Why does my float type-casting expression return two different values?

Question

Here are the expressions I'm working with:

float firstValue = (float) (5 / 2); //output is 2.0
float secondValue = (float) 5 / 2; //output is 2.5

I'm stumped here and can't figure out why this type casting is returning two different values. I understand I can just do (5f / 2f) but I wanted to experiment using the other type casting with an expression. Why is firstValue 2.0 and secondValue 2.5? Where did the .5 go?

Answer 1

As brackets have the highest precedence, they get solved first

float firstValue  = (float) (5 / 2);  // division of integers
                  = (float) (2);      // 5/2 = 2 , as Integers are being divided
                  = 2f
float secondValue = (float) 5 / 2; // division of float with integer
                  = ((float) 5) / 2;
                  = 5f / 2;          // second value is equivalent to this
                  = 2.5f             // as Float divided by Integer is Float

Answer 2

1. float firstValue = (float) (5 / 2); // division of integers

The first step is to do 5/2 calculation.Then the answer is given in float.If you explain further 5 and 2 are int numbers. After calculating the int for two int numbers, the final answer is returned by int. Here the final int answer (2) is converted to a float answer. That is, wider conversion is used here. So the final answer is the integer value(2) shown in float form(2.0).

2.float secondValue = (float) 5 / 2; //output is 2.5 Since the first value(5) is named a float number, the final answer is the decimal itself

Answer 3

The first is integer math. This

float firstValue = (float) (5 / 2); 

First divides five by two and gets two. Then it converts two to 2.0 . The second is floating point math.

float secondValue = 5f / 2; 

Which is 2.5 (and a float ). Because a float divided by an int is a float .