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将字符括在双引号中会保留引号内所有字符的字面值，但 $、`、\ 和启用历史扩展时的 ! 除外。
考虑到这一点，echo -ne "\n" 是如何产生换行符的呢？ shell 不会在传递给 echo 之前扩展“\n”吗？
我认为它可能会起作用，因为 echo 是内置的，因此 shell 足够智能，可以做正确的事情。但是，即使调用外部 /usr/bin/echo -ne "\n" 也能工作。
更奇怪的是，无论我使用双引号还是单引号 \n，以下两个命令都显示 bash 正在传递 \\n 作为参数
$ strace /usr/bin/echo "\n" 2>&1 | head -n1
execve("/usr/bin/echo", ["/usr/bin/echo", "\\n"], [/* 33 vars */]) = 0
$ strace /usr/bin/echo '\n' 2>&1 | head -n1
execve("/usr/bin/echo", ["/usr/bin/echo", "\\n"], [/* 33 vars */]) = 0
：这是怎么回事？
From bash manpage
Enclosing characters in double quotes preserves the literal value of all characters within the quotes, with the exception of $, `, \, and, when history expansion is enabled, !.
With this in mind, how is it that echo -ne "\n" produces a newline? Wouldn't the shell expand "\n" before it ever gets passed to echo?
I thought it might work because echo is a builtin and so the shell is smart enough to do the right thing. However, even calling the external /usr/bin/echo -ne "\n" works.
What's even more curious is that regardless if I double-quote or single-quote \n, the following two commands show that bash is passing \\n as the argument:
$ strace /usr/bin/echo "\n" 2>&1 | head -n1
execve("/usr/bin/echo", ["/usr/bin/echo", "\\n"], [/* 33 vars */]) = 0
$ strace /usr/bin/echo '\n' 2>&1 | head -n1
execve("/usr/bin/echo", ["/usr/bin/echo", "\\n"], [/* 33 vars */]) = 0
What's going on here?
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因为它后面只是一个字母，所以它不保留其特殊含义，而是按字面意思传递给程序。
Read on:
Since it’s just followed by a letter, it does not retain its special meaning and is passed literally to the program.