bit manipulation - Bitwise operations in java -

what piece of code mean ...can explain how work..

sum += (i & (1<<j)) != 0 ? n[j] : 0; 

full code:

  int max = (1 << n)-1;    //system.err.println(max);   string res = "no";   (int = 0; <= max; i++)   {     long sum = 0;     (int j = 0; j < n; j++)     {       sum += (i & (1<<j)) != 0 ? n[j] : 0;     }     //system.err.println(i + " " + sum);     if(sum == m){        res = "yes";       break;     } 

let's a = 0011 1100

so binary left shift operator (<<). left operands value moved left number of bits specified right operand.

a << 2 give 240 1111 0000 

so in code have loop , loop j

and line

sum += (i & (1<<j)) != 0 ? n[j] : 0; 

so second iteration of = 2 , first iteration of j = 1

first left shift operator shift left bits 1 position, resulting in 0000 0001 << 1 = 0000 0010 (or 2)

then have binary and comparison (0000 0010 in binary) & (0000 0010) = 0000 0010 (or 2)

and and result asked if it's distinct of zero. if result it's true sum increased number in n[j] array position, else not increased.

java has shortened version of if else command. use of easy, once understand it.

it’s written:

x ? y : z; 

here question mark , colon take place of commands if , else. means:

condition ? incaseoftrue : elsecase;