need help in bit field

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10-29-2010
lovesunset21

need help in bit field

Hi everyone, I have a program with bit field. However, it displays error when I try to compile. Please show me why it displays error. Thank you so much.

#include <stdio.h>
#include <stdlib.h>
typedef struct
{
unsigned b1:1;
unsigned b2:1;
unsigned b3:1;
unsigned b4:1;
unsigned b5:1;
unsigned b6:1;
unsigned b7:1;
unsigned b8:1;
} byte_s;

int main(void) {
byte_s bfbyte;
unsigned char a[]={0x11,0xff};

bfbyte= (byte_s) a[0]; //displays error: conversion to non-scalar type requested

printf ("%d %d %d %d : %d %d %d %d \n",
bfbyte.b1, bfbyte.b2, bfbyte.b3, bfbyte.b4,
bfbyte.b5, bfbyte.b6, bfbyte.b7, bfbyte.b8);

printf ("%d %d %d %d : %d %d %d %d \n",
bfbyte.b1, bfbyte.b2, bfbyte.b3, bfbyte.b4,
bfbyte.b5, bfbyte.b6, bfbyte.b7, bfbyte.b8);

return EXIT_SUCCESS;
}
10-29-2010
Salem

Guess what you've ignored 11 times now?

And no, that isn't binary, it's DECIMAL
http://cboard.cprogramming.com/c-pro...ead-first.html
10-29-2010
CommonTater

Quote:

Originally Posted by lovesunset21

Hi everyone, I have a program with bit field. However, it displays error when I try to compile. Please show me why it displays error. Thank you so much.

I take it you are trying to use a struct to extract the bits from a byte?

You need to use a union, not a struct.
Also your bitfield definition is wrong.

Check your docs, try again... and this time please post your code inside the appropriate tags.
10-30-2010
lovesunset21

Thanks for your quick reply. I am working with memcopy function, but it displays a warning "implicit declaration of function ‘memcopy’". I don't know why
10-30-2010
GReaper

Have you included "memory.h"?
10-30-2010
R.Stiltskin

The function is memcpy and it's in string.h.
10-30-2010
GReaper

Quote:

Originally Posted by R.Stiltskin

The function is memcpy and it's in string.h.

"memory.h" maps to "string.h" ...
10-30-2010
laserlight

Quote:

Originally Posted by Sipher

"memory.h" maps to "string.h" ...

That may be true, but it is non-standard. Furthermore, unless it is guaranteed that that header will always include <string.h>, you should also include <string.h> for memcpy.
10-30-2010
GReaper

Oh, ok. I didn't know that.
10-30-2010
lovesunset21

Thanks a lot. I have solved my problem. However, I still have question, please help me.

I am working with bit field. I have

Code:

typedef struct bytes{ unsigned char b8:1; unsigned char b7:1; unsigned char b6:1; unsigned char b5:1; unsigned char b4:1; unsigned char b3:1; unsigned char b2:1; unsigned char b1:1; } bytes;

Then I have an array of bytes

bytes s[10];

Then I want to copy 0x00 to s[0]. I did as followed:

Code:

unsigned char *m; *m= 0x00 s[0]= *((bytes*)(m));

I don't know why it displays segmentation fault. Please help me. Thank you so much.
10-30-2010
CommonTater

Quote:

Originally Posted by lovesunset21

Thanks for your quick reply. I am working with memcopy function, but it displays a warning "implicit declaration of function ‘memcopy’". I don't know why

Why would you need memcpy?

First you've got to straighten out your bitfield definition...

Code:

typedef struct { unsigned b1:1; unsigned b2:1; unsigned b3:1; unsigned b4:1; unsigned b5:1; unsigned b6:1; unsigned b7:1; unsigned b8:1; } byte_s;

Your struct is going to be 32 bytes long.

Try it like this...

Code:

typedef struct t_byte_s { unsigned char b0:1, b1:1, b2:1, b3:1, b4:1, b5:1, b6:1, b7:1; } byte_s;

This will be only 1 byte long (although not all compilers will do this for a byte)

Now you need some means of putting a value in there in one form and getting it out in another. For that we have to use a union...

Code:

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;

Ok, now you can do this...

Code:

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

You can feed in individual bits and read the byte or vice versa, hand it a byte and read the bits.
The reason this works is that in a union all variables occupy the same space (i.e. start at the same address).
10-30-2010
lovesunset21

Thanks so much. But i don't really understand.

Code:

typedef struct { unsigned b1:1; unsigned b2:1; unsigned b3:1; unsigned b4:1; unsigned b5:1; unsigned b6:1; unsigned b7:1; unsigned b8:1; } byte_s; Your struct is going to be 32 bytes long.

I thought that it should be 8 bits long.

And in case I want to copy its value to a element of an array. What should I do? I face error segmentation fault
10-30-2010
CommonTater

Quote:

Originally Posted by lovesunset21

Thanks so much. But i don't really understand.

Code:

typedef struct { unsigned b1:1; unsigned b2:1; unsigned b3:1; unsigned b4:1; unsigned b5:1; unsigned b6:1; unsigned b7:1; unsigned b8:1; } byte_s; Your struct is going to be 32 bytes long.

I thought that it should be 8 bits long.

And in case I want to copy its value to a element of an array. What should I do? I face error segmentation fault

You are using "unsigned" which is an int (32bits, 4 bytes on most systems)
See the semicolon at the end of each line?
That causes each bit to occupy a separate 4 byte value... 4 * 8 = 32 bytes.

Note that I specified unsigned char ... on most systems that is 8bits.
By terminating each line with a coma, except the last one, I've told it to put all 8 values in the same unsigned char... thus, only 1 byte.
10-30-2010
CommonTater

Quote:

Originally Posted by lovesunset21

Thanks so much. But i don't really understand.

Code:

typedef struct { unsigned b1:1; unsigned b2:1; unsigned b3:1; unsigned b4:1; unsigned b5:1; unsigned b6:1; unsigned b7:1; unsigned b8:1; } byte_s; Your struct is going to be 32 bytes long.

I thought that it should be 8 bits long.

And in case I want to copy its value to a element of an array. What should I do? I face error segmentation fault

Take a look at the way I showed you to do it....

If you want to copy each bit to an array element you will have to do it manually...

Code:

Int bitarray[8]; // array for bits testvar.MyByte = x; // stuff in a value // copy out the bits bitarray[0] = testvar.b0; bitarray[1] = testvar.b1; ... bitarray[7] = testvar.b7; // or to get them as ascii 1 and 0 for printing char txtarray[8]; testvar.MyByte = x; // value in // ascii out txtarray[0] = testvar.b0 + '0'; ... txtarray[7] = textvar.b7 + '0'; //or to be extra sneaky sprintf(txtarray,"%d%d%d%d%d%d%d%d", testvar.b0,testvar.b1,testvar.b2,testvar.b3,testvar.b4,testvar.b5,testvar.b6,testvar.b7);
10-30-2010
lovesunset21

Thanks so much. Can you explain the UNION keyword for me?
10-30-2010
CommonTater

Quote:

Originally Posted by lovesunset21

Thanks so much. Can you explain the UNION keyword for me?

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.

testvar.MyByte is 1 byte long at the same address as the unnamed char with the 8 bits.... Change one, you change the other.
10-30-2010
lovesunset21

actually, when I try this:

Quote:

testvar.b2 = 1;
printf("We got %d",testvar.MyByte); // = 4

It is 1, not 4. It should be 4, right?
10-30-2010
Salem

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
10-30-2010
CommonTater

Quote:

Originally Posted by lovesunset21

actually, when I try this:
It is 1, not 4. It should be 4, right?

Did you create the union exactly as I showed you?
It should work....
10-30-2010
lovesunset21

Quote:

Originally Posted by CommonTater

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; }
10-30-2010
CommonTater

Hmmm... if you're doing b3 you should get 8.

It works here... (on PellesC).
10-30-2010
lovesunset21

Quote:

Originally Posted by CommonTater

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
10-30-2010
CommonTater

Quote:

Originally Posted by lovesunset21

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...
10-30-2010
lovesunset21

What should we add into the code as Salem told? I am a dummy C. Therefore, I don't really understand his idea. Thanks
10-30-2010
Prelude

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; }
10-30-2010
CommonTater

Alrighty.... give this a try....

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; }

The problem appears to be that without the nested struct, the union was laying all the bits on top of each othe as well....
10-30-2010
CommonTater

Quote:

Originally Posted by Prelude

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:

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.
10-31-2010
CommonTater

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.