Eric Niebler writes:
On 6/16/2010 3:04 AM, Alex Dubov wrote:
Greetings.
However, if functor is wrapped with assertion, this won't work:
sregex r = (expr)[check(do_something(p_string, _))];
Compilation will fail, because do_something has no method which
accepts xpressive::placeholder<> objects.
It should if it is a lazy function. How is do_something defined? Can you
send a self-contained program that demonstrates the problem you're having?
I'm using boost-1.41 and gcc-4.4.2. The error I'm receiving compiling an
example:
--------------------------------------------------------------------
testreg1.cpp:51: instantiated from here
/usr/include/boost/proto/context/default.hpp:426: error: no match for call to
‘(const check_a_impl) (const boost::xpressive::detail::action_arg >&, const int&)’
testreg1.cpp:24: note: candidates are: bool check_a_impl::operator()(int&, int)
const
Here goes a problematic example itself:
--------------------------------------------------------------------
#include <iostream>
#include
#include
using namespace std;
using namespace boost::xpressive;
const placeholder<int> _cnt = {{}};
struct count_a_impl {
typedef void result_type;
void operator()(int &cnt) const
{
++cnt;
}
};
function::type const count_a = {{}};
struct check_a_impl {
typedef bool result_type;
bool operator()(int &cnt, int val) const
{
++cnt;
return (cnt == val);
}
};
function::type const check_a = {{}};
int main(int argc, char **argv)
{
int cnt = 0;
string a_str("a_aaaaa___a_aa_aaa_");
const sregex expr1(*(as_xpr('a')[count_a(_cnt)] | _));
const sregex expr2(*(as_xpr('a')[check(check_a(_cnt, 5))] | _));
/* ^ this check() causes a problem */
sregex_iterator iter1(a_str.begin(), a_str.end(), expr1,
let(_cnt = cnt));
cout << "expr1 matched: " << iter1->str(0) << " counted " << cnt
<< endl;
cnt = 0;
sregex_iterator iter2(a_str.begin(), a_str.end(), expr2,
let(_cnt = cnt));
cout << "expr2 matched: " << iter2->str(0) << " counted " << cnt
<< endl;
return 0;
}
