bash echo command ignored in function loop

Question

I simplified my bash script to the scope of the problem. I was wondering how come when I do this...

#!/bin/bash                                
                                            
read -p "Please enter your name: " name    
                                            
function test()                            
{                                          
    while true                             
    do                                     
        echo $name                         
    done                                   
}                                          
                                            
echo $(test)

the echo command doesn't loop the name in the terminal. However, if I were to remove the function and have the while loop by itself like so....

#!/bin/bash                                
                                           
read -p "Please enter your name: " name    
                                                 
while true                             
do                                     
   echo $name                         
done

It would work. Or if I do this, it will also work

#!/bin/bash
                            
read -p "Please enter your name: " name    
        
function test()                            
{    
    echo $name                                                           
}                                          
                                            
echo $(test)

What's causing the echo command to not to not display the name. This only happens when the echo command is inside a while loop whilst being inside a function.

Does this answer your question? Why are bash variables not global if sourced script runs in a function? — 0stone0
– 0stone0, Commented Nov 12, 2020 at 19:06

KamilCuk · Accepted Answer · 2020-11-12 19:17:03Z

4

What's causing the echo command to not to not display the name

The parent shell is waiting for the subshell to exit before expanding the command subsitution to it's contents. Because the subshell never exits (as it's in an endless loop), the echo command never gets executed.

echo $(test)
     ^^    ^  - shell tries to expand command substitution
                so it runs a subshell with redirected output
                and waitpid()s on it.
       ^^^^   - subshell executes `test` and never exits
                cause its an endless loop.
                Parent shell will endlessly wait on subshell to exit.

Note that test is already a very standard shell builtin for testing expressions. Defining such function which will cause to overwrite the builtin may cause unexpected problems.

For some reading I could recommend bash guide functions.

edited Nov 12, 2020 at 19:17

answered Nov 12, 2020 at 19:07

KamilCuk

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3 Comments

PrimRock Over a year ago

I'm new to bash, so what would be the best way to call the function and have it execute in the way I expected. Also I only used test for the example. My function was named something else in my actual program haha

KamilCuk Over a year ago

just type the name of the function... Doing echo $(something) is a useless use of echo, it's like echo $(echo $(echo $(echo $(something)))). Just something. On a side note, results from unquoted expansions undergo word splitting expansion - check your scripts with shellcheck.net and preferably quote the expansions - research when to use double quotes in bash.

PrimRock Over a year ago

My apologies...I was reading someone elses script and saw that they were calling functions that way. So I thought that was the way you were suppose to call functions. Thank you for letting me know, I learned something new today

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