int i=3,j=2,k=0,m;
m = ++i && ++j || ++k;
printf(“%d %d %d %d”, i , j, k,m);

• ++i=4(binary equivalent=100)
  ++j=3(binary equivalent =011)
  ++k=1(binary equivalent=001)
  now
  100&&011||001=001(which is nothing but 1)
• 3 years agoHelpfull: Yes(0) No(0)
• ++i , ++j,++k are pre incrment i.e the value is incrmented by 1 before executing
  now m=++i&&++j||++k i.e m=4&&3||1
  whenever && is used it will check if the both operand are non zero .if yes then it returns true ie 1
  and || is used it will check if one of the operand is non zero it returns 1
  so herby m=1
  the answer will be 4,3,1,1
• 5 Months agoHelpfull: Yes(0) No(0)