China DOS Union

There is an example like this in the help:
set VAR=before
if "%VAR%" == "before" (
set VAR=after
if "%VAR%" == "after" @echo If you see this, it worked
)
It won't display anything. The explanation in the help says this:

The message will not be displayed, because when the first IF statement is read, the %VAR% in BOTH IF statements
is replaced; the reason is: it is within the body of the IF, and IF is a
compound statement. So, the IF inside the compound statement is actually comparing "before" and
"after", and those two can never be equal.
I don't really understand what that means. Could the experts explain it again?

OP can search for the newbie tutorial thread about delayed variables that a friend posted before, then you'll understand it.
Roughly speaking, it means that because statements like if and for, before delayed variables are enabled, once they execute, changes to variable values are not judged (this is how I understand it). Here:
if "%VAR%" == "before" (
set VAR=after(not judged)
if "%VAR%" == "after" @echo If you see this, it worked(so the value of var here is still before, and as a result nothing will be displayed)
)
You'd better read that thread yourself; I'm just explaining it according to my own understanding.

My understanding is that, when delayed variable expansion is not enabled, before each statement is executed (a compound one counts as one statement), the variables inside it are first replaced and then that statement is executed. The variable will not be updated again until that statement finishes. In the example from the help: when executing this compound if statement, %VAR% is first completely replaced with before and then executed, so the statement the system executes is this:
if "before" == "before" (
set VAR=after
if "before" == "after" @echo If you see this, it worked
)

So nothing is displayed. If you've learned C, you'll find it very similar to the preprocessor directive #define.