Multiplying float with double overflow in C++

Question

I am a bit confused of the point of having this warning:
Arithmetic overflow: Using operator '' on a 4byte value and then casting the result to a 8byte value. Cast the value to the wider type before calling '' operator to avoid overflow.

#include <iostream>
using std::cin;
using std::cout;
using std::ios_base;

int main() {
    cout.setf(ios_base::fixed, ios_base::floatfield);
    double mints = 10.0 / 3.0;
    const float c_MILLION = 1e6;
    cout << "\n10 million mints: " << 10 * c_MILLION * mints;

    cin.get();
}

According to my understanding when we multiply a float value with a double value we are basicaly multiplying a 4byte value to an 8byte value and it we will hence, lose some precision according to the links that I have read:

• Cannot implicitly convert type 'double' to 'float'
• Multiply a float with a double
  

http://www.cplusplus.com/articles/DE18T05o/#:~:text=Overflow%20is%20a%20phenomenon%20where,be%20larger%20than%20the%20range

However, when I do output this, I get a double value
https://i.sstatic.net/EOQzm.png

If that is the case, why does it bother to warn me to cast c_MILLION to double value if it is automatically changing it to a double result? It cant convert an 8byte value to a 4byte value anyways. So, why does it bother to warn the programmers when it is already saving us from this trouble?
Or can it convert an 8byte value to a 4byte value as well. If so, how does it determine what type to print? This is a question that I cannot find the answer to from the links I read.
If it automatically converting the result to 8byte value, what is the point of displaying this warning?
Here is the warning:
https://i.sstatic.net/L2szy.png

Severity Code Description Project File Line Suppression State Warning C26451 Arithmetic overflow: Using operator '*' on a 4 byte value and then casting the result to a 8 byte value. Cast the value to the wider type before calling operator '*' to avoid overflow (io.2)

The problem was that I was multiplying an int value with a double value. But still the warning should not exist when it automatically converts the multiplication of an int to double to a double value.

Answer 1

The warning is because of this multiplication: 10 * c_MILLION. There can be some values of c_MILLION where some precision is lost that would not have been lost if c_MILLION was first converted to a double. Since the result of this multiplication is converted to double, a mistaken programmer might assume that no precision was lost beyond what might be expected if the operands were double in the first place. Hence the warning.