main()
{
int a=10,b=20;
a^=b^=a^=b;
printf("%d%d",a,b);

a) a=20,b=10
b) a=10,b=20
c) Syntax error
d) Unpredictable

Ans : a=20 b=10

can somebody conclude it.

• This program is to swap the contents of a and b using ex-or operator.
  ^= is short-hand assignment i.e. (a ^= b)is equivalent to (a = a ^ b)
  
  As assignments are right associative (i.e. = ,+=, -=, *=, ^= are evaluated right to left when assignments are cascaded).
  Thus statement :
  a^=b^=a^=b is equivalent to (a = a ^ (b = b ^ (a = a ^ b)))
  Performing bitwise ex-or operation contents of a and b are swapped.
• 11 years agoHelpfull: Yes(6) No(0)
• if prog is as it is... then there will be 1 error(since '{}' is not completed) and 1 warning(return value is not mention (e.g. void main())
  => [for the statement, int a=10,b=20;
  a^=b^=a^=b; ans:a=20,b=10]
  => x^=y => x=x^y (where ^ is bitwise exclusive OR)
  for statement a^=b^=a^=b;
  => (right to left) a^=b => a=a^b
  a=10^20= 01010^10100 =11110 (will save in a)
  now, a^=b^=a
  b^=a => b=b^a
  =>b=10100^11110=01010 (decimal value 10)
  again, a^=b
  (but here a=11110,b=01010) a= a^b= 11110^01010=10100(decimal value 20)
  =>a=20 & b=10
• 11 years agoHelpfull: Yes(4) No(0)
• a^=b^=a^=b;
  /*
  This can be written as follows (right to left)
  
  ->
  1.a^=b;
  2.b^=a;
  3.a^=b;
  step 1:
  01010:a=10
  10100:b=20
  -----------(performing exclusive or)
  11110:a=30 (step-1 result)
  
  10100:b=20
  11110:a=30
  -----------(xor)
  01010:b=10(step-2 result)
  
  11110:a=30
  01010:b=10
  ----------
  10100:a=20(step-3 result)
  finally:
  10100:a=20
  01010:b=10
  */
• 11 years agoHelpfull: Yes(3) No(0)
• a)a=20,b=10
• 11 years agoHelpfull: Yes(0) No(0)
• starting from right
  new_a1=a+b
  new_b=b+new_a=b+a+b=a
  new_a2=new_a1a+new_b=a+b+a=b
  thus now a and b are inverted
• 11 years agoHelpfull: Yes(0) No(0)