@Avinash: maxint is the largest possible integer. There aren't any
integers greater than it. Thus, a can't be greater than maxint. For
example, if an int is 32 bits, maxint = 2^31 - 1.

Dave

On Aug 27, 10:41 pm, "Avinash LetsUncomplicate.." <avin.
[email protected]> wrote:
> @dave i was saying if user enter a+b in which a>intmax .. A goes
> negative(if a sligtly >intmax)  a+b =no overflow which we know
> shouldnt be an answer..
>
> On 8/28/11, Dave <[email protected]> wrote:
>
>
>
>
>
> > @Kunal: You are very kind.
>
> > Dave
>
> > On Aug 27, 12:58 pm, Kunal Patil <[email protected]> wrote:
> >> @Dave: Still your approach to solve the problem remains correct.
> >> (subtracting a number from max possible value & then comparing this
> >> difference with another number). So, no need to think that you were brain
> >> dead (If you were, you would have posted a movie story here)..[?]
> >> Mathematically it is wrong, not in terms of approach..[?]
>
> >> On Sat, Aug 27, 2011 at 11:02 PM, dipit grover
> >> <[email protected]>wrote:
>
> >> > I think you just need to reverse the comparison operators in Dave's
> >> > earlier
> >> > post
>
> >> > On Sat, Aug 27, 2011 at 10:59 PM, Dave <[email protected]> wrote:
>
> >> >> @Abishek: I was brain-dead in my earlier posting. Let me try again:
>
> >> >> If either number is zero, the sum will not overflow.
> >> >> If the numbers have different signs, the sum will not overflow.
> >> >> If both numbers are positive, overflow will occur if b > maxint - a.
> >> >> If both numbers are negative, overflow will occur if b < -maxint - a -
> >> >> 1.
>
> >> >> Dave
>
> >> >> On Aug 27, 12:18 pm, Abhishek <[email protected]> wrote:
> >> >> > @Dave: i didn't understand,
> >> >> > suppose a=30000, b=31000 and MaxInt=32000;
> >> >> > you are saying if (MaxInt-a)>=b; then overflow will occur. but here
> >> >> > condition is not satisfying.
> >> >> > plz explain.
>
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> >> > --
> >> > Dipit Grover
> >> > B.Tech in CSE
> >> > IIT Roorkee
>
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> >>  361.gif
> >> < 1KViewDownload
>
> >>  360.gif
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