Why can I not get a substring of a delayed expansion variable in an if statement?

Question

Why does string-manipulation work inline in an if statement without using delayed expansion variables - but fails otherwise. For example:

set test=testString         


        

Answer 1

,;= and are delimiters for cmd.exe and in many cases are ignored if they are not in quotes and act like empty space.Probably in this case , is taken as the end of the first operand and IF is surprised that there is not valid comparison operator e.g. this will print yep :

if a   ;==;,,=a echo yep

(but will not work if there's a equal sign in the first part of the operands)

But this will not:

if "a ;" == ";,,=a" echo yep

so to make a valid IF expression when you use comma you need quotes.And this will work

setLocal enableDelayedExpansion
set test=testString
if "!test:~0,4!" == "test" echo Success

Without delayed expansion substitution is made immediately and this will work without quotes:

set test=testString
setlocal disableDelayedExpansion
if %test:~0,4% == test echo Succes
endlocal

For the same reason this will be taken as a wrong syntax expression (see jeb's comment):

set "test="
setlocal disableDelayedExpansion
if %test% == test echo Succes
endlocal

May be is not the complete answer , but should be close.As both echo !test:~0,4! and echo %test:~0,4% will work without quotes stays the question why exactly IF fails - may because IF command uses its own parser

As a conclusion - it's always good to use quotes when you compare strings with IF :

1. with delayed expansion commas and semicolons will cause troubles.
2. without delayed expansion undefined variables will cause troubles.