int OnLoad() {
cout << "Hi whats your name? ";
cin >> name;
system("cls");
cout << "Hi " << name << "." << " Are you here to Take Over the city from zombies?"<< endl;
cin >> userInput;
if (userInput == "yes" || "Yes") {
cout << "Yes" << endl;
}
else if (userInput == "no" || "No") {
cout << "No" << endl;
}
else {
cout << "I don't understand." << endl;
}
return 0;
}
int main() {
OnLoad();
system("pause");
return 0;
}
This code only returns Yes back, after the console window pops up and ask are you here to take over the city from zombies even after i type no it returns yes!
if (userInput == "yes" || "Yes")
actually means
if ((userInput == "yes") || ("Yes"))
It's logical OR between two expressions: userInput == "yes"
and "Yes"
. The first one is correct and evaluates to bool
directly. The second one is just a char*
that will be converted to bool
implicitly. Since it's a compile time string it cannot be nullptr
, which means it will always evaluate to true
. And that, in turn, means the whole condition is always true
(that's how logical OR works).
The correct code is
if (userInput == "yes" || userInput == "Yes")
P. S. This is why I always recommend compiling with the highest warning level possible (/W4
for MSVC, -Wall -pedantic-errors
for GCC and clang). Most compilers will generate a warning in this case.
that's not how the || operator works, if you just put "Yes" as a condition it will always evaluate to true
if (userInput == "yes" || userInput == "Yes") {
cout << "Yes" << endl;
}
the reason why is because of precedence
userInput == "yes"
and
userInput == "Yes"
get evaluated before || ( logical OR)