menu choices (interpretation?)

[ChrisVer]

Hi
I have been trying to create a menu with the following code. The menu here is made of 2 choices. However if I give for choice=1, I get the error message (L15) and I can't leave the inner "while"-loop.
I tried removing from the while loop's statement the << || choice!="2" >>-part on L12 and then it was able to recognize my choice "1" and get me out. So I think that the problem is in that while loop.

Any idea of how I could fix it?

#include<iostream>
#include<string>

using namespace std;

int main(){

char ans;
do{

string choice="0"; //assign a string to enter the next loop
while(choice != "1" || choice !="2" ){ //checks if string is not 1 or not 2 & enters loop
cout<< "Would you like to calculate: "<<endl <<"1. integral" <<endl <<"2. derivative";
cin>>choice;  //assign new string to choice
if(choice !="1" || choice!="2"){cout<<"Please give a legit answer";} //if choice is not 1 or not 2 print error
} //end of loop

if(choice=="1") { cout<<"You chose 1" ;}//TESTING write code to solve the integral 
if(choice=="2") { cout<<"You chose 2";} //TESTING write code to solve the derivative 

cout<<"Would you like to retry?(y/n)";
cin>>ans;
}while(ans=='y' || ans=='Y');

if(ans=='n' || ans=='N'){cout<<"Thanks for trying me";}

system("pause");
return 0;
}

[mlholder]

When reading in strings, I usually use the string function getline, because it "eats" the newline character. From the looks of it, probably the reason you have a infinite while loop is because of the newline character. On line 15, I would change the || to &&, because if choice is "1", then the boolean expression will evaluate to true. If choice is "2", then it will also evaluate to true.

[ChrisVer]

Thanks for the getline() feedback.
The program works fine when instead of the OR (||) operator I use the AND (&&) operator both in the while and the if...
Now I am trying to understand why is that the case...it's really weird since not 1 or not 2 sounds the same as not 1 and not 2...I guess it has to do with logic.

[mlholder]

It has to do with the truth tables.

When evaluating boolean expressions, C++ uses short-circuit evaluation.
For example:

if(a == 1 && b == 2)
//Do Something

If a is not equal to 1, then the expression b == 2, will not be evaluated, since according to the truth tables when using &&, the whole expression must be true.

If a is equal to 1, then the expression b == 2 will be evaluated and if b really is 2, the the whole expression is true, else it is false.

if(a==1 || b==2)
//Do Something

If a is not equal to 1, then the expression b == 2, will be evaluated, since according to the truth tables when using ||, only one expression must be true.

If a is equal to 1, then the expression b == 2 will not be evaluated, thus the expression will evaluate to true.

Hope this helps.

[ChrisVer]

Yes it does... thank you :)

May I ask one more thing? Will the perfomance become better if instead of assigning a string (L11) and a while-loop right after, I use a "do-while" at L12-16 ?
I think I'm doing an extra unneeded iteration as it is.

[mlholder]

The performance is the same either way. The boolean expression is checked first in a while loop and last in a do-while loop. So one rule to use is if there is a chance that something may never be true, use a while loop and use a do-while loop when you know that something should happen at least once. In your case,

string choice;

do
{
       cout << "Would you like to calculate\n:"
              <<"1. integral\n"
              <<"2. derivative\n";
       getline(cin, choice);

       if(choice !="1" && choice!="2")
             cout <<"Please give a legit answer";

}while(choice != "1" && choice != "2");