It seems to me the input to your function is a string, not a filehandle.

Now, with some symbolic trickery, you can create a bareword filehandle with the given name, but it seems utterly weird to write a function that gets passed the name of a variable it should use.

So, I suggest you further elaborate on the purpose of your input variable. What is $File_Handle = "File_Handle"; supposed to mean?

In reply to Re: File handle as a input variable by JavaFan
in thread File handle as a input variable by hem

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