Shorthand for checking for equality to multiple possibilities [duplicate] - c++

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Possible Duplicate:
C++ comparing bunch of values with a given one
I'm needing to check for equality in a for loop in C++, however the loop needs to work for x equaling multiple possibilities.
For example, right now I have something similar to:
if(x==a || x==b || x==c || x==d || x==e || x==f || x==g || x==h)
{
//loop body
}
But with the number I have, it looks messy and I was wondering if there was a shorthand way of saying "if (x == (any of these))" or if writing them all out was the only option.
Thanks!

Answer 1

Thank you for your question, now as I found a solution (and an elegant one I dare say), I'll use it myself.
Unlike solutions with std::find : Will a ) be unrolled to N comparisons in compile time b) work with any types which X can be compared to
struct TagAnyOf {};
template <typename... Args>
std::tuple <TagAnyOf, Args...> AnyOf (Args&&... args)
{
return std::tuple <TagAnyOf, Args...> (TagAnyOf(), std::forward<Args>(args)...);
}
template <class X, class Tuple, size_t Index, size_t ReverseIndex>
struct CompareToTuple
{
static bool compare (const X& x, const Tuple& tuple)
{
return x == std::get<Index> (tuple) || CompareToTuple<X, Tuple, Index+1, ReverseIndex-1>::compare (x, tuple);
}
};
template <class X, class Tuple, size_t Index>
struct CompareToTuple <X, Tuple, Index, 0>
{
static bool compare (const X& x, const Tuple& tuple)
{
return false;
}
};
template <typename X, typename... Args>
bool operator == (const X& x, const std::tuple<TagAnyOf, Args...>& any)
{
typedef std::tuple <TagAnyOf, Args...> any_of_type;
return CompareToTuple <X, any_of_type, 1, std::tuple_size<any_of_type>::value-1>::compare (x, any);
}
Usage
int main()
{
int x = 1;
if (x == AnyOf (1, 2, 3, 4))
{
std::cout << "Yes!" << std::endl;
}
else
{
std::cout << "No!" << std::endl;
}
if (x == AnyOf (4, 3, 2, 1))
{
std::cout << "Yes!" << std::endl;
}
else
{
std::cout << "No!" << std::endl;
}
if (x == AnyOf (2, 3, 4, 5))
{
std::cout << "Yes!" << std::endl;
}
else
{
std::cout << "No!" << std::endl;
}
return 0;
}

Answer 2

Consider using a function that takes an initializer_list (this is a c++11 feature).
the first parameter would be the left hand value(x in your case), and the rest of the parameters will be right hand values.
Here is an example that accomplishes the task using templates.
#include <iostream>
#include <cstdlib>
#include <algorithm>
template<class T>
bool Test(T const& test, std::initializer_list<T> const& values){
return std::find(std::begin(values), std::end(values), test) != std::end(values);
}
int main(){
char var1 = 'a';
char var2 = 'a';
char var3 = 'b';
char var4 = 'c';
char var5 = 'd';
if (Test<char>(var1,{var2,var3,var4,'o',var5})){
std::cout << "true. at least one is equivelent" << std::endl;
}else{
std::cout << "false. none are equivelent" << std::endl;
}
if (Test<char>(var1,{var3,var4,var5})){
std::cout << "true. at least one is equivelent" << std::endl;
}else{
std::cout << "false. none are equivelent" << std::endl;
}
return EXIT_SUCCESS;
}
If you are doing this with classes, make sure you overload the '!=' operator.
edit: mistake fixed. pointed out by GManNickG

Answer 3

I was wondering if there was a shorthand way of saying "if (x == (any
of these))"
Yes, the standard gives you std::find and std::find_if to answer exactly this question:
int a=3,b=5,c=6,d=7;
std::array<int,4> vals{{a,b,c,d}}; // or std::vector
int x=5;
bool foundit= (end(vals) != std::find_if(begin(vals), end(vals),x );
You will need
#include <array>
#include <algorithm>
You can also use std::initializer_list<int> booleans{a,b,c,d}; instead of vector or array
If your conditions where more complicated, you could use find_if:
bool foundit= (end(vals) !=
std::find_if(begin(vals), end(vals),
[&x](const int &v){return v*(v+x)<x;}));