How can I launch my application automatically on system reboot?

Rohan Pande 445 Reputation points
2024-05-02T13:09:11.53+00:00

Is there an API or method that I can use to make sure my application launches automatically when the system restarts? I would appreciate any help or guidance on this matter.

Windows development Windows API - Win32
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  1. guilherme rodrigues 390 Reputation points
    2024-05-02T13:28:49.5233333+00:00

    Hi, Rohan!

    The approach varies depending on the operating system your application is targeting.

    Windows

    For Windows, you can use the Windows Registry or the Startup folder to launch your application automatically:

    • Startup Folder Method:
      • You can place a shortcut to your application in the Startup folder. Press Win+R, type shell:startup, and press Enter to open the Startup folder. Drop your application shortcut here.
    • Windows Registry Method:
      • You can add a new key to the registry to start your application. You need to modify HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Run via a script or manually using the Registry Editor (regedit). Add a new String Value with the path to your application executable.

    Linux

    For Linux systems, depending on the distribution, you can use systemd, init.d, or cron @reboot:

    • systemd (most common for newer distributions):
      • Create a .service file in /etc/systemd/system with the specifications for your application. Example of a systemd service file:
    iniCopy code
    [Unit]
    

    Enable the service to start at boot: sudo systemctl enable service-name.service

    • cron @reboot (universal but less robust):
      • Edit the crontab with crontab -e and add the following line:
            @reboot /path/to/application
        

    Each method has its own merits and depends on your specific needs and the system configurations. If you need more detailed steps or have a specific setup in mind, feel free to ask! Linux

    For Linux systems, depending on the distribution, you can use systemd, init.d, or cron @reboot:

    • systemd (most common for newer distributions):
      • Create a .service file in /etc/systemd/system with the specifications for your application. Example of a systemd service file:
    [Unit]
    

    Enable the service to start at boot: sudo systemctl enable service-name.service

    • cron @reboot (universal but less robust):
      • Edit the crontab with crontab -e and add the following line:
            @reboot /path/to/application
        

    Each method has its own merits and depends on your specific needs and the system configurations. If you need more detailed steps or have a specific setup in mind, feel free to ask!

    1 person found this answer helpful.

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  1. RLWA32 49,536 Reputation points
    2024-05-02T13:24:29.77+00:00
    1 person found this answer helpful.

  2. Xiaopo Yang - MSFT 12,731 Reputation points Microsoft External Staff
    2024-05-03T02:12:08.4366667+00:00

    Hello @Rohan Pande,

    As @guilherme rodrigues answered, you can add a new key to the registry HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\Run to start your application. If you want to do it and enable/disable programmatically, use Registry API. For more information, see https://stackoverflow.com/questions/66903672/how-to-enable-disable-windows-startup-items-programmatically


  3. RLWA32 49,536 Reputation points
    2024-05-03T07:16:26.22+00:00

    Since it appears that your intent is to start your application every time the system starts or a user logs in you may also use a scheduled task for these purposes. A scheduled task that runs when the system boots will start the application in non-interactive session 0. A scheduled task that runs at user logon will start the application in the user's interactive session.

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