Latest C Aptitude Question SOLUTION: 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.

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(#M40019670) C question Keep an EYE

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.

Asked In C zen mano (10 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

Deepashree  (10 years ago)

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.

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Chandni Maurya  (10 years ago)

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

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vikash kumar  (10 years ago)

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
*/

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rahul rathorew  (10 years ago)

a)a=20,b=10

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mohini  (10 years ago)

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

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