Is a TextIOWrapper object automatically closed if it is not saved to a variable? [duplicate]












0
















This question already has an answer here:




  • Is there a need to close files that have no reference to them?

    6 answers



  • Does a File Object Automatically Close when its Reference Count Hits Zero?

    4 answers




I generally like to use the with open(...) as file: idiom for opening files because it avoids failing to close a file, improves code readability, and improves exception handling. However, I have seen other people's code use one-liners to open and read/write a file without saving the TextIOWrapper object:



# Read a file
contents = open('the_file.txt').read()

# Write to a file
open('the_file.txt', 'w').write('This is the thing.')

# Write an empty file
open('empty_file.txt', 'w')


These idioms are appealing because of their simplicity, but I want to understand what python does with those file objects.



I tried using the dis module to inspect the bytecode, but I'm not sure how to read the output:



def myfunc():
open('test.txt', 'w')
dis.dis(myfunc)


  2           0 LOAD_GLOBAL              0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 POP_TOP
10 LOAD_CONST 0 (None)
12 RETURN_VALUE


def withfunc():
with open('test.txt', 'w') as f:
pass
dis.dis(withfunc)



2 0 LOAD_GLOBAL 0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 SETUP_WITH 6 (to 16)
10 STORE_FAST 0 (f)

3 12 POP_BLOCK
14 LOAD_CONST 0 (None)
>> 16 WITH_CLEANUP_START
18 WITH_CLEANUP_FINISH
20 END_FINALLY
22 LOAD_CONST 0 (None)
24 RETURN_VALUE


def savefunc():
f = open('test.txt', 'w')
f.close()
dis.dis(savefunc)


  2           0 LOAD_GLOBAL              0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 STORE_FAST 0 (f)

3 10 LOAD_FAST 0 (f)
12 LOAD_ATTR 1 (close)
14 CALL_FUNCTION 0
16 POP_TOP
18 LOAD_CONST 0 (None)
20 RETURN_VALUE


I have two questions:




  1. What happens under the hood when you use open without saving the TextIOWrapper object (especially regarding whether they are properly closed)

  2. Are there reasons to NOT use the one-liners?










share|improve this question













marked as duplicate by juanpa.arrivillaga python
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Nov 20 '18 at 21:07


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
















  • dis isn't going to really help you here, you have to know the implementations of the objects themselves

    – juanpa.arrivillaga
    Nov 20 '18 at 21:07











  • Never open() without referencing the obj. The object remains in memory and cannot be freed since you don't reference to close() with.

    – Idlehands
    Nov 20 '18 at 21:10








  • 1





    @Idlehands that's not true. it will be freed (in CPython), and closed, but the output buffer won't necessarily write to disk. Well, actually, that only happens if you do have a reference around, because CPython doesn't guarantee that __del__ will be called for objects that still have references when the interpreter shuts down.

    – juanpa.arrivillaga
    Nov 20 '18 at 21:11













  • You should always just use a context-manager.

    – juanpa.arrivillaga
    Nov 20 '18 at 21:14
















0
















This question already has an answer here:




  • Is there a need to close files that have no reference to them?

    6 answers



  • Does a File Object Automatically Close when its Reference Count Hits Zero?

    4 answers




I generally like to use the with open(...) as file: idiom for opening files because it avoids failing to close a file, improves code readability, and improves exception handling. However, I have seen other people's code use one-liners to open and read/write a file without saving the TextIOWrapper object:



# Read a file
contents = open('the_file.txt').read()

# Write to a file
open('the_file.txt', 'w').write('This is the thing.')

# Write an empty file
open('empty_file.txt', 'w')


These idioms are appealing because of their simplicity, but I want to understand what python does with those file objects.



I tried using the dis module to inspect the bytecode, but I'm not sure how to read the output:



def myfunc():
open('test.txt', 'w')
dis.dis(myfunc)


  2           0 LOAD_GLOBAL              0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 POP_TOP
10 LOAD_CONST 0 (None)
12 RETURN_VALUE


def withfunc():
with open('test.txt', 'w') as f:
pass
dis.dis(withfunc)



2 0 LOAD_GLOBAL 0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 SETUP_WITH 6 (to 16)
10 STORE_FAST 0 (f)

3 12 POP_BLOCK
14 LOAD_CONST 0 (None)
>> 16 WITH_CLEANUP_START
18 WITH_CLEANUP_FINISH
20 END_FINALLY
22 LOAD_CONST 0 (None)
24 RETURN_VALUE


def savefunc():
f = open('test.txt', 'w')
f.close()
dis.dis(savefunc)


  2           0 LOAD_GLOBAL              0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 STORE_FAST 0 (f)

3 10 LOAD_FAST 0 (f)
12 LOAD_ATTR 1 (close)
14 CALL_FUNCTION 0
16 POP_TOP
18 LOAD_CONST 0 (None)
20 RETURN_VALUE


I have two questions:




  1. What happens under the hood when you use open without saving the TextIOWrapper object (especially regarding whether they are properly closed)

  2. Are there reasons to NOT use the one-liners?










share|improve this question













marked as duplicate by juanpa.arrivillaga python
Users with the  python badge can single-handedly close python questions as duplicates and reopen them as needed.

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Nov 20 '18 at 21:07


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
















  • dis isn't going to really help you here, you have to know the implementations of the objects themselves

    – juanpa.arrivillaga
    Nov 20 '18 at 21:07











  • Never open() without referencing the obj. The object remains in memory and cannot be freed since you don't reference to close() with.

    – Idlehands
    Nov 20 '18 at 21:10








  • 1





    @Idlehands that's not true. it will be freed (in CPython), and closed, but the output buffer won't necessarily write to disk. Well, actually, that only happens if you do have a reference around, because CPython doesn't guarantee that __del__ will be called for objects that still have references when the interpreter shuts down.

    – juanpa.arrivillaga
    Nov 20 '18 at 21:11













  • You should always just use a context-manager.

    – juanpa.arrivillaga
    Nov 20 '18 at 21:14














0












0








0









This question already has an answer here:




  • Is there a need to close files that have no reference to them?

    6 answers



  • Does a File Object Automatically Close when its Reference Count Hits Zero?

    4 answers




I generally like to use the with open(...) as file: idiom for opening files because it avoids failing to close a file, improves code readability, and improves exception handling. However, I have seen other people's code use one-liners to open and read/write a file without saving the TextIOWrapper object:



# Read a file
contents = open('the_file.txt').read()

# Write to a file
open('the_file.txt', 'w').write('This is the thing.')

# Write an empty file
open('empty_file.txt', 'w')


These idioms are appealing because of their simplicity, but I want to understand what python does with those file objects.



I tried using the dis module to inspect the bytecode, but I'm not sure how to read the output:



def myfunc():
open('test.txt', 'w')
dis.dis(myfunc)


  2           0 LOAD_GLOBAL              0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 POP_TOP
10 LOAD_CONST 0 (None)
12 RETURN_VALUE


def withfunc():
with open('test.txt', 'w') as f:
pass
dis.dis(withfunc)



2 0 LOAD_GLOBAL 0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 SETUP_WITH 6 (to 16)
10 STORE_FAST 0 (f)

3 12 POP_BLOCK
14 LOAD_CONST 0 (None)
>> 16 WITH_CLEANUP_START
18 WITH_CLEANUP_FINISH
20 END_FINALLY
22 LOAD_CONST 0 (None)
24 RETURN_VALUE


def savefunc():
f = open('test.txt', 'w')
f.close()
dis.dis(savefunc)


  2           0 LOAD_GLOBAL              0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 STORE_FAST 0 (f)

3 10 LOAD_FAST 0 (f)
12 LOAD_ATTR 1 (close)
14 CALL_FUNCTION 0
16 POP_TOP
18 LOAD_CONST 0 (None)
20 RETURN_VALUE


I have two questions:




  1. What happens under the hood when you use open without saving the TextIOWrapper object (especially regarding whether they are properly closed)

  2. Are there reasons to NOT use the one-liners?










share|improve this question















This question already has an answer here:




  • Is there a need to close files that have no reference to them?

    6 answers



  • Does a File Object Automatically Close when its Reference Count Hits Zero?

    4 answers




I generally like to use the with open(...) as file: idiom for opening files because it avoids failing to close a file, improves code readability, and improves exception handling. However, I have seen other people's code use one-liners to open and read/write a file without saving the TextIOWrapper object:



# Read a file
contents = open('the_file.txt').read()

# Write to a file
open('the_file.txt', 'w').write('This is the thing.')

# Write an empty file
open('empty_file.txt', 'w')


These idioms are appealing because of their simplicity, but I want to understand what python does with those file objects.



I tried using the dis module to inspect the bytecode, but I'm not sure how to read the output:



def myfunc():
open('test.txt', 'w')
dis.dis(myfunc)


  2           0 LOAD_GLOBAL              0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 POP_TOP
10 LOAD_CONST 0 (None)
12 RETURN_VALUE


def withfunc():
with open('test.txt', 'w') as f:
pass
dis.dis(withfunc)



2 0 LOAD_GLOBAL 0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 SETUP_WITH 6 (to 16)
10 STORE_FAST 0 (f)

3 12 POP_BLOCK
14 LOAD_CONST 0 (None)
>> 16 WITH_CLEANUP_START
18 WITH_CLEANUP_FINISH
20 END_FINALLY
22 LOAD_CONST 0 (None)
24 RETURN_VALUE


def savefunc():
f = open('test.txt', 'w')
f.close()
dis.dis(savefunc)


  2           0 LOAD_GLOBAL              0 (open)
2 LOAD_CONST 1 ('test.txt')
4 LOAD_CONST 2 ('w')
6 CALL_FUNCTION 2
8 STORE_FAST 0 (f)

3 10 LOAD_FAST 0 (f)
12 LOAD_ATTR 1 (close)
14 CALL_FUNCTION 0
16 POP_TOP
18 LOAD_CONST 0 (None)
20 RETURN_VALUE


I have two questions:




  1. What happens under the hood when you use open without saving the TextIOWrapper object (especially regarding whether they are properly closed)

  2. Are there reasons to NOT use the one-liners?





This question already has an answer here:




  • Is there a need to close files that have no reference to them?

    6 answers



  • Does a File Object Automatically Close when its Reference Count Hits Zero?

    4 answers








python file bytecode






share|improve this question













share|improve this question











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asked Nov 20 '18 at 21:04









reynoldsnlpreynoldsnlp

6591826




6591826




marked as duplicate by juanpa.arrivillaga python
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Nov 20 '18 at 21:07


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by juanpa.arrivillaga python
Users with the  python badge can single-handedly close python questions as duplicates and reopen them as needed.

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Nov 20 '18 at 21:07


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • dis isn't going to really help you here, you have to know the implementations of the objects themselves

    – juanpa.arrivillaga
    Nov 20 '18 at 21:07











  • Never open() without referencing the obj. The object remains in memory and cannot be freed since you don't reference to close() with.

    – Idlehands
    Nov 20 '18 at 21:10








  • 1





    @Idlehands that's not true. it will be freed (in CPython), and closed, but the output buffer won't necessarily write to disk. Well, actually, that only happens if you do have a reference around, because CPython doesn't guarantee that __del__ will be called for objects that still have references when the interpreter shuts down.

    – juanpa.arrivillaga
    Nov 20 '18 at 21:11













  • You should always just use a context-manager.

    – juanpa.arrivillaga
    Nov 20 '18 at 21:14



















  • dis isn't going to really help you here, you have to know the implementations of the objects themselves

    – juanpa.arrivillaga
    Nov 20 '18 at 21:07











  • Never open() without referencing the obj. The object remains in memory and cannot be freed since you don't reference to close() with.

    – Idlehands
    Nov 20 '18 at 21:10








  • 1





    @Idlehands that's not true. it will be freed (in CPython), and closed, but the output buffer won't necessarily write to disk. Well, actually, that only happens if you do have a reference around, because CPython doesn't guarantee that __del__ will be called for objects that still have references when the interpreter shuts down.

    – juanpa.arrivillaga
    Nov 20 '18 at 21:11













  • You should always just use a context-manager.

    – juanpa.arrivillaga
    Nov 20 '18 at 21:14

















dis isn't going to really help you here, you have to know the implementations of the objects themselves

– juanpa.arrivillaga
Nov 20 '18 at 21:07





dis isn't going to really help you here, you have to know the implementations of the objects themselves

– juanpa.arrivillaga
Nov 20 '18 at 21:07













Never open() without referencing the obj. The object remains in memory and cannot be freed since you don't reference to close() with.

– Idlehands
Nov 20 '18 at 21:10







Never open() without referencing the obj. The object remains in memory and cannot be freed since you don't reference to close() with.

– Idlehands
Nov 20 '18 at 21:10






1




1





@Idlehands that's not true. it will be freed (in CPython), and closed, but the output buffer won't necessarily write to disk. Well, actually, that only happens if you do have a reference around, because CPython doesn't guarantee that __del__ will be called for objects that still have references when the interpreter shuts down.

– juanpa.arrivillaga
Nov 20 '18 at 21:11







@Idlehands that's not true. it will be freed (in CPython), and closed, but the output buffer won't necessarily write to disk. Well, actually, that only happens if you do have a reference around, because CPython doesn't guarantee that __del__ will be called for objects that still have references when the interpreter shuts down.

– juanpa.arrivillaga
Nov 20 '18 at 21:11















You should always just use a context-manager.

– juanpa.arrivillaga
Nov 20 '18 at 21:14





You should always just use a context-manager.

– juanpa.arrivillaga
Nov 20 '18 at 21:14












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