Hello Georg,
On 05/12/13 22:22, Georg Sauthoff wrote:
> Hi,
>
> the strtok man-page does not specify what happens if `str` does not
> contain any `delim`iters.
>
> If you don't already know strtok one might assume then either:
>
> - the delimiter-less string is considered the next token and thus
> returned upon the first call to strtok
> - on the first call NULL is returned because there are not any tokens
>
> Related issue: is the non-empty `str`-suffix after the last `delim`
> returned as a token or not?
Thanks for the report, though I must say that it would have been a
little more helpful if you had answered the rhetorical questions you
ask... (It would have aided me as I reminded myself of the details.)
> I know how strtok works -
In which case a patch would have made the bug report even better still!
> but perhaps it makes sense to explicitly
> address these issues in the Linux man page.
Anyway, I agree that the page could be clearer. I've added the following text:
A sequence of calls to strtok() that operate on the same string
maintains a pointer that determines the point from which to
start searching for the next token. The first call to strtok()
sets this pointer to point to the first byte of the string.
The start of the next token is determined by scanning forward
for the next nondelimiter byte in str. If such a byte is
found, it is taken as the start of the next token. If no such
byte is found, then there are no more tokens, and strtok()
returns NULL. (A string that is empty or that contains only
delimiter will thus cause strtok() to return NULL on the first
call.)
The end of each token is found by scanning forward until either
the next delimiter byte is found or until the terminating null
byte ('\0') is encountered. If a delimiter byte is found, it
is overwritten with a null byte to terminate the current token,
and strtok() saves a pointer to the following byte; that
pointer will be used as the starting point when searching for
the next token. In this case, strtok() returns a pointer to
the start of the found token.
From the above description, it follows thatt a sequence of two
or more contiguous delimiter bytes in the parsed string is con‐
sidered to be a single delimiter, and that delimiter bytes at
the start or end of the string are ignored. Put another way:
the tokens returned by strtok() are always nonempty strings.
Thus, for example, given the string "aaa;;bbb,", successive
calls to strtok() that specify the delimiter string ";," would
return the strings "aaa" and "bbb", and then a NULL pointer.
Sufficient?
Cheers,
Michael
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