If string statement

Question

I have a problem with the following code:

int main(){

    string name;
    cout<<"Enter: ";
    cin>>name;
    if(name == "AHAH"||"AHAH"||"ADAD"||"ACAC"||"ASAS"||"AHAD"||"AHAC"||"AHAS"||"ADAH"||"ADAC"\
        ||"ADAS"||"ACAH"||"ACAD"||"ACAS"||"ASAH"||"ASAD"||"ASAC"){
        cout<<"Call! Good hand!";
    }
    else if(name == "2H2H")
        cout<<"Rase";

system("pause");
};

No matter what I input the program outputs:

Call! Good hand!

What's the problem?

Answer 1

When you have so many variants to compare with it could be better to keep them in container and then use algorithm or loop to compare:

std::set<std::string> names { "AHAH", "AHAH", "ADAD", "ACAC", 
                              "ASAS", "AHAD", "AHAC", "AHAS",
                              "ADAH", "ADAC" };
if( names.count( name ) )
     cout<<"Call! Good hand!";

or vector:

std::vector<std::string> names { "AHAH", "AHAH", "ADAD", "ACAC", 
                                 "ASAS", "AHAD", "AHAC", "AHAS",
                                 "ADAH", "ADAC" };
if( std::find( names.begin(), names.end(), name ) != names.end() ) 
    cout<<"Call! Good hand!";

The thing is that most probably you will need this list of names again anyway (when you write a real program that does something usefull)

Answer 2

The problem is that

if (name == "AHAH" || "ADAD"..)

Does not mean if name equals "AHAH" OR name equals "ADAD" but if (name == "AHAH") OR "ADAD". The latter is a char * implicitly converted to bool. Since the pointer is not null converts always to true. Hence the result.

Answer 3

Instead of:

if(name == "AHAH" || "ADAD" ...

You want something like:

if((name == "AHAH") || (name == "ADAD") ...

Answer 4

You need to write your if statement like so:

if ( name == "AHAH" || name == "ABCD" || ...etc... )
{
    cout << "Call, good hand!";
}
else if ( name == "2H2H" )
{
    cout << "Rase";
}

When you write your if statement like you did, you create pointers to char and since the char * is not NULL it has a value that is non-zero and therefore evaluates as True.