回答編集履歴
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test
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for文を使ってみました。
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```bash
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#!/bin/sh
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for xx in $aaaaA $aaaaB; do
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CONFIG_AAAA=$xx #上書き
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done
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for xx in $iiiiA $iiiiB; do
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CONFIG_BBBB=$CONFIG_BBBB$xx
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done
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for xx in $uuuuA $uuuuB; do
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CONFIG_CCCC=$xx #上書き
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done
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```
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```bash
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38
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