C ++中的位数组

Question

When working with Project Euler problems I often need large (> 10**7) bit array's.

My normal approach is one of:

bool* sieve = new bool[N];

bool sieve[N];

When N = 1,000,000 my program uses 1 MegaByte (8 * 1,000,000 bits).

Is there a more efficient way to use store bit arrays than bool in c++?

Answer 1

Use std::bitset (if N is a constant) otherwise use std::vector<bool> as others have mentioned (but dont forget reading this excellent article by Herb Sutter)

A bitset is a special container class that is designed to store bits (elements with only two possible values: 0 or 1, true or false, ...).

The class is very similar to a regular array, but optimizing for space allocation : each element occupies only one bit (which is eight times less than the smallest elemental type in C++: char).

EDIT :

Herb Sutter (in that article) mentions that

The reason std::vector< bool > is nonconforming is that it pulls tricks under the covers in an attempt to optimize for space: Instead of storing a full char or int for every bool[1] (taking up at least 8 times the space, on platforms with 8-bit chars), it packs the bools and stores them as individual bits (inside, say, chars) in its internal representation.

std::vector < bool > forces a specific optimization on all users by enshrining it in the standard. That's not a good idea; different users have different requirements, and now all users of vector must pay the performance penalty even if they don't want or need the space savings.

EDIT 2 :

And if you have used Boost you can use boost::dynamic_bitset (if N is known at runtime)

Answer 2

For better or for worse, std::vector<bool> will use bits instead of bool's, to save space. So just use std::vector like you should have been in the first place.

If N is a constant , you can use std::bitset .

Answer 3

You could look up std::bitset and std::vector<bool> . The latter is often recommended against, because despite the vector in the name, it doesn't really act like a vector of any other kind of object, and in fact doesn't meet the requirements for a container in general. Nonetheless, it can be pretty useful.

OTOH, nothing is going to (at least dependably) store 1 million bool values in less than 1 million bits. It simply can't be done with any certainty. If your bit sets contain a degree of redundancy, there are various compression schemes that might be effective (eg, LZ*, Huffman, arithmetic) but without some knowledge of the contents, it's impossible to say they would be for certain. Either of these will, however, normally store each bool/bit in only one bit of storage (plus a little overhead for bookkeeping -- but that's usually a constant, and on the order of bytes to tens of bytes at most).

Answer 4

A 'bool' type isn't stored using only 1 bit. From your comment about the size, it seems to use 1 entire byte for each bool.

AC like way of doing this would be:

uint8_t sieve[N/8]; //array of N/8 bytes

and then logical OR bytes together to get all your bits:

sieve[0] = 0x01 | 0x02; //this would turn on the first two bits

In that example, 0x01 and 0x02 are hexadecimal numbers that represent bytes.

Answer 5

是的，你可以使用bitset 。

Answer 6

You might be interested in trying the BITSCAN library as an alternative. Recently an extension has been proposed for sparseness, which I am not sure is your case, but might be.

Answer 7

试试std :: bitset

Answer 8

You can use a byte array and index into that. Index n would be in byte index n/8 , bit # n%8 . (In case std::bitset is not available for some reason).

Answer 9

如果在编译时已知N，则使用std :: bitset ，否则使用boost :: dynamic_bitset 。

Answer 10

A 'bool' type isn't stored using only 1 bit. From your comment about the size, it seems to use 1 entire byte for each bool.

AC like way of doing this would be:

uint8_t sieve[N/8]; //array of N/8 bytes

element of array is:

result = sieve[index / 8] || (1 << (index % 8)); 

or

result = sieve[index >> 3] || (1 << (index & 7));

set 1 in array:

sieve[index >> 3] |= 1 << (index & 7);