Python: why does peek(1) return 8K bytes instead of 1 byte?

Question

I'm using Python 3, and the peek() method for buffered file I/O doesn't seem to work as documented. For example, the following code illustrates the problem -- it prints 8192 as the length of the byte string returned by f.peek(1):

jpg_file = 'DRM_1851.JPG'
with open(jpg_file, 'rb') as f:
    next_byte = f.peek(1)
    print(len(next_byte))

I sometimes want to peek at the next byte without moving the file pointer, but since the above doesn't work I'm doing something this in those places instead:

next_byte = f.read(1) # read a byte
f.seek(-1,1) # move the file pointer back one byte

That works, but feels like a kludge. Am I misunderstanding something about how peek() works?

Obsequies

See #24475187 – Rosendorosene

Answer 1

From the Python docs:

peek([size])

Return bytes from the stream without advancing the position. At most one single read on the raw stream is done to satisfy the call. The number of bytes returned may be less or more than requested.

Emphasis mine.

Since the file pointer isn't moved in peek, it doesn't really matter if peek reads more than the amount you want. Just take a substring after you peek: next_byte = f.peek(1)[:1]

Wary

Shouldn't that be f.peek(1)[:1] in case there are no bytes returned? – Rosendorosene

Thanks much! I understand now. I have to admit, I've read that line in the Python docs more than once and I never fully internalized those three words "or more than." – Obsequies