简单问题!
if OpenPictureDialog1.Execute() then
begin
filename:=OpenPictureDialog1.FileName ;
Image1.Picture.LoadFromFile(filename);
a:=ExtractFilePath(OpenPictureDialog1.FileName);

b:=ExtractFilePath(Application.Name)+OpenPictureDialog1.FileName;
CopyFile(pchar(a),pchar(b),false)
图片复制我这样写的,但是程序不能复制,我哪里错 了,我就是想打开的图片复制到程序的文件里去啊!

[解决办法]
a:=OpenPictureDialog1.FileName;

b:=ExtractFilePath(Application.ExeName)+ExtractFileName(OpenPictureDialog1.FileName);