How to write a 24 bit message after reading from a 4-byte integer on a big endian machine (C)?

Question

I am constructing a message to send a 24-bit number over the network. For little endian machines, the code is (ptr is the pointer to the message buffer):

*ptr++ = (num >> 16) & 0xFF;
*ptr++ = (num >> 8)  & 0xFF;
*ptr++ = (num)       & 0xFF;

(So if num0, num1, num2 and num3 are the individual bytes making up num, the message would be encoded as num2|num1|num0.)

What should be the code for encoding num2|num1|num0 on a big endian machine?

Answer 1

The question here is, in what byte order shall the message be sent/constructed ? Because whether you are on a little or big endian machine doesn't matter with respect to num, as you're already dividing num into individual bytes in an endian-agnostic way.

The code you've posted stores 24 bits of num in big endian (aka network byte order). So if that's what you want you're already done. If you want to store it in big little instead, just reverse the order:

*ptr++ = (num)       & 0xFF;
*ptr++ = (num >> 8)  & 0xFF;
*ptr++ = (num >> 16) & 0xFF;

Answer 2

Your code is portable regardless of endianess. The shift operators >> << work with the values, not with the representation.

Answer 3

In the receiving machine, regardless of endian-ness, if you receive them in same order as they are stored in ptr, assemble them like this:

num = (ptr[0] << 16) + (ptr[1] << 8) + (ptr[2]);

Answer 4

int main(int argc, char** argv) {

int a,b; 
a=0x0f000000;                  // Contain 32 bit value
printf( "before = %d\n", a);
b= a & ( ~0xff000000 );     // It will convert last 8 bit into zero so we got only 24 bit value in number b.
printf("After = %d\n", b);
return (EXIT_SUCCESS);

}

There is number a contain 32 bit value but number b contain only 24 bit's start from least significant digit. And that's not depend on endian because bitwise operator not work with memory representation.

So you use

num = num & ( ~0xff000000 );

for get least 24 bit value.