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    Question :
    void main()
    int y=1,z;
    z=y++ + y++ + y++ + y++;
    printf(" z = %d",z);
    in the above code, the output is 4
    but when the statement z=y++ + y++ + y++ + y++; is changed to z=y++ + ++y + ++y + y++; the output changes to 12. Can somebody explain how the two statements works?

    Posted by: Shatish Kumar Pandey on 3/13/2011

    Contact Shatish Kumar Pandey Contact Shatish Kumar Pandey
    Category Computers Interview Questions
    Rating (0.2) By 220 users
    Added on 3/13/2011
    Views 423
    Rate it!

    In programing instruction, suppose, y=1 if z=y than value of z is 1 (because first it assign value of y to z and than increment in y by 1) and if z= y than value of z is 2 (because it increment in y by 1 and than assign that value to z) so that, in our case,
    z = y y y y
    Z = 1 3 4 4 at finally,
    Z = 12.

    Posted by: Sagar udhwani    

    Contact Sagar udhwani Contact Sagar udhwani

    z=y++ + ++y + ++y + y++
    = 1 + 3 + 4 + 4
    How the above values are calculated is explained below:
    The above expression hold y in 4 places.
    In first,it hold 1 and the value 1 is used in expression then incremented because it is postfix y++.
    Now y=2.
    In second term,y is incremented before it is used because it is here y=2+1=3.
    In third term, y is also incremented before it is used because it is here y=3+1=4.
    In fourth term,y is used then only it is incremented because it is the value 4 is used here.
    After that the present y hold the value 4+1=5.

    Posted by: sophiya    

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    Posted by: ganavi reddy    

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