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  <blog_title>higepon blog</blog_title>
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    <anon>Mona</anon>
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  <description>またまた実験 00000000 55 push ebp 00000001 89E5 mov ebp,esp 00000003 B878563412 mov eax,0x12345678 00000008 FFD0 call eax 0000000A C9 leave 0000000B C3 retということはきっとこんなことが出来るはず void hello() { printf(&quot;hello\n&quot;); } int main(int argc, char *argv[]) { unsigned char code[] = {0x55, 0x89, 0xE5, 0xB8, 0x00, 0x00,…</description>
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  <provider_name>Hatena Blog</provider_name>
  <provider_url>https://hatena.blog</provider_url>
  <published>2006-12-02 21:06:54</published>
  <title> DLLエントリポイント - 実験4</title>
  <type>rich</type>
  <url>https://higepon.hatenablog.com/entry/20061202/1165061214</url>
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