Does dividing float by int always give float? C++

I been programming on C and C++ for some time, and was always sure that double dividing int gives double, or double dividing int also give double (only int by int gives int) and same with adding. I was doing my assignment and was sure that 1 + (f*f + 10) / k, where f is double and k is int would always return double. I was using g++ -std=gnu++11 command on mac (so it's clang compiler probably) and I got tests passed (I indeed got float number as a result), but my teacher says that it not for sure that it will be float (he is using windows). Is that behavior platform specific? Is there any C++ standard that describes double on int division? Thank you! My code:

#include <iostream>
using namespace std;

int main() {
    int  N;
    double f, P = 0;
    cin >> N >> f;
    for (double k = 1; k <= N; k++){
      P += 1 + (f*f + 10) / k;
    }
    cout << P;
    return 0;
}

Solution

From the C++20 standard draft [expr.arith.conv]:

Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows:

(1.1) If either operand is of scoped enumeration type, no conversions are performed; if the other operand does not have the same type, the expression is ill-formed.

(1.2) If either operand is of type long double, the other shall be converted to long double.

(1.3) Otherwise, if either operand is double, the other shall be converted to double.

(1.4) Otherwise, if either operand is float, the other shall be converted to float.

(1.5) Otherwise, the integral promotions ([conv.prom]) shall be performed on both operands.56 Then the following rules shall be applied to the promoted operands: ...

This paragraph has not changed in any fundmental way since at least C++11 (which was the oldest I checked), so your teacher has some explaining to do.