What kind of datatype will hold 3000000^2 in C++?

Question

What will hold something like 3000000^2 without overflow?

I have asked a [related question here](http://stackoverflow.com/questions/10053113/is-c11s-long-long-really-at-least-64-bits). Basically, in c++(11) long long is the only portable way of holding such a number. — juanchopanza, Apr 07 '12 at 10:29

score 5 · Accepted Answer · answered Apr 07 '12 at 07:12

5

3000000^2 is less than 2^48, so an uint64_t from cstdint or stdint.h will hold your number with good margin.

answered Apr 07 '12 at 07:12

Joachim Isaksson

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For some reason when I try these datatypes, it does not output the correct number. – WhatsInAName Apr 07 '12 at 07:20
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@WhatsInAName: Please update your question with some code that demonstrates the problem you are trying to solve. – johnsyweb Apr 07 '12 at 07:21
5

`^` is the bitwise xor operator. C++ has no exponentiation operator – Keith Thompson Apr 07 '12 at 07:29
@WhatsInAName also, could you tell us what compiler you are using and whether it supports some c++11 features? – juanchopanza Apr 07 '12 at 07:36

score 2 · Answer 2 · answered Apr 07 '12 at 07:11

2

The 64 bit integer:

long long int

answered Apr 07 '12 at 07:11

unsym

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juanchopanza · Answer 3 · 2012-04-07T10:27:21.737

You can check with std::numeric_limits. In c++11, long long is a standard integer of at least 64 bits, so, for example, unsigned long long has these limits:

#include <iostream>
#include <limits>


int main() {
  std::cout << "sizeof : " << sizeof(unsigned long long) << "\n"; 
  std::cout << "min    : "<< std::numeric_limits<unsigned long long>::min() << "\n";
  std::cout << "max    : "<< std::numeric_limits<unsigned long long>::max() << "\n\n";
}

producing on my platform:

sizeof : 8
min    : 0
max    : 18446744073709551615

C++ provides various integer types, but as far as I know the 64 bit types were not mandated by the standard pre-c++11. I long long is the only standard integer to be at least 64 bits, so you would need a compiler with c++11 support. But it could be that your compiler supports some other non-standard 64 bit type.

I get the following msg: integer constant is too large for "long" type (when I try unsigned long long to store my number) — WhatsInAName, Apr 07 '12 at 07:24
The problem could be that there was no "standard" 64 bit int before c++11, and maybe your compiler doesn't support c++11? What compiler are you using? — juanchopanza, Apr 07 '12 at 07:31

score 1 · Answer 4 · answered Apr 07 '12 at 07:15

When you say "something like", it suggests that you are not actually sure how large the number is that you want to store.

Take a look here (say), and see which numeric type is best for your application.

Note that there is on 64-bit integer in that list; that's because it's not quite available everywhere.

What kind of datatype will hold 3000000^2 in C++?

4 Answers4