Shell Programming and Scripting

(Simplified example):

I have in my .zshrc the following function definition:

Code:

function foo {
    local p=${1:?parameter missing}
    echo continue ....
}

Running the function with just foo produces, as expected, the message parameter missing, but it also output continue. I had expected that the function terminates, when the :? check fails, but it continues to run. Why is this the case?

The man-page zshexpn says:

... otherwise, print word and exit from the shell. Interactive shells instead return to the prompt.

ADDITIONAL FINDINGS:

I found that this behaviour depends on the presence of the local command. If I remove local, the function works as expected. Since I need local, I can rewrite the function like this:

Code:

function foo {
    : ${1:?parameter missing}
    local p=$1
    echo continue ....
}

This works fine, but I still am curious to know, why the presence of local causes this difference in behaviour.

UPDATE: The issue can be considered solved. Zsh developers confirmed that this behaviour is a bug in Zsh which is going to be fixed. See also Zsh function does not terminate when ${name:?word} fails - Stack Overflow