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Comment by JackSlateur

1 month ago

What is the meaning of this code ?

  void on_sigchld(int sig) { (void)sig; }

If it's C code, that is the way to suppress a compiler warning about sig being unused. In C++ you can omit (or comment-out) the parameter name, e.g.:

    // C++
    void on_sigchld(int /*sig*/) {}