Is `id` a keyword in python?
My editor (TextMate) shows id
in another colour (when used as vari开发者_C百科able name) than my usual variable names. Is it a keyword? I don't want to shade any keyword...
id
is not a keyword in Python, but it is the name of a built-in function.
The keywords are:
and del from not while
as elif global or with
assert else if pass yield
break except import print
class exec in raise
continue finally is return
def for lambda try
Keywords are invalid variable names. The following would be a syntax error:
if = 1
On the other hand, built-in functions like id
or type
or str
can be shadowed:
str = "hello" # don't do this
You can also get help from python:
>>> help(id)
Help on built-in function id in module __builtin__:
id(...)
id(object) -> integer
Return the identity of an object. This is guaranteed to be unique among
simultaneously existing objects. (Hint: it's the object's memory address.)
or alternatively you can question IPython
IPython 0.10.2 [on Py 2.6.6]
[C:/]|1> id??
Type: builtin_function_or_method
Base Class: <type 'builtin_function_or_method'>
String Form: <built-in function id>
Namespace: Python builtin
Docstring [source file open failed]:
id(object) -> integer
Return the identity of an object. This is guaranteed to be unique among
simultaneously existing objects. (Hint: it's the object's memory address.)
Just for reference purposes:
Check if something is a keyword in Python:
>>> import keyword
>>> keyword.iskeyword('id')
False
Check all the keywords in Python:
>>> keyword.kwlist
['and', 'as', 'assert', 'break', 'class', 'continue', 'def', 'del', 'elif',
'else', 'except', 'exec', 'finally', 'for', 'from', 'global', 'if', 'import',
'in', 'is', 'lambda', 'not', 'or', 'pass', 'print', 'raise', 'return', 'try',
'while', 'with', 'yield']
It's a built in function:
id(...)
id(object) -> integer
Return the identity of an object. This is guaranteed to be unique among
simultaneously existing objects. (Hint: it's the object's memory address.)
精彩评论