What does mean: lParam | 0x80000000 in a WM_LBUTTONDOWN

[Aidoru]

Hello everyone,
as the title, what does lParam | 0x80000000 mean in a WM_LBUTTONDOWN?
I know that HI e LO lparam words specifies cursor coordinates, but what does it mean if I use an OR bitwise operator with 0x80000000?

Thank you

[2kaud]

Are you using multiple monitors? The lo-order word specifies the x co-ord and the hi-order word specifies the y co-ord. These are signed values so LOWORD and HIWORD shouldn't be used to extract them as these treat the extraction as unsigned - use GET_X_LPARAM() and GET_Y_LPARAM(). The extracted hi-word and lo-word need to be treated as signed short - as systems with multiple monitors can have negative x and y co-ords. See https://msdn.microsoft.com/en-us/lib...(v=vs.85).aspx

[Aidoru]

Are you using multiple monitors? The lo-order word specifies the x co-ord and the hi-order word specifies the y co-ord. These are signed values so LOWORD and HIWORD shouldn't be used to extract them as these treat the extraction as unsigned - use GET_X_LPARAM() and GET_Y_LPARAM(). The extracted hi-word and lo-word need to be treated as signed short - as systems with multiple monitors can have negative x and y co-ords. See https://msdn.microsoft.com/en-us/lib...(v=vs.85).aspx

Thank you for your answer,
the problem is that I don't understand the use of bitwise operator "or" (| 0x80000000)

[2kaud]

Why do you want to use the bitwise operator with lparam? You extract the x and y co-ords as

short xPos = GET_X_LPARAM(lParam); 
short yPos = GET_Y_LPARAM(lParam);

You don't need to directly manipulate lparam.

[Aidoru]

I think that too, but I found this operation in an example without any comment, and so I would like to know what does it mean

[Igor Vartanov]

Technically this sets the leftmost bit. But what use the code makes of it totally depends on the code. The code is at your disposal, so read it and see the use. Or show the code here and let us comment on it.

[2kaud]

It means its setting the MSB of lParam (bit 32). For a signed number, the MSB is the sign bit so it is creating a negative number. As a negative number is stored as 2's complement, what this number actually is depends upon what is in the other bit positions - and whether it is being treated as a 32 bit or 2 16 bit numbers. Alternatively, it might have nothing to do with being a negative number but is some special flag that is used by the program. How is lParam used elsewhere in the program?

[Igor Vartanov]

it might have nothing to do with being a negative number but is some special flag that is used by the program.

That exactly was my thought.