create std::bitset or QBitArray from a std::string or QString containing hexadecimal numbers

Question

Is there any way to construct a std::bitset from a hexadecimal std::string or QString and vise versa without performing binary shift operations? I know how to do it that way but I was wondering if it's possible to do this using C++ streams or something similar.

here's my code so far (trying to avoid coming under fire of moderators) :

QString data("aabbccddeeff");
QByteArray temp = QByteArray::fromHex(data.simplified().toLatin1());
QBitArray bits(temp.count()*8); 
for(int i=0; i<temp.count(); ++i) {
    for(int b=0; b<8;b++) {
        bits.setBit( i*8+b, temp.at(i)&(1<<(7-b)) );
    }
}

What's the max length of your resulting bits? Is it longer than the widest integer on your platform? — John Zwinck
– John Zwinck, Commented May 8, 2016 at 14:35

Trevor Hickey · Accepted Answer · 2016-05-08 14:31:42Z

1

You could convert the hex string to an integral, and construct a bitset from that.

#include <iostream>
#include <sstream>
#include <bitset>
#include <string>

using namespace std;

int main()
{
    string s = "0xA";
    stringstream ss;
    ss << hex << s;
    unsigned n;
    ss >> n;
    bitset<32> b(n);
    // outputs "00000000000000000000000000001010"
    cout << b.to_string() << endl;
}

answered May 8, 2016 at 14:31

Trevor Hickey

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1 Comment

max Over a year ago

ss >> n will initialize n with ascii value of the character supplied to ss by line ss << hex << s.

max · Accepted Answer · 2016-05-08 16:03:53Z

0

based on answer provided below by Trevor, I wrote the following code:

std::vector<std::bitset<8ul> > Bytes2Bits(QByteArray bytes)
{
    std::vector<std::bitset<8ul> > v;
    for(int i = 0 ; i < bytes.size(); ++i)
    {
        QByteArray temp;
        temp.append(bytes.at(i));
        std::string s = temp.toHex().toStdString();
        std::stringstream ss;
        ss << s;
        int n;
        ss >> n;
        std::bitset<8ul> b(n);
        v.push_back(b);
    }
    return v;
}

I hope it's useful to others seeking the same solution.

answered May 8, 2016 at 16:03

max

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Comments

John Zwinck · Accepted Answer · 2016-05-08 16:05:30Z

How about this?

std::string str("deadBEEF");
std::bitset<128> bits; // result

char* raw = reinterpret_cast<char*>(&bits) + sizeof(bits) - 1; // last byte                                         
assert(str.size() <= 2 * sizeof(bits));

for (size_t ii = 0; ii < str.size(); ++ii) {
    char ch = str[ii];

    if (ch >= '0' && ch <= '9') {
        ch -= '0';
    } else if (ch >= 'a' && ch <= 'f') {
        ch -= 'a' - 10;
    } else if (ch >= 'A' && ch <= 'F') {
        ch -= 'A' - 10;
    } else {
        throw std::runtime_error("invalid input");
    }

    if (ii % 2 == 0) {
        ch <<= 4; // nibble                                                                                         
    }

    *raw |= ch;

    if (ii % 2) {
        --raw;
    }
}

cout << bits << endl;

The above assumes that a std::bitset has exactly one data member: an array of integers large enough to hold the templated number of bits. I think this is a fair assumption but it certainly is not guaranteed to be portable. The good news it that it will be fast--partly because it does no dynamic memory allocation.

Omid Ardestani · Accepted Answer · 2019-10-09 07:21:33Z

0

QBitArray BitConverter::GetBits(quint64 value, bool lsb)
{
    QString binary;
    binary.setNum(value, 2);

    QBitArray bits(binary.count());
    if (lsb)
    {
        for (int i = 0; i < binary.count(); i++)
        {
            bits[i] = binary[i] == '1';
        }
    }
    else
    {
        for (int i = binary.count() - 1, j = 0; i >= 0; i--, j++)
        {
            bits[j] = binary[i] == '1';
        }
    }
    return bits;
}

answered Oct 9, 2019 at 7:21

Omid Ardestani

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2 Comments

Android Over a year ago

Please include an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes.

Prince Vegeta Over a year ago

Adding some explanation might help us understand the code written better.

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