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Load an image from a stream in C#

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Image

To load an image from a Stream, use the CreateGdPictureImageFromStream method from the GdPictureImaging class.

This method returns a non-zero GdPicture image identifier (imageID) on success. If it fails, it returns 0 — use GetStat() to view the failure reason.

Available overloads include:

  • CreateGdPictureImageFromStream(Stream)
  • CreateGdPictureImageFromStream(Stream, DocumentFormat ImageFormat)
  • CreateGdPictureImageFromStream(Stream, DocumentFormat ImageFormat, bool DirectAccess)

Notes:

  • The stream must support seeking.
  • If DirectAccess is true, only properties/metadata/thumbnail are loaded; pixel operations will fail.
  • For multipage formats (for example TIFF/GIF), keep the stream open until the image is released.

Remember to release image resources with the ReleaseGdPictureImage method.

To load an image from a stream, use the following code:

using GdPicture14;
using System;
using System.IO;


using GdPictureImaging gdpictureImaging = new GdPictureImaging();


// Create a stream object from an image file.
using Stream streamImage = new FileStream(@"C:\temp\source.jpg", FileMode.Open, FileAccess.Read);


// Create a GdPicture image from the stream object.
int imageID = gdpictureImaging.CreateGdPictureImageFromStream(
    streamImage,
    DocumentFormat.DocumentFormatJPEG);


if (imageID == 0)
{
    Console.WriteLine($"CreateGdPictureImageFromStream failed: {gdpictureImaging.GetStat()}");
    return;
}


GdPictureStatus status = gdpictureImaging.SaveAsPNG(imageID, @"C:\temp\output.png");
if (status != GdPictureStatus.OK)
{
    Console.WriteLine($"SaveAsPNG failed: {status}");
}


gdpictureImaging.ReleaseGdPictureImage(imageID);