How bash interprets control operators

Question

To run Rust program with a backtrace one should set environment variable RUST_BACKTRACE to one and run the program, so my first guess as inexperienced bash user was:

$ RUST_BACKTRACE=1 && cargo run
...
note: Run with `RUST_BACKTRACE=1` for a backtrace.

but there are no backtrace in the output. So, let's check if variable is set:

RUST_BACKTRACE=1 && echo $RUST_BACKTRACE && cargo run
1
...
note: Run with `RUST_BACKTRACE=1` for a backtrace.

and finally working solution would be:

RUST_BACKTRACE=1 cargo run

Please explain to me how it works.

Answer 1

There are two possibilities how an external program sees that variable in its environment:

export it for that command

RUST_BACKTRACE=1 cargo run

In this case the variable is not part of the shell. If it was before then its value is not changed.

export it

The shell does not automatically export all its variables to programs it runs (not all shell variables are intended as part of an environment).

1. export RUST_BACKTRACE=1
   # or
   declare -x RUST_BACKTRACE=1
   cargo run
   

2. RUST_BACKTRACE=1
   export RUST_BACKTRACE
   # or
   declare -x RUST_BACKTRACE
   cargo run
   

3. set -a # Each  variable  or  function that is created or modified is given the export attribute
   RUST_BACKTRACE=1
   cargo run