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Strategies & Market Trends : TA-Quotes Plus
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To: Bob Jagow who wrote (6841)9/29/1998 9:13:00 AM
From: Sean W. Smith  Read Replies (2) of 11149
 
for i=numdays to 0 step 1 do
if close(i) >= maxHi then
maxHi := close(i);
maxDate := i;
i := 0;
endif;
next i;
for i=numdays to 0 step 1 do
if close(i) <= minLo then
minLo := close(i);
minDate := i;
i := 0;
endif;
next i;


Bob,

the above code will give erroneous results in the context of the scan I provided. Since MaxHi is set to 0 initially and MaxLo to 9999999 then your code will always exit on the first iteration. If one were to preset maxHi and maxLo to the actual values using the builtin in QP functions you could successfully exit the loop when these conditions were met. The way I wrote it it doesn't. I try using Min/Max to predetermine the hi/lo and kick out of the loop when its done and compare speed.

Sean
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