Using Batch file SHIFT command [duplicate]

Question

I'm trying to display up to 9 parameters across the screen, and then display one fewer on each following line, until there are none left.

I tried this:

@echo off
echo %*
shift
echo %*
shift
echo %*

Actual Result:

   a b c d e f
   a b c d e f

Expected Result:

A B C D E F
B C D E F
C D E F
D E F
E F
F

Any help?

Thanks.

Answer 1

SHIFT is worthwhile if you want to get the value of %1, %2, etc. It doesn't affect %*. This script gives you the output you expect.

@echo off
setlocal enabledelayedexpansion

set args=0
for %%I in (%*) do set /a "args+=1"
for /l %%I in (1,1,%args%) do (
    set /a idx=0
    for %%a in (%*) do (
        set /a "idx+=1"
        if !idx! geq %%I set /p "=%%a "<NUL
    )
    echo;
)

Output:

C:\Users\me\Desktop>test a b c d e f
a b c d e f
b c d e f
c d e f
d e f
e f
f