Revisions to How can I determine a Python variable's type?

Second iteration. Fixed the weird syntax highlighting (as a result, the diff looks more extensive than it really is - use view "Side-by-side Markdown" to compare).

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edited Apr 25 at 11:33

Peter Mortensen

31.5k
22
110
134

Do you mean in Python or using ctypes?

In the first case, you simply cannot, because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # Unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

>>> import ctypes
>>> a = ctypes.c_uint() # Unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more references on ctypes, and its type, see the official documentation.

Do you mean in Python or using ctypes?

In the first case, you simply cannot, because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # Unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more references on ctypes, and its type, see the official documentation.

Do you mean in Python or using ctypes?

In the first case, you simply cannot, because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # Unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more references on ctypes, and its type, see the official documentation.

Active reading [<https://en.wiktionary.org/wiki/and#Conjunction>].

Source Link

edited Apr 25 at 11:25

Peter Mortensen

31.5k
22
110
134

Do you mean in Python or using ctypes?

In the first case, you simply cannot -, because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # unsignedUnsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more referencereferences on ctypes, anand its type, see the official documentation.

Do you mean in Python or using ctypes?

In the first case, you simply cannot - because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more reference on ctypes, an its type, see the official documentation.

Do you mean in Python or using ctypes?

In the first case, you simply cannot, because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # Unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more references on ctypes, and its type, see the official documentation.

Fixed ctypes link

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edited Mar 4, 2022 at 20:52

rob

37.9k
2
60
67

Do you mean in Python or using ctypes ctypes?

In the first case, you simply cannot - because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more reference on ctypes, an its type, see the official documentation.

Do you mean in Python or using ctypes?

In the first case, you simply cannot - because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more reference on ctypes, an its type, see the official documentation.

Do you mean in Python or using ctypes?

In the first case, you simply cannot - because Python does not have signed/unsigned, 16/32 bit integers.

In the second case, you can use type():

>>> import ctypes
>>> a = ctypes.c_uint() # unsigned int
>>> type(a)
<class 'ctypes.c_ulong'>

For more reference on ctypes, an its type, see the official documentation.

Copy edited.

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edited Feb 16, 2013 at 13:00

Peter Mortensen

31.5k
22
110
134

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answered Dec 31, 2008 at 8:10

rob

37.9k
2
60
67

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