Understand workflow about reading and writing files in Python

[DL Neil]

I've not gone 'back' to refer to any ComSc theory on buffer-management. 
Perhaps you might benefit from such?

I like your use of the word "shift", so I'll continue to use it.

There are three separate units of data to consider - each of which could 
be called a "buffer". To avoid confusing (myself) I'll only call the 
'middle one' that:
1 the unit of data 'coming' from the data-source
2 the "buffer" you are implementing
3 the unit of data 'going' out to a data-destination.

1 and 3 may be dictated to you, eg hardware or file specifications, code 
requirements, etc.

So, data is shifted into the (2) buffer in a unit-size decided by (1) - 
in most use-cases each incoming unit will be the same size, but remember 
that the last 'unit' may/not be full-size. Similarly, data shifted out 
from the (2) buffer to (3).

The size of (1) is likely not that of (3) - otherwise why use a 
"buffer"? The size of (2) must be larger than (1) and larger than (2) - 
for reasons already illustrated.

I recall learning how to use buffers with a series of hand-drawn block 
diagrams. Recommend you try similarly!


Now, let's add a few critiques, as requested (interposed below):-


On 19/06/19 3:53 PM, Windson Yang wrote:
> I'm trying to understand the workflow of how Python read/writes data with
> buffer. I will be appreciated if someone can review it.
> 
> ### Read n data

- may need more than one read operation if the size of (3) "demands" 
more data than the size of (1)/one "read".

> 1. If the data already in the buffer, return data

- this a data-transfer of size (3)

For extra credit/an unnecessary complication (but probable speed-up!):
* if the data-remaining is less than size (3) consider a read-ahead 
mechanism

> 2. If the data not in the buffer:

- if buffer's data-len < size (3)

>      1. copy all the current data from the buffer

* if "buffer" is my (2), then no-op

>      2. create a new buffer object, fill the new buffer with raw read which
> read data from disk.

* this becomes: perform read operation and append incoming data (size 
(1)) to "buffer" - hence why "buffer" is larger than (1), by definition.
NB if size (1) is smaller than size (3), multiple read operations may be 
necessary. Thus a read-loop!?


>      3. concat the data in the old buffer and new buffer.

= now no-op. Hopefully the description of 'three buffers' removes this 
confusion of/between buffers.


>      4. return the data

* make the above steps into a while-loop and there won't be a separate 
step here (it is the existing step 1!)


* build all of the above into a function/method, so that the 'mainline' 
only has to say 'give me data'!


> ### Write n data
> 1. If data small enough to fill into the buffer, write data to the buffer

=yes, the data coming from source (1), which in this case is 'your' code 
may/not be sufficient to fill the output size (3). So, load it into the 
"buffer" (2).

> 2. If data can't fill into the buffer
>      1. flush the data in the buffer

=This statement seems to suggest that if there is already some data in 
the buffer, it will be wiped. Not recommended!

=Have replaced the next steps, see below for your consideration:-

>          1. If succeed:
>              1. create a new buffer object.
>              2. fill the new buffer with data return from raw write
>          2. If failed:
>              1. Shifting the buffer to make room for writing data to the
> buffer
>              2. Buffer as much writing data as possible (may raise
> BlockingIOError)
>      2. return the data

After above transfer from data-source (1) to "buffer" (2):

* if len( data in "buffer" ) >= size (3): output
	else: keep going

* output:
	shift size(3) from "buffer" to output
	retain 'the rest' in/as "buffer"

NB if the size (2) of data in "buffer" is/could be multiples of size 
(3), then the "output" function should/could become a loop, ie keep 
emptying the "buffer" until size (2) < size (3).


Finally, don't forget the special cases:
What happens if we reach 'the end' (of 'input' or 'output' phase), and 
there is still data in (1) or (2)?
Presumably, in "Read" we would discard (1), but in the case of "Write" 
we MUST empty "buffer" (2), even if it means the last write is of less 
than size (3).

NB The 'rules' for the latter may vary between use-cases, eg add 
'stuffing' if the output record MUST be x-bytes long.


Hope this helps.
Do you need to hand-code this stuff though, or is there a better way?
-- 
Regards =dn