I already did... It's just a special kind of struct where instead of all the variables being separate, they lay on top of one another, literally occupying the same address space.Originally Posted by lovesunset21
testvar.MyByte is 1 byte long at the same address as the unnamed char with the 8 bits.... Change one, you change the other.
actually, when I try this:
It is 1, not 4. It should be 4, right?testvar.b2 = 1;
printf("We got %d",testvar.MyByte); // = 4
Well the entire idea of using bit-fields to pull the bits out of a byte is broken to begin with.
There is NOTHING in the standard which says that your b1 (for example) should correspond to say the lsb of a byte.
If you want to be sure of getting the lsb, then you have to use something like
byte & 0x01
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Did you create the union exactly as I showed you?
It should work....
Yes, of course, I copy and paste
Code:#include <stdio.h> #include <stdlib.h> typedef union t_byte_s { unsigned char MyByte; unsigned char b0:1, b1:1, b2:1, b3:1, b4:1, b5:1, b6:1, b7:1; } byte_s; int main(void) { byte_s testvar; testvar.MyByte = 127; if (testvar.b6) puts ("eyup, it's a 1"); testvar.MyByte = 0; // clear the bits testvar.b2 = 1; printf("We got %d",testvar.MyByte); // = 4 return EXIT_SUCCESS; }
Hmmm... if you're doing b3 you should get 8.
It works here... (on PellesC).
That's right. It should be 8, but it is alway 1. I am running it on Eclipse and Ubuntu
Ok... now something of a mystery... I copied and pasted your code... and I get 1 all the time...
WEIRD!
Maybe Salem is right...
What should we add into the code as Salem told? I am a dummy C. Therefore, I don't really understand his idea. Thanks
Each member of a union shares the same space. So it stands to reason that if you have multiple bit fields directly in the union, they represent separate members and thus refer to the same spot in memory. You can get a more meaningful result by wrapping the bit fields in a separate structure such that the bits are represented by a single member in the union:
Code:#include <stdio.h> struct bit_set { unsigned b0:1; unsigned b1:1; unsigned b2:1; unsigned b3:1; unsigned b4:1; unsigned b5:1; unsigned b6:1; unsigned b7:1; }; union byte { unsigned char source; struct bit_set bits; }; int main(void) { union byte testvar; testvar.source = 'A'; printf("%d%d%d%d %d%d%d%d\n", testvar.bits.b7, testvar.bits.b6, testvar.bits.b5, testvar.bits.b4, testvar.bits.b3, testvar.bits.b2, testvar.bits.b1, testvar.bits.b0); return 0; }
My best code is written with the delete key.
Alrighty.... give this a try....
The problem appears to be that without the nested struct, the union was laying all the bits on top of each othe as well....Code:#include <stdio.h> #include <stdlib.h> typedef union t_var { unsigned char MyByte; struct t_bits { unsigned char b0 : 1, b1 : 1, b2 : 1, b3 : 1, b4 : 1, b5 : 1, b6 : 1, b7 : 1; } bits; } tvar; int main(void) { tvar testvar; testvar.MyByte = 56; printf("%d%d%d%d %d%d%d%d = %d \n", testvar.bits.b7,testvar.bits.b6,testvar.bits.b5,testvar.bits.b4, testvar.bits.b3,testvar.bits.b2,testvar.bits.b1,testvar.bits.b0, testvar.MyByte); // output : 0011 1000 = 56 testvar.MyByte = 0; // clear the bits testvar.bits.b2 = 1; printf("We got %d\n\n",testvar.MyByte); // = 4 testvar.bits.b4 = 1; printf("We got %d\n\n",testvar.MyByte); // = 20 return EXIT_SUCCESS; }
Last edited by CommonTater; 10-30-2010 at 04:54 PM.
LOL... great timing... I was working on that very thing as you posted this.
One might reasonably expect that coming from a single char as the root variable, they wouldn't pile up like that.
Doesn't explain how I had it working here...
Live and Learn.
Last edited by CommonTater; 10-30-2010 at 05:00 PM.
Well... been playing with this a little more and WOW what a learning experience.
Only problem is that now I've got to go revise a bunch of code where I'm using bitfields incorrectly... Bummer.
