AJUDA C# TO VB.NET

Public Function Resize(source As String, target As String) As Boolean
	Using src As Image = Image.FromFile(source, True)
		[ô] Check that we have an image
		If src IsNot Nothing Then
			Dim origX As Integer, origY As Integer, newX As Integer, newY As Integer
			Dim trimX As Integer = 0, trimY As Integer = 0

			[ô] Default to size of source image
			newX = InlineAssignHelper(origX, src.Width)
			newY = InlineAssignHelper(origY, src.Height)

			[ô] Does image exceed maximum dimensions?
			If origX > MaxX OrElse origY > MaxY Then
				[ô] Need to resize image
				If TrimImage Then
					[ô] Trim to exactly fit maximum dimensions
					Dim factor As Double = Math.Max(CDbl(MaxX) / CDbl(origX), CDbl(MaxY) / CDbl(origY))
					newX = CInt(Math.Ceiling(CDbl(origX) * factor))
					newY = CInt(Math.Ceiling(CDbl(origY) * factor))
					trimX = newX - MaxX
					trimY = newY - MaxY
				Else
					[ô] Resize (no trim) to keep within maximum dimensions
					Dim factor As Double = Math.Min(CDbl(MaxX) / CDbl(origX), CDbl(MaxY) / CDbl(origY))
					newX = CInt(Math.Ceiling(CDbl(origX) * factor))
					newY = CInt(Math.Ceiling(CDbl(origY) * factor))
				End If
			End If

			[ô] Create destination image
			Using dest As Image = New Bitmap(newX - trimX, newY - trimY)
				Dim graph As Graphics = Graphics.FromImage(dest)
				graph.InterpolationMode = System.Drawing.Drawing2D.InterpolationMode.HighQualityBicubic
				graph.DrawImage(src, -(trimX / 2), -(trimY / 2), newX, newY)
				dest.Save(target, SaveFormat)
				[ô] Indicate success
				Return True
			End Using
		End If
	End Using
	[ô] Indicate failure
	Return False
End Function

Citação:

:

Public Function Resize(source As String, target As String) As Boolean
	Using src As Image = Image.FromFile(source, True)
		[ô] Check that we have an image
		If src IsNot Nothing Then
			Dim origX As Integer, origY As Integer, newX As Integer, newY As Integer
			Dim trimX As Integer = 0, trimY As Integer = 0

			[ô] Default to size of source image
			newX = InlineAssignHelper(origX, src.Width)
			newY = InlineAssignHelper(origY, src.Height)

			[ô] Does image exceed maximum dimensions?
			If origX > MaxX OrElse origY > MaxY Then
				[ô] Need to resize image
				If TrimImage Then
					[ô] Trim to exactly fit maximum dimensions
					Dim factor As Double = Math.Max(CDbl(MaxX) / CDbl(origX), CDbl(MaxY) / CDbl(origY))
					newX = CInt(Math.Ceiling(CDbl(origX) * factor))
					newY = CInt(Math.Ceiling(CDbl(origY) * factor))
					trimX = newX - MaxX
					trimY = newY - MaxY
				Else
					[ô] Resize (no trim) to keep within maximum dimensions
					Dim factor As Double = Math.Min(CDbl(MaxX) / CDbl(origX), CDbl(MaxY) / CDbl(origY))
					newX = CInt(Math.Ceiling(CDbl(origX) * factor))
					newY = CInt(Math.Ceiling(CDbl(origY) * factor))
				End If
			End If

			[ô] Create destination image
			Using dest As Image = New Bitmap(newX - trimX, newY - trimY)
				Dim graph As Graphics = Graphics.FromImage(dest)
				graph.InterpolationMode = System.Drawing.Drawing2D.InterpolationMode.HighQualityBicubic
				graph.DrawImage(src, -(trimX / 2), -(trimY / 2), newX, newY)
				dest.Save(target, SaveFormat)
				[ô] Indicate success
				Return True
			End Using
		End If
	End Using
	[ô] Indicate failure
	Return False
End Function

graph.DrawImage(src, -(trimX / 2), -(trimY / 2), newX, newY);

graph.DrawImage(src, -(trimX \ 2), -(trimY \ 2), newX, newY)

Citação:

:
O problema é um pequeno erro na conversão do KERPLUNK, o problema não é ter que usar double e sim que ele espera que seja integer mas está recebendo um doublem

No C# está assim

graph.DrawImage(src, -(trimX / 2), -(trimY / 2), newX, newY);

Veja que trimX é um inteiro e divisão de inteiro por um inteiro sempre resulta em um inteiro no C#

No VB.Net divisão feita com o operador [Ô]/[Ô] resulta em um double, para ter o mesmo comportamento do C# neste caso é necessário usar o operador de divisão de inteiros, o [Ô][Ô], ficando

graph.DrawImage(src, -(trimX  2), -(trimY  2), newX, newY)