1

I need to have an int variable go up to 4294967295 (0xFFFFFFFF), without using unsigned and long, so it can actually be used by other functions. Is that even possible?

Jonathon Reinhart
  • 132,704
  • 33
  • 254
  • 328
Maxim Khannaov
  • 41
  • 4
  • 1
    Yes, if you can find or create a compiler that support that. – MikeCAT Nov 04 '20 at 16:32
  • 2
    It is possible only if your platform has an `int` larger than 32 bit. If your platform has 32 bit `int` then it cannot represent `4294967295`. Why are you trying to avoid `unsigned`? It would fix your problem. – François Andrieux Nov 04 '20 at 16:33
  • Dirty but widely used hack: `#define int long long` ... Oops, this uses `long`. Then `#define int int64_t` – MikeCAT Nov 04 '20 at 16:33
  • 2
    What are you trying to achieve? Why you can't use `unsigned int`? What does 'used by other functions' mean? – SergeyA Nov 04 '20 at 16:34
  • 1
    @MikeCAT: Alas the behaviour of doing that is undefined. – Bathsheba Nov 04 '20 at 16:34

2 Answers2

3

You can't, unless you find a platform with an int that's larger than 32 bits (they are at the time of writing rare).

If you need to use a signed type, then std::int64_t is a good choice. Or long long is also guaranteed to be large enough.

If you're absolutely constrained by the interface to your function then you could break your number up into two parts - and adopt the convention that you need to call the function twice to get anything meaningful out of it. But that could make your program very brittle.

Another option: if your function is allowed to take the int by reference then you could have an array of two ints at the call site, and pass the first one to your function. You can then reach the other using pointer arithmetic within the function body! Again, this will make your code brittle but it would work.

Bathsheba
  • 231,907
  • 34
  • 361
  • 483
3

You can't. Because: an int value is defined as 31 bits plus one leading bit as a sign.

You can push whatever you want into the 32 bits of an int, as long as it is declared as int, the leading bit will always be interpreted as the sign.

Bathsheba
  • 231,907
  • 34
  • 361
  • 483
Gyro Gearloose
  • 1,056
  • 1
  • 9
  • 26
  • 1
    Be careful: not necessarily. Although from C++20, `int` has to be a 2's complement type, with a minimum range [-32768, 32767] – Bathsheba Nov 04 '20 at 16:46