Default values of bytes1 to bytes32?

Question

Beginner here who is trying to read https://docs.soliditylang.org/en/latest/control-structures.html#default-value about default values of data types.

While all the other default data types described in this paragraph make sense to me, I cannot get my head around the statement,

"For statically-sized arrays and bytes1 to bytes32, each individual element will be initialized to the default value corresponding to its type."

Thus, as per the title: What is the default values of bytes1 to bytes32?

And any guidance on how to wrap my head around this, beyond a direct answer, is appreciated.

Answer 1

In Solidity, the default values of bytes1 to bytes32 types are arrays of bytes where each byte is initialized to 0x00.

Here are the default values for each of these types:

bytes1: 0x00
bytes2: 0x0000
bytes3: 0x000000
bytes4: 0x00000000
...
bytes32: 0x0000000000000000000000000000000000000000000000000000000000000000

So, for any bytesN type (where N is between 1 and 32), the default value is an array of N bytes, all set to 0x00.

Answer 2

0 (zero) is the default values of bytes1 to bytes32.

There are many ways zero can be written, like 0.0 (1 decimal place) or 00 (2 digits).

Table in @Syed's answer is a way you can think of it. But they are not an "array of bytes": bytes2 is just a number that can fit in 2 bytes, and so on.

As an analogy, you can think of bytes1 to bytes32 as boxes of different sizes. But each is only 1 box.

An array of bytes would be something like bytes1[2]: here you have 2 of the smallest-sized box.