title: How can I build ARMv8 aarch32 binary?
body: <p>I'm using Mediatek X20 dev board and <code>gcc-linaro-6.3.1-2017.05</code> version
  for benchmark <code>aarch64</code> vs <code>aarch32</code>.</p>
  
  <p>So I want to build binaries as 2 types of <code>aarch32</code> abi : <code>lp64</code>, <code>ilp32</code>.</p>
  
  <p>From gcc manual, (@ [toolchain_path]/share/doc/gcc/AArch64-Options.html)</p>
  
  <p>It says I can choose <code>ilp32</code> and <code>lp64</code> with <code>-mabi</code> option.</p>
  
  <p>But when I use <code>-mabi=lp64</code> or  <code>-mabi=ilp32</code>, <code>gcc</code> makes error like below.</p>
  
  <pre><code>armv8l-linux-gnueabihf-gcc: error: unrecognized argument in option \u2018-mabi=lp64\u2019
  armv8l-linux-gnueabihf-gcc: note: valid arguments to \u2018-mabi=\u2019 are: aapcs aapcs-linux apcs-gnu atpcs iwmmxt
  </code></pre>
  
  <p>How can I make <code>ilp32</code> and <code>lp64</code> formated binary?</p>
  
  <p>And How can I check the binary's format?</p>
  
  <p>Thanks for your reply :)</p>