C++ - Where in my code should I include static_cast? - c++

Question

My static_cast does not seem to be working. I create 5 int variables at the beginning of my code and attempt to change one (the total) to a double after performing calculations on the other 4, but it keeps displaying the int.
#include <iostream>
#include <iomanip>
#include <cstdlib>
using namespace std;
int main ()
{
int grade1, grade2, grade3, grade4, avg;
cout << "Please input four grades: " << endl;
cin >> grade1;
cin >> grade2;
cin >> grade3;
cin >> grade4;
avg = (grade1 + grade2 + grade3 + grade4) / 4;
cout << "Grades: " << grade1 << ", " << grade2 << ", " << grade3 << ", " << grade4 << endl;
cout << "\n";
cout << "Average grade: " << static_cast<double>(avg) << endl;
return 0;
}

Answer 1

When you are combining the integers and assigning them to avg.
You already have integer rounding conversion, so you need to static cast before you assign the number to the avg result.
You also don't even need to use static cast as long as you divide by 4.0 and make avg a double.
Example:
double avg = static_cast<double>(grade1+grade2+grade3+grade4) / 4.0;
// Alternative without static that still works
double avg = (grade1+grade2+grade3+grade4) / 4.0;

Answer 2

I would suggest that the literal answer to your question "Where in my code should I include static_cast?" is quite simple: nowhere, if at all possible.
static_cast - like all the C++ _cast operators, and all explicit (C-style) typecasts should be avoided wherever possible. They have their uses, but it is better to minimise their usage, and certainly to avoid using them as blunt instruments to fix problems in code.
In your case, the correct results can be obtained quite simply by using appropriate types of variables, and any literal values having a suitable type.
Firstly, avg needs to be of type double. This involves changing
int grade1, grade2, grade3, grade4, avg;
to
int grade1, grade2, grade3, grade4;
double avg;
Second, the calculation of the value of avg needs to be changed so it doesn't do an integer division. This means changing
avg = (grade1 + grade2 + grade3 + grade4) / 4;
(which doesn't work because grade1, ... grade4 are all of type int, their sum is an int, the literal 4 is an int, and dividing two ints produces an int) to
avg = (grade1 + grade2 + grade3 + grade4) / 4.0; // note the 4.0
This works, because the literal 4.0 has type double. The value (grade1 + grade2 + grade3 + grade4) is calculated as an int but, because it is being divided by a double, that integral value is converted implicitly to double before performing the division - and the result is of type double.
This is the only place in which I would consider using a static_cast here, by converting the 4 to a double.
avg = (grade1 + grade2 + grade3 + grade4) / static_cast<double>(4);
which has the same net effect as the previous calculation of avg. In your code, it makes little difference. But a static_cast can be useful here in more general-purpose code, for example, if the number of input values (grade1, grade2, etc) is calculated as an integral value (e.g. the number of elements in an array).
The third thing you can do is remove the static_cast from the statement that outputs avg since it is now redundant (converting a double to double has no effect). In other words turn
cout << "Average grade: " << static_cast<double>(avg) << endl;
to
cout << "Average grade: " << avg << endl;

Answer 3

There are two issues in your code.
The first is that you declared avg as int, so the result is going to be rounded as int always.
However, declaring avg as double will not solve your problem immediately. In C++, operators are executed on objects of the same type and the result is an object of the same type.
(grade1 + grade2 + grade3 + grade4) is an int and 4 is an int, therefore the result will be an int. It doesn't matter if later is implicitly or explicitly casted to double, you will still get a rounded result.
In case the operands are not the same type, the compiler will implicitly promote one to match the other. The following lines will give you the correct result:
static_cast<double>(grade1 + grade2 + grade3 + grade4) / 4;
(grade1 + grade2 + grade3 + grade4) / 4.0;
static_cast<double>(grade1 + grade2 + grade3 + grade4) / 4.0;

Answer 4

An int cannot hold decimal places. int / results in an int, there are no decimals that result. You need to use the correct /, i.e at least one of the operands must be double to get a double result.
#include <iostream>
int main () {
std::cout<<"Please input four grades: "<<std::endl;
int grade1, grade2, grade3, grade4;
std::cin>>grade1;
std::cin>>grade2;
std::cin>>grade3;
std::cin>>grade4;
double avg=static_cast<double>(grade1+grade2+grade3+grade4)/4;
std::cout<<"Grades: "<<grade1<<", "<<grade2<<", "<<grade3<<", "<<grade4<<std::endl<<std::endl;
std::cout<<"Average grade: "<<avg<<std::endl;
}
You don't need the static cast here, as you can change the int literal 4 to a double literal 4.0, but if you are summing the elements of a vector then calling size() on it, that will be an integer, and you will have to cast somewhere.