[resolved]

Code:

---

#include <iostream.h>
typedef short int si;

int main()
{
        si theShoe = 20;
        si theTie, theWatch;
        cout << "enter the price for your watch\n";
        cin >> theTie;
        cout << "Enter a price for your tie\n";
        cin >> theWatch;
        {
        if ( (theTie >= theWatch) && ( (theTie + theWatch) > theShoe) )
                cout << "the shoe is less then the watch and the tie combined\n";
        else
                if ( (theTie == theWatch) || ((theTie + 10) == theWatch) )
                        cout << "the tie is either worth as much as the watch or the tie is 10$ more then the then watch\n";
                return 0;
        }
}

---

Ok when it asks for me to type in the price of the tie and watch if I type in 11 then it will print the shoe is less then the watch... but if I type in something else wher eit makes the statement true it will not print the shoe is less then the watch.

Can anyone tell me what I have done wrong or something I can do to fix it please

Thanks,
Bb

Quote:

---

cout << "enter the price for your watch\n";
cin >> theTie;
cout << "Enter a price for your tie\n";
cin >> theWatch;

---

shouldn't it be the other way around?

Quote:

---

cout << "the tie is either worth as much as the watch or the tie is 10$ more then the then watc

---

|| is or/and in english, in this case its not possible to have and, but exclusive or("either or" in english) is ^ in C++, nothing wrong, just so you know the difference

lol, I didnt put much thought in what it said. Just trying to get the program to do what I want :D

o wait nvm I see what your saying :\

Code:

---

#include <iostream.h>
typedef short int si;

int main()
{
        si theShoe = 20;
        si theTie, theWatch;
        cout << "enter the price for your watch\n";
        cin >> theWatch;
        cout << "Enter a price for your tie\n";
        cin >> theTie;
        {
        if ( (theTie > theWatch) && ( (theTie + theWatch) > theShoe) )
       
                cout << "the shoe is less then the watch and the tie combined\n";
       
        else
                if ( (theTie == theWatch) ^ ((theTie + 10) == theWatch) )
               
                        cout << "the tie is either worth as much as the watch or the tie is 10$ more then the then watch\n";
               
                else
               
                        cout << "the watch is worth more then everything\n";
        }
                return 0;
       
}

---

Here is what I have now... I still wouldnt mind any suggestions :)

thanks,
Bb

^ is only a bitwise operator! There is no logical XOR in C++!

You have to simulate it:
cond1 XOR cond2
can be written as
(cond1 AND NOT cond2) OR (NOT cond1 AND cond2)
in C++ syntax:
(cond1 && !cond2) || (!cond1 && cond2)

And it's
#include <iostream>
using namespace std;

not
#include <iostream.h>
which is deprecated!

You also don't need to write short int, you can just write short. Which makes the typedef unnecessary.

But in your case you don't need XOR. Here are the truth tables of OR and XOR:

Code:

---

OR 0 1  XOR 0 1
 0 0 1    0 0 1
 1 1 1    1 1 0

---

Since it is logically impossible that the watch has the same price as the tie AND has a price exactly 10 higher than the tie you can remove the lower right field from the table wich leaves you with two identical tables. So you can use the simple OR here.