This seems to work:
If (a = B) && (a = c) && (a = d) && (a = e) && (a = f) && (a = g) &&(a = g) && (a = i) && (a = j) && (a = k) && (a = l) THEN
m = 1
else
m = 0
endif
But is this the only way to do this?
Thanks
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This seems to work:
If (a = B) && (a = c) && (a = d) && (a = e) && (a = f) && (a = g) &&(a = g) && (a = i) && (a = j) && (a = k) && (a = l) THEN
m = 1
else
m = 0
endif
But is this the only way to do this?
Thanks
There are as many ways as you have imagination...
That one kinda looks pretty...Code:m=0
If (a = B) then
if (a = c) then
if (a = d) then
if (a = e) then
if (a = f) then
if (a = g) then
if (a = h) then
if (a = i) then
if (a = j) then
if (a = k) then
if (a = l) then m=1
endif
endif
endif
endif
endif
endif
endif
endif
endif
endif
Could this work?Code:IF (a*a*a*a*a*a*a*a*a*a*a) = (a + b+ c +d +e + f +g +h +i +j +k +l) THEN DO
_____________-
I initially thouhgt doing the addition thing... but it wouldn't work...
(a*11)=(b+c+d+e+f+g+h+i+j+k+l)
looks fine untill you consider the equation would still be true if one of the values was less than a, whilst another of the values was greater than a by the same amount.
Another way ...Code:Idx VAR BYTE
Vars VAR BYTE[12]
a VAR Vars[0]
b VAR Vars[1]
c VAR Vars[2]
d VAR Vars[3]
e VAR Vars[4]
f VAR Vars[5]
g VAR Vars[6]
h VAR Vars[7]
i VAR Vars[8]
j VAR Vars[9]
k VAR Vars[10]
l VAR Vars[11]
m VAR BYTE
For Idx = 1 to 11
IF a != Vars[Idx] THEN m = 0 : GOTO VarTestDone
NEXT Idx
m = 1
VarTestDone:
Hi,TaznTex
As it looks close to logics ... Could you tell what kind of variables are "a" to "l" ???
Alain