Is assignment of bit field with different Endianness implementation specific

Question

As we know, we have two types of Endianness: big endian and little endian.

Let's say that an integer takes 4 bytes, so the layout of the integer 1 should be 0x01 0x00 0x00 0x00 for little endian and 0x00 0x00 0x00 0x01 for big endian.

To check if a machine is little endian or big endian, we can code as below:

int main()
{
    int a = 1;
    char *p = (char *)&a;
    // *p == 1 means little endian, otherwise, big endian

    return 0;
}

As my understanding, *p is assigned with the first octet: 0x01 for little endian and 0x00 for big endian (the two bold parts above), that's how the code works.

Now I don't quite understand how bit field works with different Endianness.

Let's say we have such a struct:

typedef struct {
    unsigned char a1 : 1;
    unsigned char a2 : 1;
    unsigned char a6 : 3;
}Bbit;

And we do the assignment as below:

Bbit bit;
bit.a1 = 1;
bit.a2 = 1;

Will this piece of code be implementation specific? I'm asking if the values of bit.a1 and of bit.a2 are 1 on little endian and are 0 on big endian? Or are they definitely 1 regardless of the different Endianness?

Answer 1

Let's say we have a struct:

typedef struct {
    unsigned char a1 : 1;
    unsigned char a2 : 1;
    unsigned char a6 : 3;
}Bbit;

and a definition:

Bbit byte;

Suppose byte is stored in a single byte and is currently zeroed out: 0000 0000.

byte.a1 = 1;

This sets the bit called a1 to 1. If a1 is the first bit, then byte has become 1000 0000, but if a1 is the fifth bit, then byte has become 0000 1000, and if a1 is the eighth bit, then byte has become 0000 0001.

byte.a2 = 1;

This sets the bit called a2 to 1. If a2 is the second bit, then byte has (likely) become 1100 0000, but if a2 is the sixth bit, then byte has (likely) become 0000 1100, and if a2 is the seventh bit, then byte has become 0000 0011. (These are only "likely" because there is no guarantee that the bits follow some reasonable order. It's just unlikely that a compiler will go out of its way to mess up this example.)

Endianness is not a factor when it comes to the values that are stored. Only the bits representing the specified bit-field are changed with each assignment, and the value being assigned is reduced to that number of bits (with implementation-defined behavior if that value is too big for that number of bits).