C++ bit manipulation

Question

I am trying to extract character value from UTF-8 format. Suppose I have two characters, and I extract 5 bits from first character => 10111 and 6 bits from another character => 010000

so

ch1 = 10111;
ch2 = 010000;

how would I combine them to form 10111010000 and output its hex as 0x5d0? Do I need to shift or is there an easier way to do this, because checking the documentation write appear to be able to read characters sequentially, is there a similar function like this? Also, it appears I would need a char buffer since 10111010000 is 11 bits long. Does any know how to go about this?

Answer 1

You need to use shifting, plus the | or |= operator.

unsigned int ch3 = (ch1 << 6) | ch2;
// ch3 = 0000010111010000

I'm assuming here that an unsigned int is 16 bits. Your mileage may vary.

Answer 2

You will definitely need to use shift and OR.

First, declare an unsigned integer type of the right size. I like the C99 types defined in stdint.h but your C++ compiler may not have them. If you don't have uint16_t then you can use unsigned short . That is 16 bits wide and can hold 11 bits.

Then you would figure out which bits go into the high bits. It looks like it should be:

unsigned short ch1 = 0x17;
unsigned short ch2 = 0x10;
unsigned short result = (ch1 << 6) | ch2;

Answer 3

1: for combining them together:

char bytes[2] = { 0x17, 0x10 }; // for example

unsigned short result = 0;      // 00000000  00000000
result = bytes[0] << 6;         // 101 11000000
result |= bytes[1];             // 101 11010000

2: for printing it out as hex

std::cout << std::showbase << std::hex << <what you want to print>;

in this case:

std::cout << std::showbase << std::hex << result
// output: 0x5d0 if it is little-endian, it depends on your operating system

Answer 4

First, from K&R: "Almost everything about bitfields is implementation dependent".

The following works on MS Visual Studio 2008:

#include <stdio.h>
#include <string.h>

struct bitbag {
    unsigned int ch2 : 6;
    unsigned int ch1 : 6;
};

int main ()
{
    struct bitbag bits;

    memset(&bits, 0, sizeof(bits));

    bits.ch1 = 0x17;    // 010111
    bits.ch2 = 0x10;    // 010000

    printf ("0x%06x 0x%06x\n", bits.ch1, bits.ch2);
    printf ("0x%0x\n", bits);

    return 0;
}

Produces the output:

0x000017 0x000010
0x5d0

However I could not guarentee that it will work in the same way in all compilers. Note the memset which initialises any padding to zero.