shell - Bash: trick program into thinking stdout is an interactive terminal

Question

I want to run a program's output through a pipe, but it apparently behaves differently when it detects stdout is not an interactive shell.

How can I trick it into writing through the pipe just as it would in regular circumstances?

Answer 1

I assume that the program will call the glibc function isatty() to check whether stdout is a terminal or not. That's common for programs which use colorized output on terminals or other features of an ANSI terminal like cursor positioning or line erasing / redrawing.

You can trick the program using the LD_PRELOAD environment variable. LD_PRELOAD is handled by the ELF linker and tells that a dynamic library should be loaded before all others. Using this feature it is possible to override library functions, in your case the glibc function isatty(). You can follow this article for example.

I've prepared an example for you:

First create the file libisatty.c:

/**
 * Overrides the glibc function. Will always return true.
 *
 * Note: Although this should be ok for most applications it can
 * lead to unwanted side effects. It depends on the question
 * why the programm calls isatty()
 */
int isatty(int param) {
    return 1;
}

and compile it as a shared lib:

gcc -shared -o libisatty.so  libisatty.c

It should build fine.

Now it's time to test the library. :) I've used the command ls --color=auto for tests. ls calls isatty() to decide whether it should colorize it's output or not. If the output is redirected to a file or a pipe it won't be colorized. You can test this easily using the following commands:

ls --color=auto        # should give you colorized output
ls --color=auto | cat  # will give you monochrome output

Now we'll try the second command again using the LD_PRELOAD environment var:

LD_PRELOAD=./libisatty.so ls --color=auto | cat

You should see colorized output.

btw cool usename: u?op ?p?sdn !!:D