If both number are negative shouldn't be b < min_int - a
On Aug 28, 11:49 pm, "Avinash LetsUncomplicate.." wrote:
> @Saurabh being ready to run your code for unsatisfactory input doesn't
> seem more logical than trying to find out if you can do something
> about it. i would have loved a kind repl
@Saurabh being ready to run your code for unsatisfactory input doesn't
seem more logical than trying to find out if you can do something
about it. i would have loved a kind reply from you.
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if( unsigned int (a) + unsigned int (b) < 0) return "overflow" else return
"ok";
On Sun, Aug 28, 2011 at 7:23 PM, saurabh singh wrote:
> @avinash in that case the number will become negative,It wont remain a
> satisfactory input.Try to think logically and believe me once you get the
> logic
@avinash in that case the number will become negative,It wont remain a
satisfactory input.Try to think logically and believe me once you get the
logic you will relent why there is no option to delete previous
mails...,
On Sun, Aug 28, 2011 at 6:32 PM, Dave wrote:
> @Avinash: Give an exam
@Avinash: Give an example.
Dave
On Aug 28, 7:00 am, "Avinash LetsUncomplicate.."
wrote:
> @dave i was saying if user input a .. he can enter a>intmax then how
> to detect overflow?
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@dave i was saying if user input a .. he can enter a>intmax then how
to detect overflow?
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@Dave- Thank u frnd By the way, this question was asked recently in
Tejas Network.
On Sun, Aug 28, 2011 at 10:17 AM, Dave wrote:
> @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
@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.." wrote:
> @dave i was saying if user enter a+b in which a
@Kunal: You are very kind.
Dave
On Aug 27, 12:58 pm, Kunal Patil 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
@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
Oops, I just saw that Dave had already corrected it. Net problem, sorry
guys!
On Sat, Aug 27, 2011 at 11:02 PM, dipit grover 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 wrote:
>
>> @Abishek: I was brain-
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 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
@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
@Dave: i didn't understand,
suppose a=3, 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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@Dave: +1 for nice solution... :)
On Sat, Aug 27, 2011 at 10:12 PM, Dave wrote:
> @Avinash: What do you mean by "exceeding limit(already overflowed)"?
> The number maxint is the limit. Every positive number is <= maxint.
>
> Dave
>
> On Aug 27, 10:31 am, "Avinash LetsUncomplicate.." 2...@gmail.
@Avinash: What do you mean by "exceeding limit(already overflowed)"?
The number maxint is the limit. Every positive number is <= maxint.
Dave
On Aug 27, 10:31 am, "Avinash LetsUncomplicate.." wrote:
> @dave if number a is entered exceeding limit(already overflowed) then
> a goes negative(if a wa
@dave if number a is entered exceeding limit(already overflowed) then
a goes negative(if a was entred just more than limit ) /a goes
positive (if a was entred sufficiently more than limit ) maxint -a>=b
concept fails
Avinash Katiyar
NIT Allahabad
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@Prags: Pseudocode:
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 maxint - a >= b.
If both numbers are negative, overflow will occur if maxint + a >= -b.
Dave
On Aug 27, 8:
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