Re: String parsing question
From: Dan Pop (Dan.Pop_at_cern.ch)
Date: 10/14/03
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Date: 14 Oct 2003 15:29:34 GMT
In <bmh0cj$t31$1@chessie.cirr.com> Christopher Benson-Manica <ataru@nospam.cyberspace.org> writes:
>I'm wondering about the best way to do the following:
>
>I have a string delimited by semicolons. The items delimited may be in any of
>the following formats:
>1) 14 alphanum characters
>2) 5 alphanums space 8 alphanums
>3) 6 alphanums colon 8 alphanums
>4) 5 alphanums colon 8 alphanums
>
>My task is to convert items in the third format to the first format, and items
>in the fourth format to the second. Also, I need to count the number of items
>in the string, which may or may not have a trailing semicolon.
>
>My plan (which I feel is sub-optimal - hence this post), is to step through
>the initial string one character at a time to accomplish these things in one
>pass. While I could count semicolons easily with strchr(), deleting the
>colons properly means stepping through the whole string anyway (right?) and so
>I may as well count semicolons simultaneously. I'd also like to validate the
>data format (i.e., 15-character items are not allowed).
>
>int myfunc( const char *list )
>{
> int items=0;
> char *cp=strdup( idlist ); /* nonstandard */
> char *newstr=cp;
> int shifts=0;
> int chars=0;
>
> for( ; *cp ; *cp++ ) {
> if( *cp == ':' ) {
> if( chars == 6 ) {
> shifts++;
> continue;
> }
> if( chars == 5 ) {
> *(cp-shifts)=' ';
> chars++;
> continue;
> }
> return( -1 ); /* error */
> }
> if( *cp == ';' ) {
> items++;
> if( chars != 14 ) {
> return( -1 ); /* error */
> }
> chars=0;
> }
> else if( ++chars > 14 ) {
> return( -1 ); /* error */
> }
> *(cp-shifts)=*cp;
> }
> *(cp-shifts)='\0';
> if( chars == 14 ) {
> items++;
> }
> if( !items || (chars && chars != 14) ) {
> return( -1 ); /* error */
> }
> printf( "The string '%s' has %d items.", newstr, items );
> free( newstr );
> return( 0 ); /* success */
>}
>
>Is there a better way?
1. Such a code is a maintenance nightmare (imagine that you'll have to
make some changes, 5 years from now).
2. I may be missing something, but I can't find any attempt to test that
your characters really are alphanums, you're merely looking for your
separators.
I would implement this function using sscanf calls. The result would be
slower, but a lot more readable. The conversion specifier for
alphanumerics can use the following macro:
#define ALNUM "[abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789]"
Dan
-- Dan Pop DESY Zeuthen, RZ group Email: Dan.Pop@ifh.de
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