Most pythonic way to convert a string to a octal number

Question

I am looking to change permissions on a file with the file mask stored in a configuration file. Since os.chmod() requires an octal number, I need to convert a string to an o

Answer 1

Have you just tried specifying base 8 to int:

num = int(your_str, 8)

Example:

s = '644'
i = int(s, 8) # 420 decimal
print i == 0644 # True

Answer 2

Here is the soluation:

Turn octal string "777" to decimal number 511

dec_num = int(oct_string, 8) # "777" -> 511

you worry about the os.chmod()? Let's try!

os.chmod("file", 511)  # it works! -rwxrwxrwx.
os.chmod("file", 0777) # it works! -rwxrwxrwx.
os.chmod("file", int("2777",8)) # it works! -rwxrwsrwx. the number is 1535!

it proves that the chmod can accept decimal and decimal can used as octal in python !

---

it is enough for the octal, because if you try

print dec_num == 0777 # True

Then get the decimal number 511 translate to octal string "0777"

oct_num = oct(dec_num)  # 511 -> "0777" , it is string again.

Pay attention there is no number 0777 but only oct string "0777" or dec num in the python if you write number 0777, it auto translate 0777 to 511, in the process there is no 0777 number exist.

summary

dec_num = int(oct_string, 8)
oct_num = oct(int(oct_string,8))

print dec_num         # 511
print dec_num == 0777 # True ! It is actually 511, just like dec_num == 511
print oct_num         # 0777
print oct_num == 0777 # Flase! Because it is a string!