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 | Hello people !
say we, I have one BMP(A) and a neckline BMP(B).
now would like I to check on whether, B somewhere in A vorkommt.
can I with the the ProSpeed.Dll somehow release?
with freundlichem greeting,
Alexander Schönfeld |
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| Hello Alexander,
there have you got you but a gewaltige task set, I know no Program or Dll, The the could...
but a idea have I still (Yes, with ProSpeed):
Verwandle both Images into ByteArray via InitExtFX(). you have now two lever for two byte Arrays, means quasi two reaches with Grafikdaten. The zurückgegebenen lever are Adressen to a structure, on its Offset 40 the Zeiger on the actual Grafikdaten standing (LongInt). The must You in a variable Save...
arrayadresse1&=Long(bytearray1&,40)
arrayadresse2&=Long(bytearray2&,40)
and Please mind, the The Graphics of system (paradoxerweise) of left under to right supra stored go (!)
on Offset 20 standing incidentally one LongInt, the angibt, wieviele Bytes in the jeweiligen Bytearray stand. These Info do you need ditto.
now can you FindBytes() utilize. best can You the first 16 Bytes the Bildausschnitts in the large Image search. If the Ergebniss here negative was, is the Bildausschnitt nirgendwo in the Image present. If it positively was, copy ex this place in the Size the Bildausschnitts these place in a new HDC and of there can You it again into byte-aray transfiguring. now can you both Bytesarrays (The Yes now The same Size having) with CompareBytes() vergleichen.
If the Ergebniss positively was, then is the Bildausschnitt objectively in the Image.
not time so simply, or?
But still a relatively speedy Solution.
alternatively can you naturally too with GetPixel() / SetPixel() works.
I hope, You get the there...
Greeting, Frank |
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