Based on this answer (among others) it seems like f-strings is [one of] the preferred ways to convert to hexadecimal representation.
While one can specify an explicit target length, up to which to pad with leading zeroes, given a goal of an output with an even number of digits, and inputs with an arbitrary # of bits, I can imagine:
- pre-processing to determine the number of bits of the input, to feed an input-specific value in to the fstring, or
- post-processing a-la
out = "0"+f"{val:x}" if len(f"{val:x}") % 2 else f"{val:02x}"(or even using.zfill())
The latter seems like it might be more efficient than the former – is there a built-in way to do this with fstrings, or a better alternative?
Examples of input + expected output:
[0x]1 -> [0x]01 [0x]22 -> [0x]22 [0x]333 -> [0x]0333 [0x]4444 -> [0x]4444
and so on.
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Answer
Here’s a postprocessing alternative that uses assignment expressions (Python 3.8+):
print((len(hx:=f"{val:x}") % 2) * '0' + hx)
If you still want a one-liner without assignment expressions you have to evaluate your f-string twice:
print((len(f"{val:x}") % 2) * '0' + f"{val:x}")
As a two-liner
hx = f"{val:x}"
print((len(hx) % 2) * '0' + hx)
And one more version:
print(f"{'0'[:len(hex(val))%2]}{val:x}")