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  • C Question

    Neha Kochhar

    Member

    Updated: Oct 26, 2024
    Views: 1.3K
    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
    0
    Replies
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Replies

  • Kaustubh Katdare

    Administrator • Jan 17, 2010

    Is this a C programming question?
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  • Manish Goyal

    Member • Jan 17, 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
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  • Neha Kochhar

    Member • Jan 20, 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
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  • Hussanal Faroke

    Member • Jan 25, 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
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  • rishita

    Member • Jun 11, 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>
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  • vik001ind

    Member • Jun 11, 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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