### If else case and choose problem

Hello!
Today i got a question that

1. To check whether no. is even or odd
2. To check whether no. is positive or negative
depend upon users choice

Actually i am confused in if else case.. that how can i use if else In if else, hope you are getting me. so i have tried many program but i failed. here is one example that what i had done.

 ``123456789101112131415161718192021222324252627`` ``````#include int main() { int y,z; cout<<"\n Choose any option"; cout<<"\n 1. To check whether no. is even/odd"; cout<<"\n 2. To check whether no. is positive/negative \n"; cin>>y; cout<<"\n Enter your no."; cin>>z; if(y==1 & z%2==0) cout<<"\n No. is even"; else cout<<"\n No. is odd"<

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You shall use logical && (and) operator instead of bitwise operator & in statements as the following for example

if(y==1 & z%2==0) // shall be &&
I can tell you that line 23 has a semicolon at the end when it shouldn't. Also, you are using a single & when you should be using two && (which means "and").

Now, the way I would think of implementing this is to have one set of an if/else if to check which option the user picked.

Then, inside of each of those blocks have some more if statements to check for what the user wanted.
Actually i have tried that too... when i use "&&" i got combined result
even/odd and positive negative both.
can anyne of you give me any example that how can i run it?
 ``12345678910111213141516`` ``````switch ( y ) { case 1: if ( z % 2 == 0 ) cout << "\nNo. is even" << endl; else cout << "\nNo. is odd" << endl; break; case 2: if ( 0 < z ) cout << "\nNo. is positive" << endl; else cout << "\nNo. is non-positive" << endl; default: cout << "\nYou should select from menu either 1 or 2" << endl; break; } ``````

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Oh..!!! It's working thanks a lot....!! really thanks ... :)
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