In a bash script foo.sh, invoked by:
y=hi z=hello export z var1=7 x=a ./foo.sh arg1 arg2 the variables y, var1 and x are set. Is it possible to enumerate all variables that are explicitly set when invoking the script, i.e. var1 and x but not y and not z?
No, not as is, because bash is doing this seamlessly and those variables aren't really set on the command line (the command line to ./foo.sh).
When bash sees the command line:
var1=7 x=a ./foo.sh arg1 arg2 it simply runs ./foo.sh arg1 arg2 with those variables already set for it, effectively the same as if you had done:
( export var1=7; export x=a; ./foo.sh arg1 arg2 ) except possibly for the number of processes created.
The process running .foo.sh has no idea how those variables were set, it only sees the bit without the variable setting as it's command line.
If you run set without arguments in the script, you'll see var1=7 and x=a in the output, but you won't see y=hi there unless y was exported. All the exported variables will be shown, and you can't tell whether they were exported by export or by prepending the assignment to the command.
Due to the order in which variables are tested and expanded, you can use set twice to get the differences. For example:
export y=1 var1=7 x=a /path/to/script.sh $(set) # script.sh oldvars=$@ newvars=$(set) var1 and x will be in newvars, but not oldvars. (You will then need to 'diff'the two to find the ones you want).
envbetween the parent and within the script inside