In message <[EMAIL PROTECTED]> Ben Smithurst writes:
: while (path[strlen(path) - 1] == '/')
: path[strlen(path) - 1] = 0;
:
: :-) Preferably '\0' too of course like you say later.
Yes. Actually, I'd do this like:
cp = path + strlen(path) - 1;
while (cp > path && *cp == '/')
*cp-- = '\0';
Since it doesn't have the buffer overflow the above code does in the
'////' case. It also gets that case right (if my mental walkthrough
can be truested). A nice side effect is that it doesn't call strlen N
times making it a O(N) algorythm, rather than an O(N^2) algorythm.
Likely only of secondary importance since most strings don't contain
lots of // in them :-).
Warner
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