Question

I have another late night brain-dead question. It’s probably simple to do but you know how it is after 8 hours+ of programming then reaching then end of a deadline. 🙂

Here is my question..

I have a boolean array of length 10 which is logically divided into seperate pieces to hold 4 different integer bit arrays (as booleans). Here is the target array:

bool[] myArray = new bool[10]; 

And here are the 4 integers I would like to insert:

int value1 = 3;  // 2 bits, myArray[0-1], 11
int value2 = 12; // 4 bits, myArray[2-5], 1100
int value3 = 2;  // 2 bits, myArray[6-7], 10
int value4 = 1;  // 2 bits, myArray[8-9], 01

myArray should end up looking like the following (note that the first element is position 0):

{(T,T),(T,T,F,F),(T,F),(F,T)}

Then ultimately what I want to do is convert myArray to an int value:

0x1111001001 = 969

Perhaps there is an even better way of doing this without having to use booleans? So let me rephrase my question in a more general sense:

How do I concatenate N int values into a target int?

Thanks!

Answer 1

Normally, the common way to combine bits from several numbers into one is this (using your values and bitlengths as examples):

var result = value1 | (value2 << 2) | (value3 << 6) | (value4 << 8);

However, the number you printed at the end of your question has the numbers back to front. If that’s what you wanted, it’d be:

var result = value4 | (value3 << 2) | (value2 << 4) | (value1 << 8);

Of course this assumes that the code knows the desired bitlength of each value. If you don’t know the bitlength at compile-time, then you have to keep track of it at runtime, otherwise a single int like 1 won’t tell your code how many bits to use from it.