View Full Version : C++: numeric_limits

erotavlas

May 3rd, 2011, 04:35 PM

Hello,

How can I find the overflow of the basic arithmetic types using standard C++?

For instance, if two unsigned long int are multiplied together, test whether or not the result is too large to fit in a long unsigned int?

For example

#include <limits>

unsigned long a;

unsigned long b;

if (a*b > std::numeric_limit<unsigned long>::max())

c = std::numeric_limit<unsigned long>::max();

else

c = a*b;

Thank you.

PaulM1985

May 3rd, 2011, 04:40 PM

You could check the multiplier.

If you have a * b must be less than max_number, you could find max_b using max_number / a. If max_b is less than b, then the number would be greater than the data type could store.

Paul

Arndt

May 3rd, 2011, 08:22 PM

You could check the multiplier.

If you have a * b must be less than max_number, you could find max_b using max_number / a. If max_b is less than b, then the number would be greater than the data type could store.

Paul

Another way is to do the multiplication with full precision, expressing the factors as P*a1+a2 and P*b1+b2, where P has half as many bits as the type of the factors.

erotavlas

May 4th, 2011, 02:05 PM

Another way is to do the multiplication with full precision, expressing the factors as P*a1+a2 and P*b1+b2, where P has half as many bits as the type of the factors.

I'm sorry but I have not understood your method...

dwhitney67

May 4th, 2011, 02:11 PM

I'm sorry but I have not understood your method...

I Googled for "C++ detect overflow", and amongst the many results, here's the first: http://stackoverflow.com/questions/199333/best-way-to-detect-integer-overflow-in-c-c

Arndt

May 4th, 2011, 03:08 PM

I'm sorry but I have not understood your method...

No, sorry, I wasn't detailed enough. The idea is to break down the numbers into smaller units, where you don't get overflow, and then do a combined shifting and addition, where it is easy to check for overflow. I only wanted to convey the idea, you can probably work out the details.

We do the same thing when we multiply numbers by hand: 13 = 1*10+3, 87 = 8*10+7. 13*87 = 1*8*10*10 + 1*7*10 + 3*8*10 + 3*7.

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