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

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

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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);    

bools 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);


The bits of myBitfieldofmyArray are 10111010

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