a method to do bit-operation in C - c

Question

I'm trying to define a structure that can allow to set a byte value directly, and also allow to manipulate the bits of the byte without using function like bit_set(), bit_clear() etc,.
Here's my definition
typedef union FLAG_WORK {
volatile unsigned char BYTE;
struct {
volatile unsigned char bit0:1;
volatile unsigned char bit1:1;
volatile unsigned char bit2:1;
volatile unsigned char bit3:1;
volatile unsigned char bit4:1;
volatile unsigned char bit5:1;
volatile unsigned char bit6:1;
volatile unsigned char bit7:1;
}BIT;
}FLAG8;
and a sample code
int main()
{
FLAG8 i;
i.BYTE=(unsigned char)0; // initial the value of i.BYTE
i.BIT.bit0 = 1; // set bit0 of i.BYTE
i.BIT.bit1 = 1;
cout << (int)i.BYTE << endl;
cout << "Hello world!" << endl;
return 0;
}
I just wonder how to modify the structure allowing me to assign value to "i" in above code directly?
any suggestion?

Answer 1

C99 allows intializing members explicitly. Assuming I understand your question correctly, you're looking for
FLAG8 i = { .BIT = { .bit2 = 1, .bit5 = 1 } };
FLAG8 j = { .BYTE = 42 };
FLAG8 k = { 42 }; // same as j as initializing the first member is default

Answer 2

As normal you can assign like But the values are overlapped, because Here union holds only 1 byte of memory
FLAG8 i = { .BIT = { .bit2 = 1, .bit5 = 1 } , .BYTE = 42};
//result of i.BYTE is 42 depens on Byte value
FLAG8 i = { .BYTE = 42 , .BIT = { .bit2 = 1, .bit5 = 1 } };
//result of i.BYTE is 36 depends on bits values
1.As #Christoph suggested You can assign value BYTE or bit by bit directly Like this
FLAG8 i = { .BYTE = 42 };
FLAG8 j = { .BIT = { .bit2 = 1, .bit5 = 1 } };
printf("%d \n\n",(int)i.BYTE );
printf("%d \n\n",(int)j.BYTE );
2.You can assign value bit by bit
i.BIT.bit0 = 1; // set bit0 of i.BYTE
i.BIT.bit1 = 1;
i.BIT.bit4 = 1;
printf("%d \n\n",(int)i.BYTE );
3.You can assign directly a BYTE with hexa decimal value
i.BYTE=0x10;
printf("%d \n\n",(int)i.BYTE );
4.You can assign directly a BYTE with decimal value
i.BYTE=100;
printf("%d \n\n",(int)j.BYTE );
5.You can assign directly with the Identifier of same type
j.BYTE=0x30;
printf("%d \n\n",(int)j.BYTE );
i=j;
printf("%d \n\n",(int)i.BYTE );
6.Assign a decimal value and check it in printing both decimal and hexa decimal format
j.BYTE=100;
printf("%d \n\n",(int)j.BYTE );
printf("%x \n\n",(int)j.BYTE );
j.BIT.bit4=0;
j.BIT.bit5=0;
printf("%d \n\n",(int)j.BYTE );
printf("%x \n\n",(int)j.BYTE );