Is there a default type for numbers in Java

Question

If I write something like this ``` System.out.println(18); ``` Which type has the '18'? Is it int or byte? Or doesn't it have a type yet? It can't be int, because something like this is correct: ``` byte b = 3; ``` And this is incorrect: ``` int i = 3; byte bb = i; //error! ``` EDIT: I think I found the right part in the spec at Assignment Conversion : The compile-time narrowing of constants means that code such as: byte theAnswer = 42; is allowed. Without the narrowing, the fact that the integer literal 42 has type int would mean that a cast to byte would be required: byte theAnswer = (byte) 42; // cast is permitted but not required

Accepted Answer

This

```
18
```

is known as an integer literal. There are all sorts of literals, floating point, `String`, character, etc.

In the following,

```
byte b = 3;
```

the literal `3` is an integer literal. It's also a constant expression. And since Java can tell that `3` fits in a `byte`, it can safely apply a narrowing primitive conversion and store the result in a `byte` variable.

In this

```
int i = 3;
byte bb = i; //error!
```

the literal `3` is a constant expression, but the variable `i` is not. The compiler simply decides that `i` is not a constant expression and therefore doesn't go out of its way to figure out its value, a conversion to `byte` may lose information (how to convert `12345` to a `byte`?) and should therefore not be allowed. You can override this behavior by making `i` a constant variable

```
final int i = 3;
byte bb = i; // no error!
```

or by specifying an explicit cast

```
int i = 3;
byte bb = (byte) i; // no error!
```

Is there a default type for numbers in Java

Solution

Problem