Passing percent sign in a batch parameter CALL

Question

On Windows 7....

This SO Answer seems to be in error for me: Use of Percent Sign as a CMD Parameter

In Called.bat

echo 1=%1

Caller Bat

call called.bat "percent%%sign"

Results

percent%%sign = percent

"percent%%sign" = "percent"

percent^%sign = percent

"percent^%sign" = "percent^%sign"

percent%%%%sign= percent

percent%%%sign= percent

"percent%%%%sign" = "percent"

"percent%%%sign" = "percent"

Answer 1

caller

@ECHO OFF
SETLOCAL

CALL qq22658604 "percent%sign"
CALL qq22658604 "percent%%sign"
CALL qq22658604 "percent%%%sign"
CALL qq22658604 "percent%%%%sign"
CALL qq22658604 percent%sign
CALL qq22658604 percent%%sign
CALL qq22658604 percent%%%sign
CALL qq22658604 percent%%%%sign

GOTO :EOF

qq22658604

@ECHO OFF
ECHO 1=%1
GOTO :EOF

results

1="percentsign"
1="percentsign"
1="percentsign"
1="percent%sign"
1=percentsign
1=percentsign
1=percentsign
1=percent%sign

NFF.

Answer 2

When I use a second.bat with

@echo off
echo %~1

There are two cases now.
If you call second.bat from a batch file, the CALL line in the caller batch needs to be
call called.bat "percent%%%%sign"
You could omit the quotes, as long as there are no spaces inside the parameter

If you call the called.bat directly from the cmd-prompt you can use
call called.bat "percent%sign"

Answer 3

Ok, this one is my bad. The issue is caused by cmd shell software called "Take Command" v13 by JP Software. The echo command is behaving differently. I have no insight on how to get it working correctly at this time.