How to make an 8bit (byte) from two 4bits (nibbles)?

Suppose we have 2 ints 5 and C(hexidecimal representation of 12) and we want to get a char, which is the concatenation of 4 bits of the first number and 4 bits of the second number. As a result in my example, i have to get a char with bits: 0101 1100. Can somebody help me? Thank you!

This issue arose from the following scenario: in my program I work directly with bits. To work with bits I use the char data type. Initially, I need to work with 4-bit numbers, but the char size is 8 bits. I have 16 int numbers, which are 4-bit numbers (0, 1, ..., 15), I need to write them using a char, and for this I need to put two ints in one char.

Caveat: OP states values of 'first' and 'second' are between 0 to 15 inclusive.

unsigned char hi = 0x5, lo = 0xC, result;

result = (hi << 4) | lo; // low 4 bits of 'first' OR'd with low 4 bits of 'second'

That should do it..

You can define a Macro for this purpose:

#include <stdio.h>
 
#define CONCAT(x,y) (x) << 4 | (y)
 
int main(void)
{
    unsigned char val=0;
    val = CONCAT(5,0xc);
    printf("%x\n", val);
 
    return 0;
}

Output:

Success #stdin #stdout 0.01s 5460KB 
5c

You can use bit-fields and union to do the same thing. If you are using C11 with anonymous union, then one might do it more easily like,

#include <stdio.h>
#include <inttypes.h>

union two_nibbles {
	uint8_t byte;
	struct { uint8_t nibble2 : 4, nibble1 : 4; };
};

int main(void) {
	const union two_nibbles n = { .nibble1 = 0x05, .nibble2 = 0x0C };
	printf("byte 0x%" PRIX8 "\n", n.byte);
}

This is very nice, but less well-defined. The only guaranteed bit-field mandated by the standard is int. "An implementation may allocate any addressable storage unit large enough to hold a bit-field." (C99 §6.7.2.1) The nibbles might conceivably might be in the wrong order, I think. On the other hand, shifting is well defined, but inconvenient and hard to read at times.