Uwe Stöhr <[EMAIL PROTECTED]> writes:
+void delete_opt(vector<string> & opts, char const * const * what)
+{
+ if (opts.empty())
+ return;
+
+ // remove found options from the list
+ // do this after handle_opt to avoid potential memory leaks and to be
able
+ // to find in every case the last language option
+ vector<string>::iterator it;
+ for (; *what; ++what) {
+ it = find(opts.begin(), opts.end(), *what);
+ if (it != opts.end())
+ opts.erase(it);
+ }
+}
Can't you use std::remove multiple times and then perform only one erase?
Should be much more efficient, no?
See http://www.sgi.com/tech/stl/remove.html (it doesn't actually remove
anything, just moves those iterators matching the predicate to the end...)
typedef vector<string>::iterator iterator;
iterator opts_end = opts.end();
iterator const opts_begin = opts.begin();
for (; *what; ++what)
{
opts_end = std::remove(opts_begin, opts_end, *what);
if (opts_end == opts_begin)
break;
}
std::erase(opts_end, opts.end();
Better still, use a predicate that itself loops over the what. Why are you
using a c-string anyway? It's been a long time (and I get lost with pointers
to pointers), but I think that the code below, or something like it should do
the trick. Minimal re-working of your opts vector...
struct ContainsCharInCString
{
private char const * const * cstring;
ContainsCharInCString(char const * const * data) : this.cstring(data) {}
bool operator(char input)
{
for (char const * const * ptr = cstring; *ptr; ++ptr)
{
if (input == *ptr)
return true;
}
return false;
}
}
iterator new_end = std::remove_if(
opts.begin(), opts.end(), ContainsCharInCString(what));
std::erase(new_end, opts.end();
Angus
