英文:
Why does process substitution streams can be read only once in bash?
问题
以下是翻译好的内容：
这是我编写的一个脚本，用于说明这个问题：
#!/bin/bash
cat_then_tac()
{
echo cat
cat "$1"
echo tac
tac "$1"
echo end
}
seq 3 > file1
echo
echo 使用正确的文件：按预期工作
cat_then_tac file1
echo
echo 使用进程替代：流在第一次使用时似乎被消耗掉，tac 没有可用的内容
cat_then_tac <(seq 3)
rm file1
这是它的输出：
使用正确的文件：按预期工作
cat
1
2
3
tac
3
2
1
end
使用进程替代：流在第一次使用时似乎被消耗掉，tac 没有可用的内容
cat
1
2
3
tac
end
显然，可以通过首先保存流（在文件或变量中）并多次使用它来解决这个问题。但为什么会发生这种消耗呢？
后来，我在Unix-StackExchange上找到了与相同问题相关的这个问题。它还包含值得一看的信息：
Bash 重复使用进程替代文件
英文:
Here's a script I wrote to illustrate the issue:
#!/bin/bash
cat_then_tac()
{
echo cat
cat "$1"
echo tac
tac "$1"
echo end
}
seq 3 > file1
echo
echo using proper file: works as expected
cat_then_tac file1
echo
echo using process substitution: stream is somehow consumed upon first use and tac has nothing to work on
cat_then_tac <(seq 3)
rm file1
And here's its output:
using proper file: works as expected
cat
1
2
3
tac
3
2
1
end
using process substitution: stream is somehow consumed upon first use and tac has nothing to work on
cat
1
2
3
tac
end
Obviously one can work around this by first saving the stream (in a file or variable) and using it as many times as needed. But why does this sort of consumption happen?
Later, I found this question on Unix-StackExchange related to the same matter. And it also contains info worth taking a look:
Bash Reuse Process Substitution File
答案1
得分: 5
Process substitution使用管道实现。替代进程的输出由操作系统缓冲并在读取后丢弃。因此，cat_then_tac <(seq 3)与以下命令相同：
seq 3 | cat_then_tac -
英文:
Process substitution is implemented using pipes. The output of the substituted process is buffered by the operating system and discarded after it is read. So cat_then_tac <(seq 3) is the same as
seq 3 | cat_then_tac -
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