Answer 1

The output will be 1 because :
Your code does the same thing as :
$bar = 0;
$bar++;
echo $bar; // remove the dot at the end of YOUR code and replace the “” by "" as they will result in an error

and if you call the function foo at the end of the code (after definition) like this : foo(3)
The result output would be :
1
5
Because the $bar argument is considered as another variable as it is in another scope