How to find out the magic number for the .pyc header in Python 3

Python bytecode (.pyc) files have a header that starts with a magic number that changes between Python versions. How can I (programmatically) find out that number for the current Python version in order to generate a valid header? I’m currently hard-coding the one for Python 3.7.1, but that means I now depend on a specific Python version.

This answer does exactly what I want using py_compile.MAGIC, but that does not seem to exist anymore in Python 3. How can I do the equivalent in Python 3?

Here’s an example of what I’m trying to do:

import dis
import marshal

PYC_HEADER = b'x42x0dx0dx0ax00x00x00x00x00x00x00x00x00x00x00x00'

def f():
    print('Hello World')

with open('test.pyc', 'wb') as pyc:
    pyc.write(PYC_HEADER)
    marshal.dump(dis.Bytecode(f).codeobj, pyc)

This should create a file test.pyc, which can then be run, using the same Python interpreter as the script, and should print “Hello World!”. And it does, but only when using Python 3.7. I’m looking for a way that generates the header for whichever version of Python 3 is used to run the script, rather than hard-coding 3.7.

For context:

I’m compiling a simple programming language to different bytecode formats (LLVM, Java bytecode, Web Assembly and now Python bytecode) as part of a planned tutorial series on compiler construction.

I can generate the Python bytecode using the byteasm library, which gives me a function as a result. But in order to write the contents to a .pyc file, I need a valid header. By hard-coding the header, the code will only work if the people following the tutorial are running the same version of Python 3 as I am (3.7) or they’d have to manually find out the magic number for their version.