@Neha Kochhar • 17 Jan, 2010
What is the value of the expression (3^6) + (a^a)?

a) 3 b) 5 c) 6 d) a+18 e) None



I am Unable to solve this Out?
Ans is 5 but how?
Explain the Process
@Kaustubh Katdare • 17 Jan, 2010 Is this a C programming question?
@Manish Goyal • 17 Jan, 2010
Neha Kochhar
What is the value of the expression (3^6) + (a^a)?

a) 3 b) 5 c) 6 d) a+18 e) None



I am Unable to solve this Out?
Ans is 5 but how?
Explain the Process
I think you want to solve this expression using c right?
ok
If you don't input the value of a then it will take garbage value..now in order to obtain output 5 in the expression it should be like this
(3^6)+(a^-a)
ie a's power is a negative value
@Neha Kochhar • 21 Jan, 2010
goyal420
I think you want to solve this expression using c right?
ok
If you don't input the value of a then it will take garbage value..now in order to obtain output 5 in the expression it should be like this
(3^6)+(a^-a)
ie a's power is a negative value
but how the value 5 is obtained?
Please explain it .I did not get what you said
@Hussanal Faroke • 25 Jan, 2010 is the 'a' defined, if no try it in another system? or u defined pls tell the value of a! u know in c 32767+1 is -32765
@rishita • 11 Jun, 2011 Nice Goyal. But, we want detail solution. How we get the answer. Can you give detail step by step process of solution.
Thanks for your solution.

Regards,
<link removed>
@vik001ind • 11 Jun, 2011 It doesn't matter whether a is defined or not.
^ stands for XOR. Xoring same bit return 0, different values return 1. Eg, 1^0 == 1 , 1^1 == 0
Any int variable in C is 16 bit (16 bit compiler) or 32 bit (32 bit compiler). So, in any case whether it is defined or not, a will be a 16/32 bit pattern.
Considering 16 bit compiler
Bit pattern of 3 is 0000 0000 0000 0000 0011
XOR
Bit pattern of 6 is 0000 0000 0000 0000 0110
Result is --> 0000 0000 0000 0000 0101 ---> 5

a^a will always be equal to 0. Since we have bit pattern same in both cases.
Therefore (3^6) + (a^a) = 5.

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