--- In c-prog@yahoogroups.com, "py2akv" <g...@...> wrote:
>
> John,
>
> Very good
Thanks!
> but you worked with an array of char's & a pointer to char's: two strings &
> one a copy of the other, a sort of a crutch, he, he, he.
Eh? That's just one string (str) and one pointer to char (s). I could always
rewrite
const char str[] = "string";
as:
const char *const str = "string";
but then you have an anonymous array of chars ("string") AND a pointer to char
(str).
> Came to a different solution using only the original string & the
> fundamentals of the language.
With better layout:
> char *s=" C Programming Language ";
Should be const char *
> int f1=1, f2=1;
>
> while (*s) {
> if (*s!=' ') {
> if (f1)
> printf("%c", *s);
> f1=0;
Mistake - the f1=0 should be conditional on the if?
(more obvious when code is laid out correctly)
> }
> else if (!f1) {
> printf("%c", *s);
> f2=0;
> }
> if (!f1 & !f2)
> f1=f2=1;
> ++s;
> }
My code was 3 lines; yours is 11 (not counting brace-only lines). Do you get
paid by the line? :-)
> > for (n = 0, s = str; *s; s++)
> > if ((*s != ' ') && ((s == str) || (s[-1] == ' ')))
> > printf(" %c" + !n++, *s);
> The backspace at the end is optional,
Sorry - don't understand what you mean by this.
