Can someone explain this

Question

Can someone explain to me how the output is 15

class Exe
{
public static void main(String args[])
{
float x=9;
float y=5;
int z= (int)(x/y);
switch(z)
{
case 1: x+=2;
case 2: x+=3;
default: x+=1;
}
System.out.println(x);
}
}

Answer 1

x/y returns 1.8 (float)

(int) 1.8 returns 1 (taking only integral part)

in switch, falls into the first case.

x = 9 + 2 = 11

no break, so second handling also executed

x = 11 + 3 = 14

still no break, so third handling also executed

x = 14 + 1 = 15

Answer 2

Amir01 Thanks for your question.

- - -

Because it is a double instead of a float.

quote:

A floating-point literal is of type float if it ends with the letter F or f; otherwise its type is double and it can optionally end with the letter D or d. It means that it's a single precision floating point literal rather than double precision.

source:

https://stackoverflow.com/questions/13704028/why-is-the-letter-f-used-at-the-end-of-a-float-no/13704077

- - -

Double has more precision and is more accurate, but it also takes up more memory.

Here is a reading about the difference between double and float :

https://javarevisited.blogspot.com/2016/05/difference-between-float-and-double-in-java.html?m=1

Answer 3

Remember:
If you donot use break statement at end of each case in switch, then all the cases that comes after matching case will also be executed.

Answer 4

For your case,
put 'break;' at the end of case
It will solve your problem like this,
switch(z)
{
case 1: x+=2;
break;
case 2: x+=3;
break;
default: x+=1;
}

The output should be 11.0

Good Luck~

Answer 5

nice!

Answer 6

Gordon ,you are right!
It makes a question in my mind. I think float variables needs to f at the end,but why above code hasn't any f?

Answer 7

Gordon thanks