CODEWITHSUNDEEP
bash
PrimRock
PrimRock

Reputation: 1156

bash echo command ignored in function loop

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.

Upvotes: 0

Views: 1300

Answers (1)

KamilCuk
KamilCuk

Reputation: 140940

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.

Upvotes: 4

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