Set a variable to the contents of another variable

Question

I am attempting to create a program that will essentially "Emulate" variables.

I tried testing this issue with this code:

@echo off
set a=b
set b=c
:: The program should echo %b% because it is expanded with double percentages
echo %%a%%

the Program only echoes "%%a%%", instead of expanding to %b%, which should result in the program echoing C.

What am I doing wrong?

This management is described at [this answer](https://stackoverflow.com/questions/10166386/arrays-linked-lists-and-other-data-structures-in-cmd-exe-batch-script/10167990#10167990), although the topic is different... — Aacini, Dec 07 '18 at 14:19

score 5 · Accepted Answer · answered Dec 07 '18 at 06:03

5

You can use call and use a doubled set of percent signs to get a value of c.

call echo %%%a%%%

Alternatively, you can put setlocal enabledelayedexpansion at the top of the script and use echo !%a%!

answered Dec 07 '18 at 06:03

SomethingDark

That still doesn't do what I am asking. When I echoed it, it echoes !b!, instead of expanding !b! into c, which is the contents of b. – ZeekPlayz Dec 07 '18 at 06:29
2

Then you did it wrong. https://i.imgur.com/ALbagx1.png You remembered to enable delayed expansion, right? – SomethingDark Dec 07 '18 at 06:32
Yes. I did enable expansion. – ZeekPlayz Dec 08 '18 at 07:37
If your code looks exactly like it does in the picture I posted, it will work and `c` will get displayed. – SomethingDark Dec 08 '18 at 12:42

score -2 · Answer 2 · answered Dec 07 '18 at 05:43

-2

It appears the formal definition is quite vague:

May not always be required [to be escaped] in doublequoted strings, just try

Therefore, try unescaped:

 echo %a%

answered Dec 07 '18 at 05:43

Shep

I am attempting to echo %b% by echoing the value of %a% like %%a%%, where it will expand the INNER %a% to b, but keep the out percent symbols, which are then expanded to the contents of B. – ZeekPlayz Dec 07 '18 at 05:55

2 Answers2