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I have two vector a, b. I wanna compare the size of them. I know I can use if (a.size() > b.size()). But my question is if the size are too big out of the type int for a or/and b. For example, a.size() is 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999, etc. How can I compare the sizes of a and b? sorry for my English.

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  • A vector's size, while technically unbounded, is practically limited to the amount of available RAM. Commented May 17, 2013 at 14:49
  • the size of a vector isn't stored in an int it's stored in a size_t, which i s a special int type. A vector can only store max value of size_t elements, if you get to this maximum you have more issues than comparing vector sizes, it's likely only bounded by address space, not the limitations of they size_t type, but I've never bothered to look into it. Commented May 17, 2013 at 14:50
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    the value you list is 10^134 - 1 (ignoring the 'etc.'). Given that there are an estimated 10^80 atoms in the observable universe, it seems unlikely you'll ever find enough storage for such a vector. Commented May 17, 2013 at 15:26

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There's no way the size of a vector (or any object) can be greater than representable by std::size_t. That type can, by definition, store the size of the largest object a program can create.

As std::size_t is the type returned by a std::vector's (or any other standard library container's) size() function, you're safe in comparing them.

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std::size_t is not unlimited (in a finite computer, nothing is). But it's guaranteed to be large enough to hold the size of any object possible. Note ion particular than an array is also an object. std::size_t is also not limited by RAM exactly, more by the platform you're building for. Examples:

  • On a normal 32-bit system (e.g. Win32), the address space is 32-bit and the largest object can (in theory) occupy 232 bytes, so std::size_t is at least 32 bits long.

  • On a normal 64-bit system (e.g. Win64), the address space is 64-bit and the largest object can (in theory) occupy 264 bytes, so std::size_t is at least 64 bits long.

  • In x86 real mode, the address space is 20-bit, but no object can occupy more than one segment. As segments are 216 bytes, 16 bits are enough for std::size_t when the target platform is x86 real mode.

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Do you mean size_t is unlimited, only RAM can limit the size of vector?
@kaji331 I've added a reply to the answer.
Thanks. You mean object and size_t is limited by platform, so I can't create a unlimited vector even if I have "unlimited" RAM, then size_t is enough to be compared?
@kaji331 Yes, that's pretty much it. Note that on today's non-embedded platforms, the platform limitation will most likely be the size of the address space, which is an upper limit on the amount of 'unlimited' RAM you can have.

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