Primitive Assignments - V. Important

The equal ( = ) sign is used for assigning a value to a variable, and it's cleverly named the assignment operator. There are actually 12 assignment operators, but only the five most commonly used

You can assign a primitive variable using a literal or the result of an expression.

Take a look at the following:

int x = 7; // literal assignment
int y = x + 2; // assignment with an expression
// (including a literal)
int z = x * y; // assignment with an expression

The most important point to remember is that a literal integer (such as 7) is always implicitly an int and an int is a 32-bit value. No big deal if you're assigning a value to an int or a long variable, but what if you're assigning to a byte variable? After all, a byte-sized holder can't hold as many bits as an int-sized holder. Here's where it gets weird. The following is legal,

byte b = 27;

but only because the compiler automatically narrows the literal value to a byte. In other words, the compiler puts in the cast. The preceding code is identical to the following:

byte b = (byte) 27; // Explicitly cast the int literal to a byte

It looks as though the compiler gives you a break, and lets you take a shortcut with assignments to integer variables smaller than an int. (Everything we're saying about byte applies equally to char and short, both of which are smaller than an int.)

We know that a literal integer is always an int, but more importantly, the result of an expression involving anything int-sized or smaller is always an int. In other words, add two bytes together and you'll get an int—even if those two bytes are tiny. Multiply an int and a short and you'll get an int. Divide a short by a byte and you'll get…an int. OK, now we're at the weird part. Check this out:

byte b = 3; //No problem, 3 fits in a byte
byte c = 8; // No problem, 8 fits in a byte
byte d = b + c; // Should be no problem, sum of the two bytes
// fits in a byte

The last line won't compile! You'll get an error something like this:

TestBytes.java:5: possible loss of precision
found : int
required: byte
byte c = a + b;
^

We tried to assign the sum of two bytes to a byte variable, the result of which (11) was definitely small enough to fit into a byte, but the compiler didn't care. It knew the rule about int-or-smaller expressions always resulting in an int. It would have compiled if we'd done the explicit cast:

byte c = (byte) (a + b);