Message147205
It is already possible to write a wrapper function that does it:
def create(file):
fd = os.open(file, os.O_EXCL | os.O_CREAT | os.O_WRONLY)
return os.fdopen(fd)
The point it not that it can't be done, but that it is not straight forward. The docs say this about os.open(): "This function is intended for low-level I/O. For normal usage, use the built-in function open()"
I wouldn't call creating a new file low-level I/O, but it is normal usage. |
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Date |
User |
Action |
Args |
2011-11-07 06:53:30 | David.Townshend | set | recipients:
+ David.Townshend, amaury.forgeotdarc, pitrou, vstinner, benjamin.peterson, Arfrever, neologix, docs@python, Julian |
2011-11-07 06:53:30 | David.Townshend | set | messageid: <1320648810.3.0.643250137336.issue12760@psf.upfronthosting.co.za> |
2011-11-07 06:53:29 | David.Townshend | link | issue12760 messages |
2011-11-07 06:53:29 | David.Townshend | create | |
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