operation to flip the bits, use a bit-mask to remove the unwanted 1 high-end bits.

Say for example your bit-string has a fixed 10 bits. Then you can mask off the unwanted high-end bits with: value & 0x2ff .

If the number of bits in the bit-string is variable:

value & ((1 << nBits) - 1)

Answer 3

来自common-lang的StringUtils.replaceChars可能会有所帮助：

StringUtils.replaceChars("0111000000", "01", "10");

Answer 4

You should use string builder so you are able to change individual bits without creating many many garbage strings. Also you can flip single bits using XOR:

b ^= 1;

Which works on both binary and ASCII values of digits.

Answer 5

maybe this will work:

String result = Integer.toBinaryString(~(Integer.parseInt("0111000000",2)));

converts binary String to int, use bitwise not operator to invert, then convert back to binary string.

Answer 6

You can do this using XOR operation

public String xorOperation(String value) {
    String str1 = value;
    long l = Long.parseLong(str1, 2);
    String str2 = "";
    for (int i = 0; i < str1.length(); i++) {
        str2 = str2 + "1";
    }
    long n = Long.parseLong(str2, 2);
    long num = l ^ n;
    String bininaryString = Long.toBinaryString(num);
    System.out.println(bininaryString);
    return bininaryString;
}