Easier code in 8.5 K&R book, only logical operator instead bit operation and logical operator

Question

In section 8.5, in fopen example, K&R write:

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

But I see this easier:

if (fp->flag != _READ)

Is my code bad or K&R is simple obscure... Thanks

Possible duplicate of [k&r fopen and fillbuf](https://stackoverflow.com/questions/50598201/kr-fopen-and-fillbuf) — tbejos, Jul 22 '19 at 18:46

score 2 · Answer 1 · answered Jul 22 '19 at 22:14

The bits of the flag member are assigned more meanings than just _READ (01), _EOF (010, which is octal for 8), and _ERR (020, 16). There are also _WRITE (02) and _UNBUF (04).

Kernighan and Ritchie’s code, (fp->flag & (_READ | _EOF | _ERR )) != _READ, is true if and only if _READ is clear, _EOF is set, or _ERR is set.

Your code, fp->flag != _READ, is true if and only if _READ is clear, _EOF is set, _ERR is set, _WRITE is set, or _UNBUF is set.

Thus, for example, if _UNBUF is set while _READ is set, _EOF is clear, and _ERR is clear, the Kernighan and Ritchie code will evaluate to false but your code will evaluate to true. The Kernighan and Ritchie code asks only about three flags and does not care about the other flags. Your code is affected by all the flags.

Thanks Eric, now I understand. – llantones Jul 24 '19 at 14:54 — llantones, Jul 24 '19 at 14:54

Easier code in 8.5 K&R book, only logical operator instead bit operation and logical operator

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