Question
You have a server named Host1 that runs Windows Server 2022 and is configured as a container host. Host1 stores a container image named image1 that is based on Windows Server 2019.
You need to start a container from image1 on Host1.
How should you complete the command? To answer, select the appropriate options in the answer area.
NOTE: Each correct selection is worth one point.
Command:
[Selection 1] run -d [Selection 2] image1
Explanation
Docker: containers on a Windows Server container host are started with the Docker CLI (docker run -d ...). MsiExec installs MSI packages, Start launches processes, and Start-VM starts Hyper-V virtual machines; none of them start a container from an image.
-isolation=hyperv: Process-isolated Windows Server containers share the host kernel, so the image's OS build must match the host's. On a Windows Server 2022 host, a Windows Server 2019 image supports only Hyper-V isolation. Hyper-V isolation runs the container in a lightweight utility VM with its own matching kernel, so the version mismatch does not matter. With -isolation=process, the container fails with the error 'The operating system of the container does not match the operating system of the host'. 'container' is not an isolation value.
