Bash script variable placement

Question

I started to learn Bash scripting and I'm using Bash scripting tutorial

There it says

Before Bash interprets (or runs) every line of our script it first checks to see if any variable names are present. For every variable it has identified, it replaces the variable name with its value. Then it runs that line of code and begins the process again on the next line.

So does Bash first run through the whole script to find variables? I'm not sure whether this is what the author tried to say but if yes I guess it is not correct?

when I execute:

#!/bin/bash echo "hello $USERR" USERR=John 

I get helloas result.

If I run:

#!/bin/bash USERR=John echo "hello $USERR" 

then i get hello John as result.

Answer 1

So does Bash first run through the whole script to find variables?

Nope. As you yourself discovered in your example, Bash scripts are executed from top to bottom.

A good practice is to define all variables that you need at the top of your script.

Answer 2

That's a very sloppy way of saying that the shell scans each command for expansions, such as for example variables (but also command substitutions etc.)

The text could be interpreted to imply that the shell reads the whole script and looks for variables on every line before execution. That is not so. It processes the script command by command, not line by line. A command could span several lines. A command is not processed until the shell interpreter gets to it in its execution of the script.

The bash shell does the following with each command before executing it:

1. brace expansion
2. tilde expansion
3. parameter and variable expansion
4. arithmetic expansion
5. command substitution (done in a left-to-right fashion)
6. word splitting
7. pathname expansion
8. quote removal