Piping not working with echo command [duplicate]

This question already has an answer here:

How to pass command output as multiple arguments to another command 4 answers

When I run the following Bash script, I would expect it to print Hello. Instead, it prints a blank line and exits.

echo 'Hello' | echo

Why doesn't piping output from echo to echo work?

回答1:

echo prints all of its arguments. It does not read from stdin. So the second echo prints all of its arguments (none) and exits, ignoring the Hello on stdin.

For a program that reads its stdin and prints that to stdout, use cat:

$ echo Hello | cat
Hello

回答2:

You don't seem to understand pipes. In this case they are more correctly known as anonymous pipes, because they have no name (there are also named pipes). Anonymous pipes only work between related processes, for example processes with the same parent.

Pipes are part of the IO system resulting from the C runtime-library. These streams are buffered (there is an exception) by default. Basically a pipe is just connecting the output buffer from one process to the input buffer of another.

The first three streams used (called file descriptors) are numbered 0, 1, and 2. The first, 0, is known as standard input, or stdin (the name used in C). By default this is connected to the keyboard, but it can be redirected either using the < symbol or the program name being on the right side of a pipe.

The second, 1, is known as standard output, or stdout. By default this is connected to the terminal screen, but can be redirected by using the > symbol or the program name being on the left side of a pipe.

So:

echo 'Hello' | echo

takes the standard output from echo and passes it to the standard input of echo. But echo does not read stdin! So nothing happens.

Filter programs process the filenames specified on the command-line. If no filenames are given then they read stdin. Examples include cat, grep, and sed, but not echo. For example:

echo 'Hello' | cat

will display 'Hello', and the cat is useless (it often is).

echo 'Hello' | cat file1

will ignore the output from echo and just display the contents of file1. Remember that stdin is only read if no filename is given.

What do you think this displays?

echo 'Hello' | cat < file1 file2

and why?

Finally, the third stream, 2, is called standard error, or stderr, and this one is unbuffered. It is ignored by pipes, because they only operate between stdin and stdout. However, you can redirect stderr to use stdout (see man dup2):

myprog 2>&1 | anotherprog

The 2>&1 means "redirect file descriptor 2 to the same place as fie descriptor 1".

The above is normal behaviour, however a program can override all that if it wants to. It could read from file descriptor 2, for example. I have omitted a lot of other detail, including other forms of redirection such as process substitution and here documents.

回答3:

Piping can be done only for commands taking inputs from stdin. But echo does not takes from stdin. It will take input from argument and print it. So this wont work. Inorder to echo you can do something like echo $(echo 'hello')

回答4:

It is because echo (both builtin and /bin/echo) don't read anything from stdin.

Use cat instead: