Bit Manipulation in Integer Array

  #define SET_BIT(byte, bit) (byte |= (1 << bit))
  #define CLEAR_BIT(byte,bit) (byte &= ~(1 << bit))
    uint8_t data [5];
    for (int i = 0; i < 5; ++i)
    {
        for(int j = 7; j >= 0; --j)
        {
            if (some condition)          
        -->    CLEAR_BIT(data[i],j);
            else if (some condition)  
        -->    SET_BIT(data[i],j); 
        }
}

I want to understand how is bit manipulation taking place in arrowed lines?

When I declare uint8_t data [5]; does it mean... An array of name data and I can store 5 uint8_t value(char basically) in it and in the location of each array index there will be 8 bits.Like This

Solution

Because you use define the line SET_BIT(data[i],j) is being replaces during pre-processing with data[i] |= (1 << j) (consider using inline instead and have the benefit of type safety).

The best way to understand is to make an example of a simple case (avoid these nested loops).

(lets assume you're data array is initialized to 0, which is not currently the case)

for instance: i = 2, j = 3:

1 << 3 = 00001000 shift the value 1 three times to the left , data[2] = 00000000

|= is a bit-wise operator, there for - a logical OR between each corresponding bits will take place and data[2] will be equal to 00001000

now, for i = 2, j = 6:

1 << 6 = 01000000, data[2] = 00001000

A bit-wise OR will take place and yield 1 for bits 3,6.

data[2] will be equal to 01001000.

With this example you can understand the more complex example.