CODEWITHSUNDEEP
java
Krishna Teja
Krishna Teja

Reputation: 3

convert integers back after converting them into bitsets and OR -ing them in java

I perform the operation for two integers... consider I do

A = (1 << 6) OR (1 << 7)

Is there a way that I can get the integers 6 and 7 from A after doing the OR operation?

Please let me know if the answer has been answered anywhere else.

Upvotes: 0

Views: 52

Answers (4)

Anders R. Bystrup
Anders R. Bystrup

Reputation: 16050

Aside from the fact that your expression isn't valid Java and not knowing what you want to do, but assuming that you wish to ascertain which bits are "set", you can do this with Integer.numberOfTrailingZeros(), eg.:

Integer.numberOfTrailingZeros( A & 64 );   // 1<<6 == 64
Integer.numberOfTrailingZeros( A & 128 );  // 1<<7 == 128

This would get you the ints 6 and 7 respectively.

Cheers,

Upvotes: 1

Danny
Danny

Reputation: 7518

If you are always bitshifting 1 then you can do the following.

int a = (1 << 6) ^ (1 << 7);
int temp = a;
for(int i=0;i<32;i++) {
    if((temp&1)==1) 
        System.out.println("number: " + i);
    temp = temp >> 1;
}

Output:

 number: 6
 number: 7

Upvotes: 0

Holger
Holger

Reputation: 298153

int value=(1<<7) | (1<<6);

StringBuilder sb=new StringBuilder();
if(value==0) sb.append("0");
else while(value!=0) {
  int b=Integer.highestOneBit(value);
  if(b==1) sb.append("1");
  else
    sb.append("(1<<").append(Integer.numberOfTrailingZeros(b)).append(')');
  value-=b;
  if(value>0) sb.append(" | ");
}
System.out.println(sb);

Upvotes: 0

Henry
Henry

Reputation: 43738

You can fetch the individual bits by testing them:

bit6 = (A >> 6) & 1;
bit7 = (A >> 7) & 1;

The integer 6 is in the bitset if bit6 == 1.

Upvotes: 0

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