使用bash进程替换拒绝文件描述符权限？

# ...
# ...

echo "Hello user,behold a cleared gnuplot window"

# pass the string "clear" to gnuplot via file descriptor 3
echo "clear" >&3

所以我认为我可以通过首先启动子进程来“设置它”：

#!/bin/bash

# Initiate(?) file descriptor 3,and let it direct to a newly 
# started gnuplot process:
exec >3 >( gnuplot )

但这会产生错误：

/dev/fd/63: Permission denied

这是预期的吗？

我不明白发生了什么. (我做错了什么？可能是我的系统有一些特殊的安全设置,不允许我正在尝试做什么？(运行Ubuntu Linux 12.10.))

“解决方法” – 以下似乎与我正在尝试的内容相同,并且可以正常工作：

#!/bin/bash

# open fd 3 and direct to where fd 1 directs to,i.e. std-out 
exec 3>&1 

# let fd 1 direct to a newly opened gnuplot process
exec 1> >( gnuplot ) 

# fd 1 now directs to the gnuplot process,and fd 3 directs to std-out.
# I would like it the other way around. So we'll just swap fd 1 and 3 
# (using an extra file descriptor,fd 4,as an intermediary)

exec 4>&1 # let fd 4 direct to wherever fd 1 directs to (the gnuplot process)
exec 1>&3 # let fd 1 direct to std-out 
exec 3>&4 # let fd 3 direct to the gnuplot process
exec 4>&- # close fd 4

或者,作为一个班轮：

#!/bin/bash
exec 3>&1 1> >( gnuplot ) 4>&1 1>&3 3>&4 4>&-

为什么这样可行,但初始版本不是？

任何帮助非常感谢.

$bash --version
GNU bash,version 4.2.37(1)-release (x86_64-pc-linux-gnu)
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