C gurus, how to understand these expressions?

Code:

---

      if ((fp->flag&(_READ|_EOF_ERR)) != _READ)
          return EOF;

---

Code:

---

      { 0, (char *) 0, (char *) 0, _WRITE, | _UNBUF, 2 }

---

plz pay attention to the part bolded and colored.

The above is excerpted from the below.
Hope this helps.

Code:

---

Once the buffer is established, _fillbuf calls read to fill it,
sets the count and pointers, and returns the character
at the beginning of the buffer. Subsequent calls to _fillbuf
will find a buffer allocated.

  #include "syscalls.h"

  /* _fillbuf:  allocate and fill input buffer */
  int _fillbuf(FILE *fp)
  {
      int bufsize;

      if ((fp->flag&(_READ|_EOF_ERR)) != _READ)
          return EOF;
      bufsize = (fp->flag & _UNBUF) ? 1 : BUFSIZ;
      if (fp->base == NULL)    /* no buffer yet */
          if ((fp->base = (char *) malloc(bufsize)) == NULL)
              return EOF;      /* can't get buffer */
      fp->ptr = fp->base;
      fp->cnt = read(fp->fd, fp->ptr, bufsize);
      if (--fp->cnt < 0) {
          if (fp->cnt == -1)
              fp->flag |= _EOF;
          else
              fp->flag |= _ERR;
          fp->cnt = 0;
          return EOF;
      }
      return (unsigned char) *fp->ptr++;
  }

The only remaining loose end is how everything gets
started. The array _iob must be defined and initialized
for stdin, stdout and stderr:

  FILE _iob[OPEN_MAX] = {    /* stdin, stdout, stderr */
      { 0, (char *) 0, (char *) 0, _READ, 0 },
      { 0, (char *) 0, (char *) 0, _WRITE, 1 },
      { 0, (char *) 0, (char *) 0, _WRITE, | _UNBUF, 2 }
  };

---

I believe the second bolded spot you have is a typo. You seem to have added a comma after _WRITE, which shouldn't be there. Try and compile that and you'll find out what I mean.

What you're seeing here is the bitwise OR operator. For a full explanation, consider reading this FAQ entry. Basicly it's OR-ing two values on a bit by bit basis, giving you a resulting value which has all set bits of both values combine.

Quzah.

basically _WRITE and _UNBUF are, more than likely, macros #defined somewhere. whatever their values are, they are bitwise OR'd together; _WRITE | _UNBUF.

if _WRITE = 00000001
and _UNBUF = 00000010
then (to my knowledge) ORing them will result
00000011

Read up on bitwise operations.

thanks for replys from you both.

I know some about bitwise, and here what that confuses me is whether a "|" has been missed by typo.

and yes, they've been defined before being used.

Code:

---

  enum _flags {
      _READ  = 01,  /* file open for reading */
      _WRITE  = 02,  /* file open for writing */
      _UNBUF  = 04,  /* file is unbuffered */
      _EOF    = 010,  /* EOF has occurred on this file */
      _ERR    = 020  /* error occurred on this file */
  };

---

so I'm guessing:

Code:

---

if ((fp->flag&(_READ|_EOF_ERR)) != _READ)

---

is:

Code:

---

if ((fp->flag&(_READ|_EOF|_ERR)) != _READ)

---

and:

Code:

---

{ 0, (char *) 0, (char *) 0, _WRITE, | _UNBUF, 2 }

---

is:

Code:

---

{ 0, (char *) 0, (char *) 0, _WRITE | _UNBUF, 2 }

---

Do you agree? Or you have the same book, K&R - The C programming language, please refers to chapter 8.5-Examples to see whether they are same as that in my book.

...

Im trying to figure out what they exactly mean.

1) (_READ|_EOF|_ERR) set corresponding bits to 1;
fp->flag AND with it, to see whether result is _READ, namely _EOF and _ERR are gone.

2) like what quazh and sand_man said.

sigh! so hard!

They look like typos to me but you can only guess.

if it compiles then i would assume _EOF_ERR == 0011
given _EOF = 0010 //made up value
and _ERR = 0001 //made up value

i don;t know where _EOF_ERR would be defined though