我有代码可以从 std::vector<int>
中删除所有元素少于一些 int limit
.我编写了一些部分应用 lambda 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> v{1,2,3,4,5};
测试它时,我得到了预期的结果:
for(auto e: v) {
std::cout << less_than_three(e) << " ";
}
// 1 1 0 0 0
我可以轻松删除所有少于三个的元素:
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? ?
我尝试包装 less_than_three
在std::not1
,但出现错误:
/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))
,但出现了这些错误:
/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)); }
^
如何取反 std::remove_if
中的函数不改变我的 lambda 的逻辑?换句话说,我如何模拟 remove_unless
?
最佳答案
not1
有点过时(并且要求仿函数提供某些成员类型定义,而 lambda 显然不提供)。
你必须自己写一个否定器:
auto negate = [] (auto&& f) {return [f=std::forward<decltype(f)>(f)]
(auto&&... args) {return !f(std::forward<decltype(args)>(args)...);};};
Demo .
关于c++ - 如何模拟 remove_unless,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34750561/