bash - Interchanging two programs inputs and outputs -

the title of question rather misleading, not find better title it.

the rephrased title "i want program x's output program y's input , program y's output program x's input. program x start giving output, while program y start accepting input."

program x         stdout --> <program y>stdin         stdin  --> <program y>stdout 

any help?

you can named pipe:

mkfifo xy_pipe ./program_x < xy_pipe | ./program_y > xy_pipe 

a regular pipe used connect x's stdout y's stdin.

to connect y's stdout x's stdin create second, named pipe using mkfifo. named pipe explicit way connect 2 processes way | does. whenever process writes named pipe blocks until process reads pipe. although xy_pipe appears file, no data written disk.

example:

$ cat program_x #!/bin/bash echo foo read line && echo "program_x: read '$line'" >&2  $ cat program_y #!/bin/bash read line && echo "program_y: read '$line'" >&2 echo bar  $ mkfifo xy_pipe $ ./program_x < xy_pipe | ./program_y > xy_pipe program_y: read 'foo' program_x: read 'bar'  

don't forget delete xy_pipe when you're done!

$ rm xy_pipe 

if want see both programs' output on screen can adding tee mix.

$ mkfifo xy_pipe $ ./program_x < xy_pipe | tee /dev/stderr | ./program_y | tee xy_pipe foo program_x says: foo bar program_y says: bar