Question

#include <stdio.h>
typedef struct _octet
{
char tft_op:3;
char e_bit:1;
char no_of_pkt:4;
}octet_t;

int main()
{
octet_t st_octet;
st_octet.tft_op = 0x8;
st_octet.e_bit = 0x1;
st_octet.no_of_pkt = 0x10;
char *ptr = (char*)&st_octet;
printf("%#x\n",*ptr);
return 0;
}

here i want to store 3 bits in tft_op, 1 bit e_bit, 4 bits in no_of_pkt, but i want the o/p as a combined 1 byte i.e is want the o/p as FA but the o/p i get is 0x8 what should i do to get the o/p as FA

Answer 1

Your values are too big for the allocated space:

0x8 = 1000b - four bits, not three
0x1 = 1b - but if char is signed you can only have 0 and -1!
0x10 = 16 = 10000b - five bits, you'be allocated four

So, the effective result is:

(0x8 | (0x1 << 3) | (0x10 << 4)) & 0xff

=> (0x8 | 0x8 | 0x100) & 0xff

=> 0x8

which is what you’re getting. 0xfa is:

11111010b

but you assume that the first item in the structure is at bit position 7 (the most significant) but the standard (as far as I know) doesn’t specify where the bits should start (it could be bit0, the least significant) and looking at your output I guess the first field is the least significant bit, so you have:

ccccbaaa

where a is tft_op, b is e_bit and c is no_of_pkt. To get 0xfa it’s either:

tft_op = 2
e_bit = 1
no_of_pkt = 15

or, if the order is the other way round:

tft_op = 7
e_bit = 1
no_of_pkt = 10

Also, rather than doing char *ptr = (char*)&st_octet; look into using a union:

typedef union
{
  struct _octet
  {
    char tft_op:3;
    char e_bit:1;
    char no_of_pkt:4;
  };
  char data;
} octet_t;