Powershell: how get exception line number when using ScriptBlock::Invoke - powershell-4.0

Question

How can I get the line number where the exception is thrown in this example? The example below only gives me the line number where I invoke the script block (i.e. $sb.Invoke()). I want the line number where I throw "Error in FooBar".
function Foobar {
throw "Error in FooBar"
}
function Bar {
FooBar
}
function Foo {
Bar
}
try {
$sb = ${function:Foo}
$sb.Invoke()
}
catch {
$ErrorMessage = $_.Exception.Message
$line = $_.InvocationInfo.ScriptLineNumber
$script_name = $_.InvocationInfo.ScriptName
write-host "<--Error: Occurred on line $line in script $script_name."
Write-host "<--Error: $ErrorMessage"
}
UPDATE:
If I change:
$sb = ${function:Foo}
$sb.Invoke()
to this:
$func_info = get-command Foo
Invoke-Command $func_info.ScriptBlock
it works.

Answer 1

This seems like a limitation of Invoke()'s error handling - it's passing on the error message but throwing away the invocation info.
One workaround is to catch and rethrow inside the scriptblock itself, but with more information, e.g.
$sb = {
try {
Foo
}
catch {
throw "Error: $($_.Exception.Message). Stack trace:`n$($_.ScriptStackTrace)"
}
}
$sb.Invoke()
This outputs:
<--Error: Occurred on line 22 in script .
<--Error: Exception calling "Invoke" with "0" argument(s): "Error: Error in FooBar. Stack trace:
at Foobar, <No file>: line 2
at Bar, <No file>: line 6
at Foo, <No file>: line 10
at <ScriptBlock>, <No file>: line 16
at <ScriptBlock>, <No file>: line 22"