Why isn't the exit code of the `if` statement a failure if the command failed?-CodePudding

I thought I know the shell, but I got bitten with unexpected behavior.

Consider this code:

#!/bin/bash
foo()
{
    if false
    then
        echo success
    else
        return
    fi
}

foo || echo failure

Now at first glance one would think the else return part is superfluous, but it is not. The code as it is outputs failure, but when the cited part is removed, then nothing is output.

The bash manual explains (for if):

The exit status is the exit status of the last command executed, or zero if no condition tested true.

Somehow I'd expect the if to fail when the command had failed. Imagine where echo success stands there are actually dependent commands that make only sense to execute when the main command (false here) succeeded.

What is the logic behind that unexpected behavior?

Seen for bash-4.4. Related: https://stackoverflow.com/a/63884458/6607497

CodePudding user response：

The if statement never actually completes with the return statement; the function returns immediately, with the exit status provided by return. That exit status is the status of the last command prior to return to exit, which was false.

Without the return, the if does complete, and as documented, the exit status of an if command is defined to be 0 if none of the conditions in the if clause or any elif clause evaluated to true. This makes sense, because the if statement successfully determined that none of the if or elif blocks needed to be executed.

CodePudding user response：

The bash manual explains (for return):

return
return [n]
Cause a shell function to stop executing and return the value n to its caller. If n is not supplied, the return value is the exit status of the last command executed in the function.

So:

if false
then
    echo success
else
    # The LAST command is `false` with exit status 1 !
    return  # Returns 1 - returns failure!
fi