What is happening behind bash forward process substitution? -

according wikipedia, "process substitution can used capture output go file, , redirect input of process."(http://en.wikipedia.org/wiki/process_substitution).

so, in own words, means process substitution, can take output of command , use input of command b. in other words, it's pipe(is correct?).

so if true, , if this:

echo "test" >(wc) 

then should expect following:

1 1 5 

because understanding of above command similar following:

$echo "test" > tmp $wc tmp 1 1 5 tmp 

except don't make tmp file process substitution.

but instead following output:

test /dev/fd/63 

this indicates somehow mental model incorrect. wrong?

i understand <(command). example

$diff <(head file1) <(head file2)  

perfectly makes sense. not >(command).

from process substitution

the process list run its input or output connected to fifo or file in /dev/fd. the name of file passed argument current command result of expansion. if >(list) form used, writing file provide input list.

what happens echo "test" >(wc)?

the file /dev/fd/63 opened connecting echo test wc. wc started it's input connected /dev/fd/63. name of file (/dev/fd/63) passed argument current command (echo "test"), resulting in echo "test" /dev/fd/63. why see

test /dev/fd/63 

as output. wc waits input, since echo doesn't write /dev/fd/63, count 0.

if want work, must create script, takes last argument , echos first n-1 arguments last

#! /bin/bash echo "${@:1:$(($# - 1))}" >${@: -1} 

when call

bash script.sh test >(wc) 

you see expected output

1 1 5