title: c++ how to initialize open file dialog (GetOpenFileName)
body: <p>my code is as follows.</p>
  
  <pre><code>OPENFILENAMEA open;
  ZeroMemory(&amp;open, sizeof(open));
  
  open.lStructSize = sizeof(LPOPENFILENAMEA);
  open.lpstrFilter = "Képek\0*.jpg;*.jpeg;*.gif;*.png;*.bmp\0\0";
  open.nFileOffset = 1;
  open.lpstrFile[0] = '\0';
  open.nMaxFile = 2048;
  open.lpstrTitle = "Képek kiválasztása..";
  open.Flags = OFN_ALLOWMULTISELECT | OFN_FILEMUSTEXIST;
  
  OPENFILENAME selected = GetOpenFileName(open);
  </code></pre>
  
  <p>My problem is, that I get the following error when trying to build:
  <code>error: cannot convert 'OPENFILENAME {aka tagOFNA}' to 'LPOPENFILENAMEA {aka tagOFNA*}' for argument '1' to 'BOOL GetOpenFileNameA(LPOPENFILENAMEA)'</code> when I call <code>GetOpenFileName</code></p>
  
  <p>If I call it with the open parameter as a ptr <code>GetOpenFileName(&amp;open)</code> I get the following <code>error: conversion from 'BOOL {aka int}' to non-scalar type 'OPENFILENAME {aka tagOFNA}' requested</code></p>
  
  <p>Question: what do?</p>