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Home/ Questions/Q 980045
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T04:19:02+00:00 2026-05-16T04:19:02+00:00

I am attempting to open the current process’s executable file for read-write operations (I

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I am attempting to open the current process’s executable file for read-write operations (I have additional data attached to the executable), however std::fstream will not open the file in ios::in | ios::out | ios::binary mode, even though open() will (with O_RDWR flag set).

Does anyone know why std::fstream will not open the file, while open() will, and if so, how I can get std::fstream to open the file?

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  1. Editorial Team
    Editorial Team
    2026-05-16T04:19:02+00:00Added an answer on May 16, 2026 at 4:19 am

    It most probably has to do with file sharing semantics. See this thread which deals with a similar question – and the answer there is “The concept of file protection, file sharing, file permissions is OS-specific, which is why it is not covered by standard C++”.

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