C++ Simple Calculator (no if, Switch, Case, and its kind) - c++

Question

I tried to make Simple calculator. There are 3 inputs :
number 1
number 2
operators
The operator is selected using number :
1 = '+'
2 = '-'
3 = '*'
4 = '/'
Example :
Number1 = 1, number 2 = 1, operator = 1, then the equation goes 1 + 1 = 2.
I didn't know how to make this equation possible. Any help will be appreciated. Thanks!
*note : not using array or string, not using case, not using switch, not using if, while, for.
Here's my code, but it's not yet completed and I need to change the operator selector.
#include <iostream>
using namespace std;
int main (){
int a, b, c, hasil;
cout << "Masukkan Bilangan 1 : ";
cin >> a;
cout << "Masukkan Bilangan 2 : ";
cin >> b;
cout << "Masukkan Operator ['(0 = +)' , '(1 = -)' , '(2 = *)', '(3 = /)' ] : ";
cin >> c;
hasil = a + (c * -2 + 1 )* b;
cout << "Hasilnya = " << hasil;
}

Answer 1

You can use different functions for the different operators and then select them by using the "c" value as the index to a table of functions.
#include <iostream>
using namespace std;
typedef int (*OperatorFunction)(int a, int b);
static int OperatorPlus(int a, int b)
{
return a + b;
}
static int OperatorMinus(int a, int b)
{
return a - b;
}
static int OperatorMultiply(int a, int b)
{
return a * b;
}
static int OperatorDivide(int a, int b)
{
return a / b;
}
OperatorFunction operators[] = { OperatorPlus, OperatorMinus, OperatorMultiply, OperatorDivide };
int main()
{
int a, b, c, hasil;
cout << "Masukkan Bilangan 1 : ";
cin >> a;
cout << "Masukkan Bilangan 2 : ";
cin >> b;
cout << "Masukkan Operator ['(0 = +)' , '(1 = -)' , '(2 = *)', '(3 = /)' ] : ";
cin >> c;
hasil = operators[c](a, b);
cout << "Hasilnya = " << hasil << "\n";
}

Answer 2

you basically have to calculate all four of them at the same time (warning, "1+0" will then crash, because it will divide by zero during calculation, even if "+" operation was chosen), and filter out the unneeded ones.
To filter out unneeded ones you need to turn c into one of the four value sets:
c==1: [1, 0, 0, 0]
c==2: [0, 1, 0, 0]
c==3: [0, 0, 1, 0]
c==4: [0, 0, 0, 1]
Let's call those inner four values ci, i=1..4
c1 = ((c-2) * (c-3) * (c-4)) / -6;
c2 = ((c-1) * (c-3) * (c-4)) / 2;
c3 = ((c-1) * (c-2) * (c-4)) / -2;
c4 = ((c-1) * (c-2) * (c-3)) / 6;
Then:
result = c1 * (a+b) + c2 * (a-b) + c3 * (a*b) + c4 * (a/b);
This is "pure math" solution, without exploiting C++ bool implicit conversion to 0/1, which would be better in production code, but I think it's a bit "cheating" in this exercise, that's why I'm calculating the 0/1 coefficients without the bool conversions in such complex manner by using the polynomials.
Few updates, to steer away from "pure math" a bit toward something more practical.
unwanted division by zero solution: instead of (a/b) you can calculate (a/(b|(c4-1))) (for c4 being 0 or 1). This will make the divisor equal to -1 whenever the c4 is zero, so it will neutralise "division by zero" for inputs like "a = 1, b = 0, c = 1" (i.e. "1+0"), and the division by zero may happen only when operation "division" is selected and b == 0.
if you will flip ci values from [0, 1] to [0, -1] (just change sign of the fixed constants at the final normalizing division), and everything is int, then you can replace the ci multiplications in final formula with bitwise AND like: result = (c1 & (a+b)) + (c2 & (a-b)) + (c3 & (a*b)) + (c4 & (a/b)); - which will be marginally faster on modern CPU (and considerably faster on historic CPU). {then of course the division by zero fix has to be flipped too => b|(-1-c4)}
note: uh, I aimed for -1 for the division fix, thinking about making whole division going to 0 in unsigned math for most of the inputs, then I figured out this is signed int and overall it makes little sense, but I kept -1 as target value, while anything non-zero will work, even simple 1. And the bitwise AND works only on platforms where negative integers are implemented by two's complement logic, so -1 is full bitmask then (as you didn't specify platform, this may break on some weird ones... the original answer with multiplication will work even on those).

Answer 3

The only thing I can think of with this assignment is to make use of that in C++ there's an implicit conversion between boolean type and integer. To take advantage of it you can do :
#include <iostream>
using namespace std;
int main (){
int a, b, c, hasil;
cout << "Masukkan Bilangan 1 : ";
cin >> a;
cout << "Masukkan Bilangan 2 : ";
cin >> b;
cout << "Masukkan Operator ['(0 = +)' , '(1 = -)' , '(2 = *)', '(3 = /)' ] : ";
cin >> c;
// if your input is a=10, b=5, c=0 :
// a + b == 15
// !c == 1
// 15 * 1 == 15
// a - b == 5
// !(c - 1) == !(0 - 1) == 0
// 5 * 0 == 0
// and so on...
hasil = ((a + b) * !c) + ((a - b) * !(c - 1)) + ((a * b) * !(c - 2)) + ((a / b) * !(c - 3));
cout << "Hasilnya = " << hasil;
}
explanation :
If your c input is 0 and you're doing !c it will return 1 as a int representation of TRUE but then for each other value eg. 3 it will return 0 as a int representation of FALSE. This is then multiplied by the calculated value of each possible operator and return either calculated value or 0 ( because 0 * 99 == 0 ).
Try this online