process-substitution

I want to run this script: #!/bin/bash echo <(true) I run it as: sh file.sh And I get " Syntax error: "(" unexpected " . I found some similar situations but still can't solve this. I'm a beginner at shell scripting , but as I understand: the shebang I use is correct and chooses the bash shell , so the process substitution syntax should work I try the same from the command line and it works. I checked with echo $0 and it gives me " bash " , so what's the difference from running the command in the command line and from a script that invokes the same shell? Maybe it's something simple, but I

I've the following shell script: cat <(echo foo) source <(echo bar=bar) echo $bar However it works differently in GNU bash 3.2 and 4.3 as shown below: $ /bin/bash foo.sh foo 3.2.53(1)-release $ /usr/local/bin/bash foo.sh foo bar 4.3.33(1)-release Why this works only on one version? Is it a bug or added feature? It seems the process substitution works fine, however problem lay when sourcing the file. If this is expected behaviour, what other syntax should I use instead to source something from the standard input to be compatible between different bash versions? This is a known limitation in

I have a bash script which calls several long-running processes. I want to capture the output of those calls into variables for processing reasons. However, because these are long running processes, I would like the output of the rsync calls to be displayed in the console in real-time and not after the fact. To this end, I have found a way of doing it but it relies on outputting the text to /dev/stderr. I feel that outputting to /dev/stderr is not a good way of doing things. VAR1=$(for i in {1..5}; do sleep 1; echo $i; done | tee /dev/stderr) VAR2=$(rsync -r -t --out-format='%n%L' --delete -s