What version of Visual Studio is Python on my computer compiled with?
Question:
I am trying to find out the version of Visual Studio that is used to compile the Python on my computer
It says
Python 2.6.2 (r262:71605, Apr 14 2009, 22:40:02) [MSC v.1500 32 bit (Intel)] on win32
What I do not understand is this MSC V.1500 designation. Does it mean it is compiled with Visual Studio 2005? I cannot find this information on http://python.org.
Answers:
MSC v.1500 appears to be Visual C++ 2008 according to this thread on the OpenCobol forums (of all places).
The MSDN page on Predefined Macros indicates 1500 to be the result of the _MSC_VER macro.
This other forum post mentions that
(For reference, Visual Studio 2003 has _MSC_VER = 1310; Visual Studio 2005 has _MSC_VER = 1400; Visual Studio 2008 has _MSC_VER = 1500.)
The above MSDN link said that 1600 indicates VS2010.
Strangely, I wasn’t able to find that info about the earlier _MSC_VER values on MSDN.
Visual C++ version
_MSC_VER
Visual C++ 4.x
1000
Visual C++ 5
1100
Visual C++ 6
1200
Visual C++ .NET
1300
Visual C++ .NET 2003
1310
Visual C++ 2005 (8.0)
1400
Visual C++ 2008 (9.0)
1500
Visual C++ 2010 (10.0)
1600
Visual C++ 2012 (11.0)
1700
Visual C++ 2013 (12.0)
1800
Visual C++ 2015 (14.0)
1900
Visual C++ 2017 (15.0)
1910
Visual C++ 2017 (15.3)
1911
Visual C++ 2017 (15.5)
1912
Visual C++ 2017 (15.6)
1913
Visual C++ 2017 (15.7)
1914
Visual C++ 2017 (15.8)
1915
Visual C++ 2017 (15.9)
1916
Visual C++ 2019 RTW (16.0)
1920
Visual C++ 2019 (16.1)
1921
Visual C++ 2019 (16.2)
1922
Visual C++ 2019 (16.3)
1923
Visual C++ 2019 (16.4)
1924
Visual C++ 2019 (16.5)
1925
Visual C++ 2019 (16.6)
1926
Visual C++ 2019 (16.7)
1927
Visual C++ 2019 (16.8)
1928
Visual C++ 2019 (16.9)
1928
Visual C++ 2019 (16.10)
1929
Visual C++ 2019 (16.11)
1929
Visual Studio 2022 RTW (17.0)
1930
Source: the documentation for the _MSC_VER predefined macro
I am trying to find out the version of Visual Studio that is used to compile the Python on my computer
It says
Python 2.6.2 (r262:71605, Apr 14 2009, 22:40:02) [MSC v.1500 32 bit (Intel)] on win32
What I do not understand is this MSC V.1500 designation. Does it mean it is compiled with Visual Studio 2005? I cannot find this information on http://python.org.
MSC v.1500 appears to be Visual C++ 2008 according to this thread on the OpenCobol forums (of all places).
The MSDN page on Predefined Macros indicates 1500 to be the result of the _MSC_VER macro.
This other forum post mentions that
(For reference, Visual Studio 2003 has
_MSC_VER= 1310; Visual Studio 2005 has_MSC_VER= 1400; Visual Studio 2008 has_MSC_VER= 1500.)
The above MSDN link said that 1600 indicates VS2010.
Strangely, I wasn’t able to find that info about the earlier _MSC_VER values on MSDN.
| Visual C++ version | _MSC_VER |
|---|---|
| Visual C++ 4.x | 1000 |
| Visual C++ 5 | 1100 |
| Visual C++ 6 | 1200 |
| Visual C++ .NET | 1300 |
| Visual C++ .NET 2003 | 1310 |
| Visual C++ 2005 (8.0) | 1400 |
| Visual C++ 2008 (9.0) | 1500 |
| Visual C++ 2010 (10.0) | 1600 |
| Visual C++ 2012 (11.0) | 1700 |
| Visual C++ 2013 (12.0) | 1800 |
| Visual C++ 2015 (14.0) | 1900 |
| Visual C++ 2017 (15.0) | 1910 |
| Visual C++ 2017 (15.3) | 1911 |
| Visual C++ 2017 (15.5) | 1912 |
| Visual C++ 2017 (15.6) | 1913 |
| Visual C++ 2017 (15.7) | 1914 |
| Visual C++ 2017 (15.8) | 1915 |
| Visual C++ 2017 (15.9) | 1916 |
| Visual C++ 2019 RTW (16.0) | 1920 |
| Visual C++ 2019 (16.1) | 1921 |
| Visual C++ 2019 (16.2) | 1922 |
| Visual C++ 2019 (16.3) | 1923 |
| Visual C++ 2019 (16.4) | 1924 |
| Visual C++ 2019 (16.5) | 1925 |
| Visual C++ 2019 (16.6) | 1926 |
| Visual C++ 2019 (16.7) | 1927 |
| Visual C++ 2019 (16.8) | 1928 |
| Visual C++ 2019 (16.9) | 1928 |
| Visual C++ 2019 (16.10) | 1929 |
| Visual C++ 2019 (16.11) | 1929 |
| Visual Studio 2022 RTW (17.0) | 1930 |
Source: the documentation for the _MSC_VER predefined macro