Weird C++ Number Conversion

While learning the C++ programming language, I found out that there are three ways to convert number types. For instance, suppose you want to convert an int to a float, then you can do either of the followings:

int myInt = 10;
float myFloat1 = (float) myInt;
float myFloat2 = float(myInt);
float myFloat3 = static_cast<float>(myInt);

As explained everywhere on the Internet, the first method is a C-style method, while the second and the third are C++ methods. But, are they completely interchangeable? Try this:

float myFloat1 = (float) 1/2;
float myFloat2 = float(1/2);
float myFloat3 = static_cast<float>(1/2);

That is, we are doing a division of two ints, and we want the result to be a float. However, try to print each variables that we just created!

std::cout << myFloat1 << std::endl; // prints 0.5

std::cout << myFloat2 << std::endl; // prints 0

std::cout << myFloat3 << std::endl; // prints 0

Now, why do myFloat2 and myFloat3 equal to zero? I think this is due to how the difference in the mechanism of the conversion that we just did. This is my explanation of what happens:

1. When we do (float) 1/2, the compiler will first convert each number to a float, and then execute the division. In this case, 1 is converted to 1.0, 2 is converted to 2.0, and finally the compiler divides 1.0 with 2.0. The result is a float that equals to 0.5 since both numbers are floats.
2. When we do float(1/2) or static_cast<float>(1/2), the compiler will first execute the divison, and then convert the result to float. In this case, the compiler will divide 1 (an int) with 2 (an int), and the result will be 0 (since 0.5 rounded down to the nearest integer is 0). After that, the number 0 is converted to float. So, the final result is still 0.