Why does `(33).to_bytes(2, 'big')` return `b'\x00!'` instead of `b'\x00\x21'`?

Question

I'm having a problem converting int to bytes in Python.

This works -

>>> (1024).to_bytes(2, 'big')
b'\x04\x00'  

However this does not work as I would expect -

>>> (33).to_bytes(2, 'big')
b'\x00!'

What am I not understanding?

Answer 1

The decimal value 33 maps into the character ! by the ASCII standard, so the interpreter can show it without using escape codes:

>>> b'\x21' * 3
b'!!!'

---

When printing a bytes object, python treats it as a sequence of characters (every character is saved as a byte, with each byte using normally a memory of 8 bits that maps into 2 hexadecimal digits value, e.g. 0x21 => 0b 0010 0001 => 33), so values with corresponding printable ASCII characters are shown as their ASCII characters, and the rest are being represented by their hexadecimal values (in the format of \xDD).

Answer 2

According to documentation -> https://docs.python.org/3.3/library/stdtypes.html

 >>> (1024).to_bytes(2, byteorder='big')
 b'\x04\x00'
 >>> (1024).to_bytes(10, byteorder='big')
 b'\x00\x00\x00\x00\x00\x00\x00\x00\x04\x00'
 >>> (-1024).to_bytes(10, byteorder='big', signed=True)
 b'\xff\xff\xff\xff\xff\xff\xff\xff\xfc\x00'
 >>> x = 1000   
 >>> x.to_bytes((x.bit_length() // 8) + 1, byteorder='little')
 b'\xe8\x03'

Answer 3

You're not understanding that ! is ASCII character 33, equivalent to \x21. This bytestring is exactly the bytestring you asked for; it just isn't displayed the way you were expecting.