Concatenate three 8-bit integers to one 16-bit integer in C

Question

I have three integers, one integer contains data of 7 bits, another integer contains data of 4 bits and the third integer contains data of 5 bits.

I want to concatenate all the three integers into one 16-bit integer side by side without changing the value of each of the integers.

Example:

int a = 1; //7 bit data (uint8)
int b = 2; //4 bit data (uint8)
int c = 3; //5bit data (uint8)

where, 0< a <100 , 0< b < 13, 0< c <32 The result should be as follows:

result = 123;

Example 2:

int a = 99; //7 bit data (uint8)
int b = 12; //4 bit data (uint8)
int c = 31; //5bit data (uint8)

result = 991231; //expected

How can this be achieved using bit wise operators?

Answer 1

What about something like the following? You modify it to works for three integers and for the size you want.

unsigned concatenate(unsigned x, unsigned y) {
    unsigned pow = 10;
    while(y >= pow)
        pow *= 10;
    return x * pow + y;        
}
 int main(){

    int a=1,b=2,c=3;
    printf("%d\n", concatenate(concatenate(a,b),c));
 }

credit to: How to concatenate two integers in C

If you feel brave and you are using c++ you can extend it in order to work with a variable number of arguments.

template <typename T>
inline T concatenate(T N){
     return N;
}
template <typename T,typename...NS>
T concatenate(const T N, NS&&... ns){
    const T rest = concatenate(std::forward<NS>(ns)...);
    unsigned pow = 10;
    while(rest >= pow)
        pow *= 10;
    return N * pow + rest;      
}

int main()
{
    int a=1,b=2,c=3;
    printf("%d\n", concatenate(a,b,c,987,a));
}