Write a C program to check whether a given integer is positive even, negative even, positive odd or negative odd.

#include <iostream>
using namespace std;
int main()
{
    int n;
    cout<< "Enter the number you want to check: \n";
    cin>>n;

    if(n>0 &&n%2 ==0){
        cout<< "number is positive even \n";
    }
    else if(n<0 &&n%2 ==0){
        cout<< "number is negative even \n";
    }
    else if(n<0 &&n%2 !=0){
        cout<< "number is negative odd \n";
    }
    else (n>0 && n%2 !=0){
            cout<< "number is positive odd \n";
    }
    
    return 0;
}

This is my code. When I don't add a semicolon after else (n>0 && n%2 !=0). It shows an error: expected ';' before '{' token. When there is semi colon there is a logic error where else output is shown no matter what the number is.

CodePudding user response：

You can not write else (condition) without using if. the correct way is write else if (condition) like you are doing or you can simply write else { cout << "Number is positive"; }.

CodePudding user response：

You can't leave out the if when you have a condition - the compiler interprets n>0 && n%2 !=0 as the body of the else branch, equivalent to this:

else { 
    (n>0 && n%2 !=0)
}
{
        cout<< "number is positive odd \n";
}       

which is missing a semicolon - and if you add one, you get an expression that does nothing and an unconditional output, equivalent to:

else 
    n>0 && n%2 !=0;
cout<< "number is positive odd \n";

Since none of your conditions cover the case where n is zero, you probably want

else if (n>0 && n%2 !=0){
    cout<< "number is positive odd \n";
}
else {
    cout << "number is zero\n";
}

CodePudding user response：

This line: else (n>0 && n%2 !=0)

else means that if all the above if and if-else conditions are false, then the logic in the else condition is executed, no condition need. But you add a condition n>0 && n%2 !=0 which is wrong syntax.

You can fix it by changing:

else (n>0 && n%2 !=0){
    cout<< "number is positive odd \n";
}

To:

else if (n>0 && n%2 !=0){
    cout<< "number is positive odd \n";
}

Or:

else{
    cout<< "number is positive odd \n";
}