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Home/ Questions/Q 97767
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Asked: May 11, 20262026-05-11T00:04:04+00:00 2026-05-11T00:04:04+00:00

Using Win32-specific APIs, is there an easy way to start an external application to

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Using Win32-specific APIs, is there an easy way to start an external application to open a file simply by passing in the path/name of the file?

For example, say I have a file called C:\tmp\image.jpg. Is there a single API that I can call to tell Windows to open the application associated with .jpg files? Without having to do a bunch of registry lookups?

I thought I remembered doing this many years ago, but I cannot find it.

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  1. 2026-05-11T00:04:05+00:00Added an answer on May 11, 2026 at 12:04 am

    ShellExecute

    Performs an operation on a specified file.

    Syntax

    C++

    HINSTANCE ShellExecute(   _In_opt_ HWND    hwnd,   _In_opt_ LPCTSTR lpOperation,   _In_     LPCTSTR lpFile,   _In_opt_ LPCTSTR lpParameters,   _In_opt_ LPCTSTR lpDirectory,   _In_     INT     nShowCmd ); 

    Parameters

    …

    nShowCmd [in]

    Type: INT

    The flags that specify how an application is to be displayed when it is opened. If lpFile specifies a document file, the flag is simply passed to the associated application. It is up to the application to decide how to handle it. These values are defined in Winuser.h…

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