Converting a four character string to a long

Question

I want to convert a four character string (i.e. four characters) into a long (i.e. convert them to ASCII codes and then put them into the long).

As I understand it, this is done by writing the first character to the first byte of the long, the second to the adjacent memory location, and so on. But I don't know how to do this in C++.

Can someone please point me in the right direction?

Thanks in advance.

Answer 1

Here's your set of four characters:

const unsigned char buf[4] = { 'a', '0', '%', 'Q' };

Now we assemble a 32-bit unsigned integer:

const uint32_t n = (buf[0]) | (buf[1] << 8) | (buf[2] << 16) | (buf[3] << 24);

Here I assume that buf[0] is the least significant one; if you want to go the other way round, just swap the indices around.

Let's confirm:

printf("n = 0x%08X\n", n); // we get n = 0x51253061
                           //               Q % 0 a

Important: Make sure your original byte buffer is unsigned, or otherwise add explicit casts like (unsigned int)(unsigned char)(buf[i]); otherwise the shift operations are not well defined.

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Word of warning: I would strongly prefer this algebraic solution over the possibly tempting const uint32_t n = *(uint32_t*)(buf), which is machine-endianness dependent and will make your compiler angry if you're using strict aliasing assumptions!

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As was helpfully pointed out below, you can try and be even more portable by not making assumptions on the bit size of a byte:

const unsigned very long int n = buf[0] |
              (buf[1] << (CHAR_BIT)     |
              (buf[2] << (CHAR_BIT * 2) |
              (buf[3] << (CHAR_BIT * 3)   ;

Feel free to write your own generalizations as needed! (Good luck figuring out the appropriate printf format string ;-) .)

Answer 2

If your bytes are in the correct order for a long on your machine then use memcpy, something like this -

#include <cstdlib>
#include <iostream>

int main()
{
    char data[] = {'a', 'b', 'c', 'd'};
    long result;

    std::memcpy(&result, data, 4);
    std::cout << result << "\n";
}

Note that this will be platform dependent for byte ordering in the long which may or may not be what you need. And the 4 is hard coded as the size in bytes of the long for simplicty. You would NOT hard code 4 in a real program of course. All the compilers I've tried this on optimize out the memcpy when optimization is enabled so it's likely to be efficient too.

EDIT: Go with the shift and add answer someone else posted unless this meets your specific requirements as it's much more portable and safe!

Answer 3

#include <string>
#include <iostream>

std::string fourCharCode_toString ( int value )
{
    return std::string( reinterpret_cast<const char*>( &( value ) ), sizeof(int) );
}

int fourCharCode_toInt ( const std::string & value )
{
    return *( reinterpret_cast<const int*>( value.data() ) );
}

int main()
{
    int a = 'DROW';
    std::string str = fourCharCode_toString( a );
    int b = fourCharCode_toInt( str );

    std::cout << str << "\n";
}