## Set Last Element of boolean array to as Most Significant Bit within an Bitfield

let's suppose I have an Boolean array with 7 elements like

boolean myArray = {1,0,1,1,1,0,1}

this array I want to represent within an bitfield:

uint8 myBitfieldofmyArray;

It is important that the last element of array is set as the Most Significant Bit in *myBitfieldofmyArray*.
The Least Significant Bit (Bit 0) is not considered and is by default always 0.

With the following solution the 1.element is set as the Most Significant Bit:

uint8_t myBitfieldofmyArray= 0; for (int i = 0; i < 7; i++) { myBitfieldofmyArray|= myArray[i] ? 1 : 0; myBitfieldofmyArray<<= 1; }

How to adapt it so that the last element is set as the Most Significant Bit?

The Solution above was taken from: Pushing boolean values of an array to uint8 bitfield

You can use a "reverse loop":

#include <stdio.h> #include <stdint.h> #include <stdbool.h> int main() { bool myArray[7] = {1, 0, 0, 1, 1, 0, 0}; uint8_t myBitfieldofmyArray= 0; for (int i = 6; i >= 0; --i) { myBitfieldofmyArray |= myArray[i]; myBitfieldofmyArray <<= 1; } printf("%d\n", myBitfieldofmyArray); }

`bool`

s evaluate to either 0 or 1, so I think the ternary `? 1 : 0`

is not necessary.

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Well, if you want to store your bitfield starting from the MSB you can change your loop to:

for(int i=0; i<7; ++i) myBitfieldofmyArray |= (myArray[i]<<(7-i))

That being said, are you sure you want to need bitfields and not int/bool arrays?.

If you have 256 field storing them in int/bool array will take 1KB, and 32B with a bitfield.

So you gain less than 1KB by using a bitfield but you lose compactness, your code gets more complex and each read/write requires extra computation to get the right bit to set/clear so you lose performances (you can check the assembly code).

Thus, unless you have good reasons to use a bitfields (for exemple sending it through the network) you should definitely avoid them.

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A simple way is to use two counters to loop shifting the bits from the array starting at `array[0]`

to the bit in your `uint8`

starting with the MSB and incrementing the array index while decrementing the bit position, e.g.

uint8_t myBitfieldofmyArray = 0, i = 7, j = 0; while (i--) myBitfieldofmyArray |= myArray[j++] << (i + 1);

**Result**

The bits of `myBitfieldofmyArray`

are `10111010`

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