Gfilui

Sorry if it is an extremely stupid noob question, but I am not an expert and I cannot find any adequate response on this or other forums.

I am running Ubuntu 18.04 x64 and I'm working with Intel Compilers. In order to keep my system and the compilation of my software consistent I've set my .bashrc to set the Intel compilers as default.

source /opt/intel/parallel_studio_xe_2018.2.046/bin/psxevars.sh
export CC='/opt/intel/bin/icc'
export CXX='/opt/intel/bin/icpc'
export FC='/opt/intel/bin/ifort'
export F77='/opt/intel/bin/ifort'
export CFLAGS='-O3 -xHost -ip'
export CXXFLAGS='-O3 -xHost -ip'
export FCFLAGS='-O3 -xHost -ip'

Now, if I do this I cannot start MATLAB R2017b. Every time I start it it clashes with the Intel compilers and aborts the startup with the error:

symbol lookup error: /opt/intel/compilers_and_libraries_2018.2.199/linux/compiler/lib/intel64/libirng.so: undefined symbol: __intel_skx_avx512_memcpy

The only way I have to get MATLAB to work is to comment out the above variables in my .bashrc file, source it and start MATLAB.

This of course crazy and I was wondering if it was possible to actually have, for example, a .desktop files that ignores .bashrc or if there's any other way to have those environmental variables up and running, but have a software ignoring them.

Thanks!

It is indeed possible to start bash with ignoring the ~/.bashrc file. Specifically, via --no-rc flag:

 bash --noprofile --norc

You can make a wrapper .desktop file for that in a fairly simple way:

[Desktop Entry]
Version=1.0
Name=No-rc shell
Comment=Starts Bash with --no-rc flag
Exec=bash --noprofile --norc
Icon=/usr/share/icons/gnome/48x48/apps/terminal.png
Terminal=true
Type=Application
Categories=Utility;Development;

Alternatively, you could modify the ~/.bashrc in such way, that would allow you sourcing it dynamically. For instance, I'd personally do something like this:

if [ -z "$no_intell_compilers" ][ "$no_intell_compilers" = "false" ]; then
 source /opt/intel/parallel_studio_xe_2018.2.046/bin/psxevars.sh
 export CC='/opt/intel/bin/icc'
 export CXX='/opt/intel/bin/icpc'
 export FC='/opt/intel/bin/ifort'
 export F77='/opt/intel/bin/ifort'
 export CFLAGS='-O3 -xHost -ip'
 export CXXFLAGS='-O3 -xHost -ip'
 export FCFLAGS='-O3 -xHost -ip'
fi

What this does is that it checks for environment variable no_intell_compilers; if it is unset or set to false, ~/.bashrc will export those variables. When you want to start over without those variables in the environment, do

$ no_intell_compilers="true"
$ source ~/.bashrc

Once you're done developing in that shell, just unset no_intell_compilers and source the file again to have the variables back.

This doesn't exactly answer your question about ignoring .bashrc but I figured this could be a reasonable alternative:

Remove the statements needed for development from your ~/.bashrc and put them in a separate file instead, e.g. ~/dev-env:

#
# Set environment for icc compiler
#
source /opt/intel/parallel_studio_xe_2018.2.046/bin/psxevars.sh
export CC='/opt/intel/bin/icc'
export CXX='/opt/intel/bin/icpc'
export FC='/opt/intel/bin/ifort'
export F77='/opt/intel/bin/ifort'
export CFLAGS='-O3 -xHost -ip'
export CXXFLAGS='-O3 -xHost -ip'
export FCFLAGS='-O3 -xHost -ip'

Then, when you want to develop something with the icc compiler, simply source that file manually in the affected terminal window(s):

source ~/dev-env

The file doesn't need to be executable because you won't execute it but source it instead. This means the statements therein are processed by the current shell and hence affect the current shell. If you would make it executable and then execute it, a subshell would be spawned, the settings were set there and then the subshell would close. Not what you want.