What is the output of the following program:

void main()
{
printf("%d %d %d %d",65,065,0x65,0X65);
}

• O/p: 65, 53, 101, 101
  
  %d always print the decimal form of any number...
  65 -> decimal no
  065 => octal no. So it ll convert to decimal no.
  065=> 000 110 101 => 1*2^0 + 1*2^2+ 1*2^4 + 1*2^5 = 53
  0x65 => hexa-decimal number
  65 => 0110 0101 => 1*2^0 + 1*2^2 + 1*2^5 + 1*2^6 = 101
  0X65 => hexa decimal no.
  
  Hence it ll print 65, 53, 101, 101
• 10 years agoHelpfull: Yes(46) No(2)
• octal to decimal conversion
  Octal = 065
  Decimal= 0×8²+6×8¹+5×8⁰ = 53
  
• 9 years agoHelpfull: Yes(3) No(1)
• printf() is predefined method in stdio
  but not included
• 9 years agoHelpfull: Yes(1) No(4)
• saraswathy..!!! can u tell me hw this 1*2^0 + 1*2^2+ 1*2^4 + 1*2^5 = 53 and 1*2^0 + 1*2^2 + 1*2^5 + 1*2^6 = 101 written??
• 9 years agoHelpfull: Yes(0) No(3)
• priyanka.....
  065 is octal no ..convert it to binary so the binary value is 000 110 101 ...now convert this binary no to decimal ..so 1*2^0 + 1*2^2+ 1*2^4 + 1*2^5 = 53...
• 9 years agoHelpfull: Yes(0) No(5)
• 65
  53 => 065 is in octal notation ..after converting it into decimal 53
  101 => from hex to decimal
  101
• 9 years agoHelpfull: Yes(0) No(1)
• ans:65 53 101 101
• 9 years agoHelpfull: Yes(0) No(0)
• error in program , there is no declaration
• 9 years agoHelpfull: Yes(0) No(0)
• it shows error.."compile error:variable not defined"
• 7 years agoHelpfull: Yes(0) No(0)
• 65,53,101,101
• 6 years agoHelpfull: Yes(0) No(0)