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(base) C:\Users\Abj>python
Python 3.6.5 |Anaconda, Inc.| (default, Mar 29 2018, 13:32:41) [MSC v.1900 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> exit()

(base) C:\Users\Abj>py
Python 3.7.0 (v3.7.0:1bf9cc5093, Jun 27 2018, 04:06:47) [MSC v.1914 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> exit()

What is the difference between these two lines, MSC v.1900 64 bit (AMD64) and MSC v.1914 32 bit (Intel)

Damodara Sahu
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1 Answers1

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The MSC v.1900 64 bit (AMD64) and MSC v.1914 32 bit (Intel) are the compiler versions used to compile the respective Python interpreters.

MSC stands for Microsoft C compiler (of Visual Studio); and Intel is a code name for 32-bit Intel Architecture, whereas AMD64 is the code name for 64-bit version of the same architecture since AMD was the first to introduce that - calling it "Intel64" at the time would not have been unambiguous because Intel had introduced an incompatible line of processors called IA-64 / Intel Itanium architecture, which no one uses because it is not compatible with the 32-bit architecture :D. Itanium is also less affectionally called Itanic which quite well describes its success.

The compiler versions usually don't matter that much but the bitness does matter. The 32-bit Python is not able to to use more than maybe 4 gigabytes of RAM in one program, the 64-bit one is.

Glenn Slayden
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  • Great answer; although I've long known about the failed IA-64 saga, I hadn't heard the hilarious joke. A follow up... is Python commonly built for Windows using non Microsoft compiler(s)? – Glenn Slayden Oct 20 '22 at 21:23