Java-位操作

Question

我有以下 Java 代码:

long a = Long.parseLong("11001100", 2);
long b = Long.parseLong("11000000", 2);
int npos = 0 ;
int pos = 0 ;
long n = ~(a ^ b) ;
int cnt = 0;
while (n != 0) {
  pos++ ;
  if ((n & 3) == 3) cnt++; // count how many matches
  else{
    npos = pos ;  // position that not matched also giving wrong which should be 2 nd position.
  }
  n >>>= 2;
}

System.out.println(npos + "  " + cnt) ; // also print which two bits are not matched i.e. 00 and 11

我试图找出两个整数中有多少个两位序列匹配。我还想找出哪两位不匹配。有人可以帮我怎么做吗？

PS:我的原始代码中没有字符串，只有整数。因此，我无法进行字符串操作。

编辑:

long a = Long.parseLong("11000100", 2);
long b = Long.parseLong("11000000", 2);
long mask = 0x03;
int npos = 0 ;
int cnt = 0;
long p1 = 0;
long p2 = 0;
for (int pos = 0; pos < 64; pos++, mask <<= 2) {

  if ((a & mask) == (b & mask)) {
     cnt++; // count how many matches
  } else {
    npos = pos ;  // *last* position that did not match
    p1 = (a & mask) ; // two bits that not matched
    p2 = (b & mask) ; // two bits that not matched
  }
}

  System.out.println(npos + "  " + cnt + " " + p1 + " " + p2) ; // also print which two bits are not matched i.e. 00 and 01

Answer 1

您将整数解析为以 10 为基数的数字，您可能希望将它们解析为二进制整数，为此，请使用具有基数参数的方法:

long a = Long.parseInt("11001100", 2);
long b = Long.parseInt("11000000", 2);

使用掩码运行循环比较 2 个值可能会更容易:

long mask = 0x03;
int npos = 0 ;
int cnt = 0;

for (int pos = 0; pos < 32; pos++, mask <<= 2) {

  if ((a & mask) == (b & mask)) {
     cnt++; // count how many matches
  } else {
    npos = pos ;  // *last* position that did not match
  }
}