bash printf : how to understand zero dot s ("%0.s") syntax

Question

使用 Bash printf，我不知道如何理解以下代码的语法。

echo $(printf "%0.s../"  2  ); 
# ../
echo $(printf "%0.s../"  );   
# ../
echo $(printf "%0.s../" 1 );
# ../

从 man printf 读取，

...

s
Bytes from the string argument are printed until the end is reached or until the number of bytes indicated by the precision specification is reached; however if the precision is 0 or missing, the string is printed entirely.

...

我还是不知道上面的代码。

为什么会重复打印以下代码？

# repeating character 
echo $(printf "%0.s../" 1 2  );
# ../../

从https://wiki.bash-hackers.org/commands/builtin/printf看

...

. The dot: Together with a field width, the field is not expanded when the text is longer, the text is truncated instead. "%.s" is an undocumented equivalent for "%.0s", which will force a field width of zero, effectively hiding the field from output

...

还没看懂。我需要一些解释。

Answer 1

这个例子可以阐明宽度字段的工作原理

printf '%.0s../\n' ABCDEFG
../
printf '%.1s../\n' ABCDEFG
A../
printf '%.2s../\n' ABCDEFG
AB../
printf '%.7s../\n' ABCDEFG
ABCDEFG../

当宽度为0时，则根本不打印该字段。

然后，如果参数多于字段，printf 重复格式

printf '%.7s../\n' ABCDEFG abcdefg
ABCDEFG../
abcdefg../