shell - What is the difference between an inline variable assignment and a regular one in Bash?

Question

What is the difference between:

prompt$   TSAN_OPTIONS="suppressions=/somewhere/file" ./myprogram

and

prompt$   TSAN_OPTIONS="suppressions=/somewhere/file"
prompt$   ./myprogram

The thread-sanitizer library gives the first case as how to get their library (used within myprogram) to read the file given in options. I read it, and assumed it was supposed to be two separate lines, so ran it as the second case.

The library doesn't use the file in the second case, where the environment variable and the program execution are on separate lines.

What's the difference?

Bonus question: How does the first case even run without error? Shouldn't there have to be a ; or && between them? The answer to this question likely answers my first...

Answer 1

The format VAR=value command sets the variable VAR to have the value value in the environment of the command command. The spec section covering this is the Simple Commands. Specifically:

Otherwise, the variable assignments shall be exported for the execution environment of the command and shall not affect the current execution environment except as a side-effect of the expansions performed in step 4.

The format VAR=value; command sets the shell variable VAR in the current shell and then runs command as a child process. The child process doesn't know anything about the variables set in the shell process.

The mechanism by which a process exports (hint hint) a variable to be seen by child processes is by setting them in its environment before running the child process. The shell built-in which does this is export. This is why you often see export VAR=value and VAR=value; export VAR.

The syntax you are discussing is a short-form for something akin to:

VAR=value
export VAR
command
unset -v VAR

only without using the current process environment at all.