24bit unsigned integer

[akitta]

If I want to convert 3 bytes to a '24bit unsigned integer' Is this how I would do it ?

  #include <iostream>

using namespace std;

int main()

{
   int id3 = 11000111+(1001101<<8)+(11001010<<16);

}

And then I need to Display the Results.!

[Chervil]

11000111 is a decimal number, but it looks like you intended it to be a binary value. It's probably easier to handle if you use hex rather than binary. Or use a bitset.

#include <iostream>
#include <iomanip>
#include <string>
#include <bitset>

    using namespace std;

int main()
{
    // construct integer from three 8-bit values
    int id3 = 0xC7  + (0x4D<<8) + (0xCA<<16);

    // display result in hex and decimal
    cout << "id3 = "
         << hex << uppercase << id3 << " (hex) "
         << dec << id3  << " (decimal) "
         << endl;

    // construct bitset from string
    bitset<24> bitsa (string("110010100100110111000111"));
    // convert it to unsigned long and output result
    unsigned long a2 = bitsa.to_ulong();
    cout << "a2 = " << a2;

    //construct another bitset from unsigned long
    bitset<24> bitsb(a2);
    // convert to string and output bits
    cout << "  a2 (binary) = " << bitsb.to_string() << endl;

    return 0;
}

id3 = CA4DC7 (hex) 13258183 (decimal)
a2 = 13258183  a2 (binary) = 110010100100110111000111

[akitta]

Thankyou Chervil..
I was doing a Challenge. I had to Convert (199), (77) and (202). Into a byte each(199) Must go first.
Then from 'binary' to 'Hex' and combine the 'hex' together to get a 'decimal'.
I was trying to write a small code to do it.

Ok , So the '0x' tells it that i want to read a 'hex' value.!

[Chervil]

Well, I only used hex as it was the easiest way to deal with the pseudo-binary values originally presented. You could just use the decimal values 199, 77 and 202 directly if that was the actual data given in the question.
int id3 = 199 + (77<<8) + (202<<16);

[Chervil]

There may be more to this question than this, I don't think the full details of the original challenge have been stated clearly.

A bit of googling has turned up this, which I think might be the actual question:

Let us take the following three decimal numbers:
199, 77, 202
Convert each one into a byte. (Even though 77 does not require all 8 bits to express itself, when dealing with a group of data, we usually keep it in a consistent form.) Now, take those three bytes and combine them to form a 24-bit unsigned integer. The 199 byte is the high byte (most significant) and so forth. Please enter that 24-bit integer in decimal form, and that is your answer. (Hint: your answer will not be '19977202'!)

If that is the actual question, then the solutions I gave so far have been wrong, for two reasons. Firstly, it used integer rather than byte values, and secondly the byte order in the result was incorrect.

For a byte value one might use an unsigned char or uint8_t type. Similarly the result might be an unsigned int or uint32_t type.
Include header <cstdint> for the latter types, which are guaranteed to occupy the specified number of bits.