>      echo $$(( 4 < 2 ) && ( 5 < 6 ))

For this, You do not really need the inner parentheses, so throw them away try: echo $$(4<2 && 5<6)

HTH,
Michael.

Well, the permutations attempting to reverse-engineer a not-intuitive system approaches infinity.

Tried this:

	echo "Logic test follows"
	echo $(or 1,0)
	echo $(or 0,0)
	echo "Or 0 1"
	echo $(or 0,1)
	echo "Or 1 1"
	echo $(or 1,1)
	echo $(and 0,0)
	echo $(and 1,0)
	echo $(and 0,1)
	echo $(and 1,1)
	echo $(xor 0,0)	
	echo $(xor 0,1)	
	echo $(xor 1,0)	
	echo $(xor 1,1)	

Get this:
echo "Logic test follows"
Logic test follows
echo 1
1
echo 0
0
echo "Or 0 1"
Or 0 1
echo 0
0
echo "Or 1 1"
Or 1 1
echo 1
1
echo 0
0
echo 0
0
echo 1
1
echo 1
1
echo

echo

echo

echo


Note:
Or 0 1
echo 0
0
echo "Or 1 1"
Or 1 1
echo 1
1

So whatever or does, it isn't or.

xor doesn't do anything other than nullify input apparently.


Tried or without comma:

	echo $(or 1 1)	
-->
echo 1 1
1 1

These aren't errors, nor are they boolean operations.

In lieu of that, trying the daisy-chain route :(



QCCMAJORVER := 6
M3 = $$(( $(QCCMAJORVER) > 2 ))
...
(stuff)
	echo "Hello, world!"
	echo $(QCCMAJORVER)
	echo "$(M3)"
	echo $(M3)
ifeq ($(M3),1)
	echo "New"
else
	echo "Old"
endif
	echo "End"


Gives

echo "Hello, world!"
Hello, world!
echo 6
6
echo "$(( 6 > 2 ))"
1
echo $(( 6 > 2 ))
1
echo "Old"
Old
echo "End"
End



Why does 1 not equal 1, causing it to print Old instead of New?  Is it comparing something like a string of "$(( 6 > 2 
))" to 1 instead of evaluating it?  What forces it to evaluate in an ifeq lline?