Win form opening position change on show

Question

I am trying to open winform on secondary screen with code below

ff.FormBorderStyle = FormBorderStyle.None;
ff.Left = sc[1].Bounds.Left; value 1366
ff.Top = sc[1].Bounds.Top;
ff.Height = sc[1].Bounds.Height;
ff.Width = sc[1].Bounds.Width;

Console.WriteLine(ff.Left); // output 1366

ff.Show();

Console.WriteLine(ff.Left); // output 50

Why value got changed on show and because of that form open in primary screen. Even in Visual studio property window when i try to change x,y parameter of location, values updated to old values once i leave property cell. What can be the issue.

duplicate to http://stackoverflow.com/questions/1363374/showing-a-windows-form-on-a-secondary-monitor — AsfK, Oct 21 '15 at 10:37
If you care about the position of the window on the screen then you *must* use the Load event. When it fires, the small changes to the window size due to user preferences are applied. And the changes due to the video DPI setting, forcing the form to rescale. Big changes. Only then can you align the window correctly. The StartPosition property is irrelevant. One of the few good reasons to use the Load event btw. — Hans Passant, Oct 21 '15 at 11:02

Huntt · Answer 1 · 2015-10-21T11:03:38.343

1

Try setting StartUpLocation parameter as "manual" for the ff form.

edited Oct 21 '15 at 11:03

answered Oct 21 '15 at 10:25

Huntt

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i am not able to find windowstartuplocation property in winform. – Hot Cool Stud Oct 21 '15 at 10:32

score 1 · Accepted Answer · answered Oct 21 '15 at 10:38

1

Set the StartPosition property of the form ff to Manual

answered Oct 21 '15 at 10:38

Ramesh Babu

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Win form opening position change on show

2 Answers2