dynamic_bitset | Simple Useful Libraries : C17/20 header-only dynamic bitset | File Utils library
kandi X-RAY | dynamic_bitset Summary
kandi X-RAY | dynamic_bitset Summary
Simple Useful Libraries: C++17 header-only dynamic bitset.
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of dynamic_bitset
dynamic_bitset Key Features
dynamic_bitset Examples and Code Snippets
Lines of Code : 26License : Permissive (MIT)
Copy#include
#include
#include
int main()
{
// predefined bitset
sul::dynamic_bitset<> bitset1(12, 0b0100010110111);
std::cout << "bitset1 = " << bitset1 << std::endl;
// random bitset
std::minstd_rand rand(std::rand add_subdirectory()
target_link_libraries( PRIVATE sul::dynamic_bitset)
cmake_minimum_required(VERSION 3.10)
project(CoolProject LANGUAGES CXX)
add_executable(CoolProject main.cpp)
target_compile_features(CoolProject PRIVATE cxx_std_20)
add_subdir $ git clone --recursive https://github.com/pinam45/dynamic_bitset.git
$ git submodule update --init --recursive
$ mkdir cmake-build
$ cd cmake-build
$ cmake ..
$ cmake --build .
$ ctest --output-on-failure
Community Discussions
Trending Discussions on dynamic_bitset
QUESTION
Is there a clean way to return the reverse ordering of a boost::dynamic_bitset object?
For example:
01001100
becomes
00110010.
The simplest solution I can think of is to convert the bitset to a string, reverse the string and convert it back to a bitset, but this seems a rather slow method that nullifies the speed of bitstring operations.
Thank you in advance!
...ANSWER
Answered 2021-Mar-30 at 13:41boost::dynamic_bitset doesn't have iterators, so a long range of comfy STL solutions like, off the top of my head, std::reverse or std::swap or their boost counterparts are not available, I reckon that a good way would be to make your own trivial reverse method:
QUESTION
I am using Visual Studio 2017 to build LucenePlusPlus which in turn uses boost/dynamic_bitset.
The following code in LucenePlusPlus
...ANSWER
Answered 2021-Feb-10 at 21:58I have traced the code for my boost 1_75_0 and I see it has a define to conditionally reserve access to friends:
QUESTION
I am trying to send raw bits from a boost::dynamic_bitset with the Winsock2 sendto function. MS documentation shows that sendto uses const char * type for the buffer parameter. How do I send only the raw bits stored in the dynamic_bitset? I don't know how to cast or manipulate the dynamic_bitset so that it can be used in the sendto function.
Here are some relevant lines from my program:
...ANSWER
Answered 2021-Jan-27 at 01:16I've implemented serialization for dynamic_bitset before: How to serialize boost::dynamic_bitset?
You can use that or a similar technique based on the same interfaces:
QUESTION
I am writing a C++ program and I need to estimate the memory occupied by a boost::dynamic_bitset<>. The sizeof() operator always returns 32 as the value, but I'm doubtful whether thats always the case, especially when the number of bits exceeds 1 million.
What is the better way to estimate this?
Thanks in advance.
...ANSWER
Answered 2021-Jan-12 at 08:39You can expect the actual memory usage to be (in bytes):
sizeofthe object itself (which you've said is 32 bytes in your case) plus- however-many bits you've constructed the
dynamic_bitsetto divided by 8, probably rounded up somewhat by your dynamic memory allocation library, but at worst probably to the next power-of-two in bytes
If you push_back or append to the dynamic_bitset to increase its length gradually, it'll probably double the memory usage periodically, same way e.g. std::vector and std::unordered_map do, but I haven't checked the code or docs for that. That's generally considered the sane compromise between excessive copying and being wasteful of memory. If you want to check, look at the source code.
At runtime, you can get a more accurate idea of the bytes current allocated using by calling .capacity() and dividing by 8 (but there can still be a little overhead from the allocation library, and there's the fixed number of bytes per sizeof - 32 in your case):
size_type capacity() const;
Returns: The total number of elements that *this can hold without requiring reallocation. Throws: nothing.
See: https://www.boost.org/doc/libs/1_75_0/libs/dynamic_bitset/dynamic_bitset.html
You can divide by 8 because you can fit 8 bits in a single byte. (Strictly speaking it's better to use CHAR_BIT instead of hard-coding 8).
QUESTION
So I have a vector which has three numbers. 65, 66, and 67. I am converting these numbers from int to binary and appending them in a string. the string becomes 100000110000101000011 (65, 66, 67 respectively). I am writing this data into a file through dynamic_bitset library. I have BitOperations class which does the reading and writing into file work. When I read the data from file instead of giving the above bits it gives me these 001100010100001000001 bits.
Here is my BitOperations class:
...ANSWER
Answered 2021-Jan-05 at 19:25You are facing two different issues:
The boost function
to_block_rangewill pad the output to the internal block size, by appending zeros at the end. In your case, the internal block size issizeof(unsigned char)*8 == 8. So if the bit sequence you write to the file inwriteToFileis not a multiple of8, additional0s will be written to make for a multiple of8. So if you read the bit sequence back in withreadFromFile, you have to find some way to remove the padding bits again.There is no standard way for how to represent a bit sequence (reference). Depending on the scenario, it might be more convenient to represent the bits left-to-right or right-to-left (or some completely different order). For this reason, when you use different code pieces to print the same bit sequence and you want these code pieces to print the same result, you have to make sure that these code pieces agree on how to represent the bit sequence. If one piece of code prints left-to-right and the other right-to-left, you will get different results.
Let's discuss each issue individually:
Regarding issue 1I understand that you want to define your own block size with the bitSize variable, on top of the internal block size of boost::dynamic_bitset. For example, in your main method, you construct BitOperations c(7, "bits2.bin");. I understand that to mean that you expect the bit seqence stored in the file to have a length that is some multiple of 7.
If this understanding is correct, you can remove the padding bits that have been inserted by to_block_range by reading the file size and then rounding it down to the nearest multiple of your block size. Though you should note that you currently do not enforce this contract in the BitOperation constructor or in writeToFile (i.e. by ensuring that the data size is a multiple of 7).
In your readFromFile method, first note that the inner loop incorrectly takes the blockSize into account. So if blockSize is 7, this incorrectly only considers the first 7 bits of each block. Whereas the blocks that were written by to_block_range use the full 8 bit of each 1-byte block, since boost::dynamic_bitset does not know anything about your 7-bit block size. So this makes you miss some bits.
Here is one example for how to fix your code:
QUESTION
I am attempting to convert a std::vector into a boost::dynamic_bitset. I can achieve the inverse of this using the following code, where values is a class member function defined as
boost::dynamic_bitset values.
ANSWER
Answered 2020-Jul-01 at 13:41Your decode method is marked const, but you are attempting to modify class member variable values.
Either remove the const or mark values mutable
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install dynamic_bitset
Support
Reuse Trending Solutions
Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items
Find more librariesStay Updated
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page
