我有代码从std :: vector< int>中删除所有元素小于某个int限制.我写了一些部分应用lambdas的函数：
auto less_than_limit = [](int limit) {
return [=](int elem) {
return limit > elem;
};
};
auto less_than_three = less_than_limit(3);
当我用std :: vector< int>进行测试时v {1,2,3,4,5} ;,我得到了预期的结果：
for(auto e: v) {
std::cout << less_than_three(e) << " ";
}
// 1 1 0 0 0
我可以轻松删除少于三个的所有元素：
auto remove_less_than_three = std::remove_if(std::begin(v),std::end(v),less_than_three);
v.erase(remove_less_than_three,v.end());
for(auto e: v) {
std::cout << e << " ";
}
// 3 4 5
如何使用less_than_three删除大于或等于3的元素？
我尝试在std :: not1中包装less_than_three,但是出现了错误：
/usr/local/Cellar/gcc/5.3.0/include/c++/5.3.0/bits/stl_function.h:742:11: error: no type named 'argument_type' in 'struct main()::<lambda(int)>::<lambda(int)>'
class unary_negate
^
/usr/local/Cellar/gcc/5.3.0/include/c++/5.3.0/bits/stl_function.h:755:7: error: no type named 'argument_type' in 'struct main()::<lambda(int)>::<lambda(int)>'
operator()(const typename _Predicate::argument_type& __x) const
/usr/local/Cellar/gcc/5.3.0/include/c++/5.3.0/bits/predefined_ops.h:234:30: error: no match for call to '(std::unary_negate<main()::<lambda(int)>::<lambda(int)> >) (int&)'
{ return bool(_M_pred(*__it)); }
^
然后我尝试了std :: not1(std :: ref(less_than_three)),但是遇到了这些错误：
/usr/local/Cellar/gcc/5.3.0/include/c++/5.3.0/bits/stl_function.h:742:11: error: no type named 'argument_type' in 'class std::reference_wrapper<main()::<lambda(int)>::<lambda(int)> >'
class unary_negate
^
/usr/local/Cellar/gcc/5.3.0/include/c++/5.3.0/bits/stl_function.h:755:7: error: no type named 'argument_type' in 'class std::reference_wrapper<main()::<lambda(int)>::<lambda(int)> >'
operator()(const typename _Predicate::argument_type& __x) const
^
/usr/local/Cellar/gcc/5.3.0/include/c++/5.3.0/bits/predefined_ops.h:234:30: error: no match for call to '(std::unary_negate<std::reference_wrapper<main()::<lambda(int)>::<lambda(int)> > >) (int&)'
{ return bool(_M_pred(*__it)); }
^
如何在不改变lambdas逻辑的情况下否定std :: remove_if中的函数？换句话说,我怎样才能模仿remove_unless？
