Double showing a lot of zeros with printf - c++

Question

I wanted to write a little calculator, which I have already done by using cout and cin, and I used double instead of int to not only get integers as an output.
At school, I then saw that we're going to use printf() and scanf(). To practice the new commands, I wanted to rewrite my program, but when I run my program I only see a lot of zeros after the comma as an output. Does anybody know why?
I wanted to rebuild a calculator with double instead of int to not only get integers as a result.
This is the code:
#include <stdio.h>
using namespace std;
int main(){
printf ("Taschenrechner\n\n");
int zahl1, zahl2;
char rechop;
double erg;
printf ("Gib die Rechnung ein: ");
scanf ("%d", &zahl1);
scanf ("%c", &rechop);
scanf ("%d", &zahl2);
if (rechop == '+'){
erg = zahl1+ zahl2;
printf ("Ergebnis: ");
printf ("%f", erg);
}
else if (rechop == '-'){
erg = zahl1 - zahl2;
printf ("Ergebnis: ");
printf ("%f", erg);
}
else if (rechop == '*'){
erg = zahl1 * zahl2;
printf ("Ergebnis: ");
printf ("%f", erg);
}
else if (rechop == '/'){
erg = zahl1 / zahl2;
printf ("Ergebnis: ");
printf ("%f", erg);
}
else {
printf ("Keine gültige Rechenoperation!");
}
return 0;
}

Answer 1

To decide how many decimal numbers show, you can use %.2f (where ".2" stands for "2 decimals after the comma") in the printf instead of simply %f.
Edit: forgot to say that, by doing that, your result get rounded (and not simply "cut").
There is your code updated (and translated):
#include <stdio.h>
using namespace std;
int main(){
printf ("Calculator\n\n");
int num1, num2;
char simbol;
double res;
printf ("Insert the expression: ");
scanf ("%d", &num1);
scanf ("%c", &simbol);
scanf ("%d", &num2);
if (simbol == '+'){
res = num1+ num2;
printf ("Result : ");
printf ("%.2f", res);
}
else if (simbol == '-'){
res = num1 - num2;
printf ("Result : ");
printf ("%.2f", res);
}
else if (simbol == '*'){
res = num1 * num2;
printf ("Result : ");
printf ("%.2f", res);
}
else if (simbol == '/'){
res = num1 / num2;
printf ("Result : ");
printf ("%.2f", res);
}
else {
printf ("Not a valid expression!");
}
return 0;
}

Answer 2

Your output has zeros after the decimal point, because that's actually what is in the variable (erg) that you are printing.
Your problem is that this line uses integer division:
erg = zahl1 / zahl2;
When the compiler is picking which operation to do, it only looks at the data type of the inputs (which are both int). The result is then converted to be stored into erg, but that doesn't affect the division behavior chosen.
Change the line to:
erg = zahl1 * 1.0 / zahl2;
or
erg = zahl1;
erg /= zahl2;
so that one of the inputs will be type double, and integer division will no longer be involved.

Answer 3

The problem is that you’re doing operations between int (num1 and num2 are declared as int). In order to have decimal numbers you must do operations between float (or double as in your case).
You can declare num1 and num2 as double directly (and change %d to %lf in the scanf and printf).
Another possibility is to do an explicit type cast: type cast is basically a conversion from a type to another and it is explicit because you’re writing it instead of letting the compiler do it for u (in your og program a cast is done by the compiler, after the operation, to fit the int result in the double type variable erg).
So an example is:
result = num1 + num2;
Becomes:
result = (double)num1 + (double)num2;