问题
Let's say that I want to write something like this (the {1, 3, 7, 42, 69, 550123} set is known before compilation):
int x;
...
if (x == 1 || x == 3 || x == 7 || x == 42 || x == 69 || x == 5550123)
{
...
}
The condition looks ugly because we have 9 extra symbols ("|| x ==") for each possible value. How can I rewrite it in a more C++ way?
My best guess is:
int x;
...
const std::unordered_set<int> v = {1, 3, 7, 42, 69, 5550123};
if (v.count(x))
{
...
}
It has O(1) average complexity with some memory and time overhead, but still looks kinda ugly.
回答1:
Edit: I just noticed c++14 tag. Note that my implementation of in relies on C++17. It can be done in C++14 as well using recursion, but that involves much more boilerplate, and a bit slower compilation.
One could use a template to generate a function with a logical operator sequence, such as the one in nvoigt's answer:
template<auto... ts, class T>
constexpr bool
in(const T& t) noexcept(noexcept(((t == ts) || ...))) {
return ((t == ts) || ...);
}
// usage
if (in<1, 3, 7, 42, 69, 5550123>(x))
That said, hiding the set of magic numbers behind a named function probably makes a lot of sense:
constexpr bool
is_magical(int x) noexcept {
return in<1, 3, 7, 42, 69, 5550123>(x);
}
回答2:
The only clean way to do this is to simply move it to a method. Name the method appropriately and it really does not matter what you do inside.
bool is_valid_foo_number(int x)
{
return x == 1
|| x == 3
|| x == 7
|| x == 42
|| x == 69
|| x == 5550123;
}
The method looks good enough for me, because all I will ever see of it is
if (is_valid_foo_number(input))
{
// ...
}
Should a technical detail change (like the sheer amount of valid numbers requiring another lookup approach or maybe a database instead of hardcoded values) you change the method's internals.
The point is that I think it only looks ugly... because you need to look at it while you look at your logic. You shouldn't have to look at the details anyway.
回答3:
How can I rewrite it in a more C++ way?
Suggestion: make a variadic template function for it, make it generic (non only for int values) and make it constexpr; this way you can check, the presence of the value in the set, compile time.
You tagged C++14 but I show first the recursive way for C++11, with a recursive case and a ground case
template <typename T>
constexpr bool is_in_list_11 (T const &)
{ return false; }
template <typename T, T t0, T ... ts>
constexpr bool is_in_list_11 (T const & t)
{ return (t == t0) || is_in_list_11<T, ts...>(t); }
Starting from C++14, a constexpr function can be a lot more complex, so recursion isn't necessary anymore and you can write something as
template <typename T, T ... ts>
constexpr auto is_in_list_14 (T const & t)
{
using unused = bool[];
bool ret { false };
(void)unused { false, ret |= t == ts ... };
return ret;
}
Starting from C++17 you can use also auto template type and template folding, so (as user2079303 showed before) the function become very very simple
template <auto ... ts, typename T>
constexpr auto is_in_list_17 (T const & t)
{ return ( (t == ts) || ... ); }
The following is a full working example with all versions
#include <iostream>
template <typename T>
constexpr bool is_in_list_11 (T const &)
{ return false; }
template <typename T, T t0, T ... ts>
constexpr bool is_in_list_11 (T const & t)
{ return (t == t0) || is_in_list_11<T, ts...>(t); }
template <typename T, T ... ts>
constexpr auto is_in_list_14 (T const & t)
{
using unused = bool[];
bool ret { false };
(void)unused { false, ret |= t == ts ... };
return ret;
}
template <auto ... ts, typename T>
constexpr auto is_in_list_17 (T const & t)
{ return ( (t == ts) || ... ); }
int main ()
{
constexpr auto b11a { is_in_list_11<int, 1, 3, 7, 42, 69, 5550123>(7) };
constexpr auto b11b { is_in_list_11<int, 1, 3, 7, 42, 69, 5550123>(8) };
constexpr auto b14a { is_in_list_14<int, 1, 3, 7, 42, 69, 5550123>(7) };
constexpr auto b14b { is_in_list_14<int, 1, 3, 7, 42, 69, 5550123>(8) };
constexpr auto b17a { is_in_list_17<1, 3, 7, 42, 69, 5550123>(7) };
constexpr auto b17b { is_in_list_17<1, 3, 7, 42, 69, 5550123>(8) };
std::cout << b11a << ' ' << b11b << std::endl;
std::cout << b14a << ' ' << b14b << std::endl;
std::cout << b17a << ' ' << b17b << std::endl;
}
回答4:
As the other answers say, move this to a function.
As the other answers say, you can consider adding constexpr / throw as required.
As the other answers don't say, use a switch case statement for this; which allows you to replace all the || x == with case - a few characters less, which might not seem significant (and it kinda isn't); but most importantly removes the chance of a mistake with either the variable name or a |.
When you've 300 items in this list, and it doesn't work as expected, trust me, you'll be glad to not have to check that every || is correct.
回答5:
Try this:
#include <iostream>
#include <vector>
#include <algorithm>
int main()
{
std::vector<int> v = {1, 3, 7, 42, 69, 5550123};
auto is_present = [&v](int x)->bool{
return std::find(v.begin(),v.end(),x) != v.end();
};
std::cout << (is_present(1)? "present" :" no present") <<std::endl;
}
来源：https://stackoverflow.com/questions/51496410/whats-the-most-c-way-to-check-if-value-belongs-to-certain-static-set