How to write multiple if conditions - c++

Question

I have two variables A and B and I want to write a code where if one of the two variables is equal to
151 or 156 or 720
and the other is not equal to one of these numbers then a third variable C = 0 is equal to one.
So for example
1) if A = 151 and B = 700 then C = 1
2) if A = 151 and B = 720 then C = 0
3) if A = 140 and B = 700 then C = 0
This is the code
int A = 0
cin >> A;
int B = 0
cin >> B;
int C=0;
int DECKlist[3] = {151,156,720}
for(int d=0; d<3;d++){
if(A== DECKlist[d]){
for(int w=0; w<3;w++){
if(B==DECKlist[w]) C=0;
else C=1;
}
}
if(B== DECKlist[d]){
for(int w=0; w<3;w++){
if(A==DECKlist[w]) C=0;
else C=1;
}
}
}
Is this OK? There are other better ways of doing it?

Answer 1

This is an exclusive OR, XOR. There is no logical XOR in C++, but you can use the bit-wise XOR for your case and exploit the fact that the result of a logical operator is a bool which will map to 0 or 1:
#include <iostream>
int main()
{
int A, B, C;
std::cin >> A;
std::cin >> B;
A = (A == 151 || A == 156 || A == 720);
B = (B == 151 || B == 156 || B == 720);
C = A ^ B;
std::cout << C << std::endl;
}
I've used a simple expression here to check whether a number is one of three supplied numbers. For larger sets of numbers to check against, you could use a, well, std::set.

Answer 2

You can use standard algorithms. For example you could use standard algoritnm std::binary_search along with bitwise XOR operator because array DECKlist is sorted
#include <algorithm>
#include <iterator>
//...
int DECKlist[] = { 151, 156, 720 };
//...
if ( std::binary_search( std::begin( DECKlist ), std::end( DECKlist ), A ) ^
std::binary_search( std::begin( DECKlist ), std::end( DECKlist ), B ) )
{
C = 1;
}
In this case you may add new values in the array and the approach will work as usual correctly. It does not depend on "magic numbers" and their quntity.:)

Answer 3

How about
int c=0;
for(int i=0; i<3; i++){
if(A == DECKlist[i]) c++;
if(B == DECKlist[i]) c++;
}
c = c%2;
Basically, count the number of matches, then make it zero if it was 2.

Answer 4

The first thing I'd do is hide the searching code. Here is a bit of C++ that will search an array (or any other container) linearly for if a given value is there. It uses a two C++11 features: std::begin and std::end, and auto typed variables:
#include <algorithm>
#include <utility>
#include <iterator>
template<class Array, class T>
bool find_in( Array const& arr, T const& t ) {
using std::begin; using std::end;
const auto b = begin(arr);
const auto e = end(arr);
const auto it = std::find( b, e, t ); // search the range for t
return !(e != it);
}
find_in is passed an array and a value, and returns true if that value is in the array.
We then use the same preamble. I avoid using namespace std; because it is a bad habit -- just name things with std:: or import individual symbols within a function scope.
#include <iostream>
int main() {
int A = 0
std::cin >> A;
int B = 0
std::cin >> B;
int C=0;
int DECKlist[3] = {151,156,720};
bool A_in_list = find_in(DECKlist, A);
bool B_in_list = find_in(DECKlist, B);
if (A_in_list != B_in_list)
C = 1;
std::cout << C << "\n";
}
The searching code, because I put it in a helper function, is really clean in the "core" logic of the program.
As we want to know when one is in the list, but not the other, we just compare the results using !=.