Expected Input: Start=1, End=500, Interval=100
Expected Output:
1, 100
101, 200
201, 300
301, 400
401, 500
@echo off
set start=1
set end=500
set interval=100
for /L %%g in (%start%, %interval%, %end%) do (
set first=%%g
set /a last=%first% -1 + %interval%
echo %first% , %last%
)
Actual output:
401 , 401 -1 + 100
401 , 401 -1 + 100
401 , 401 -1 + 100
401 , 401 -1 + 100
401 , 401 -1 + 100
Not sure why first is assigned with 401 value, Appreciate any help
This is a classic delayed expansion issue. Type help set or set /? from the command prompt for a discussion about delayed expansion (a bit past half way down the full help).
Normal expansion using %var% occurs when the line is parsed. The problem is your entire FOR loop block is parsed in one pass. So you are seeing a constant value that existed before the loop started. The value of 401 is probably left over from a prior run.
The solution is delayed expansion. First you must enable it by using setlocal enableDelayedExpansion. Then you use !var! instead of %var%. This will give the value at execution time instead of at parse time.
Also, there is no need to expand numeric variables when using set /a.
@echo off
setlocal enableDelayedExpansion
set start=1
set end=500
set interval=100
for /L %%g in (%start%, %interval%, %end%) do (
set /a first=%%g, last=first-1+interval
echo !first!, !last!
)
@dbenham is right on. I'll throw out an alternative. I find the delayed expansion a bit ugly and confusing for longer subroutines, so I'll tend to have a lone call as the "body" of a for loop. Inside the call variable expansion happens "normally". Observe:
@echo off
setlocal
set start=1
set end=500
set interval=100
for /L %%g in (%start%, %interval%, %end%) do call :_d %%g
endlocal
goto :EOF
:_d
set /a last=%1 - 1 + interval
echo %1, %last%
For this particular case it's more code, to be sure, but for less trivial batch files I think it's a "win".
As an aside: You have no setlocal / endlocal in there, so the values for first and last will persist across executions, assuming you keep running them in the same shell, and give you different results the second time you run your code. I tend to wrap my batch files in a setlocal / endlocal pair to prevent variables I use in the script from "leaking" out into the parent shell's environment.
Finally, I'll echo what @VasiliSyrakis says: If you can avoid using cmd.exe then, by all means, avoid it. I tend to write a lot of stuff in batch but, admittedly, it's a bit like hammering nails with a screwdriver for a lot of problems.
