How do I change the terminal inside Visual Studio code to use the non-Rosetta one i.e. have it use the arm64 one?

Question

I am new to python and am trying to run a python 2.7 script. Got pip for python 2.7 and installed a dependency of pyCrypto from the mac terminal shell.

The downloaded python script, I want to try, runs fine in the terminal app when I execute it using python2.

Now I open it in vscode and try to run the script in its terminal and I get

ImportError: dlopen(/Users/xxx/Library/Python/2.7/lib/python/site-packages/Crypto/Cipher/_DES3.so, 2): no suitable image found.  Did find:
        /Users/xxx/Library/Python/2.7/lib/python/site-packages/Crypto/Cipher/_DES3.so: mach-o, but wrong architecture
        /Users/xxx/Library/Python/2.7/lib/python/site-packages/Crypto/Cipher/_DES3.so: mach-o, but wrong architecture

When I run uname -m inside a vscode terminal(zsh) on an M1 Mac, I see an output of x86_64, implying the terminal is running under Rosetta and looking for the intel version of the library.

And when I run uname -m in the regular mac terminal app, I see arm64

How do I change the terminal inside vscode to use the non rosetta one? Or how do I get the script to run from within vscode?

Answer 1

I'm not familiar with VSCode, but you can manually force the chosen architecture slice of anything you launch with the arch command (see man arch).

If you have a script that you'd normally launch like:

./script.py

Then you can force either architecture like so:

arch -x86_64 ./script.py
arch -arm64 ./script.py