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
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
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"If the Earth were a single state, Istanbul would be its capital." Napoleon Bonaparte
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:
DT
Hi,TaznTex
As it looks close to logics ... Could you tell what kind of variables are "a" to "l" ???
Alain
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Why insist on using 32 Bits when you're not even able to deal with the first 8 ones ??? ehhhhhh ...
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IF there is the word "Problem" in your question ...
certainly the answer is " RTFM " or " RTFDataSheet " !!!
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