how bash one line variable affect variable scope

Question

x=1

function f()
{
    echo $x
}

x=2 f #2

function g()
{
    x=3
    a=4
    echo $x
}
a=5
x=2 g  #3
echo $x  #1
echo $a  #4

Why the output is 2 3 1 4 ? Why function f access the one line variable x rather than the global one? Why function g create global variables x and a, but the x does not override outside global one?

see https://stackoverflow.com/questions/48165818/understanding-lexical-scoping-is-wikipedia-correct and https://stackoverflow.com/questions/13998075/setting-environment-variable-for-one-program-call-in-bash-using-env — pynexj, Jul 12 '22 at 04:51
While we are on the topic, do also check the [`local`](https://tldp.org/LDP/abs/html/localvar.html) keyword. — anishsane, Jul 12 '22 at 06:25
`function h() { local x=3; echo $x; } x=2 h` This output is 3.So I can say the order is LEG: local -> environ -> global? — kingkong, Jul 12 '22 at 06:48

score 3 · Accepted Answer · answered Jul 12 '22 at 04:54

Why function f access the one line variable x rather than the global one?

Because specifying a variable as part of a command sets it in the dynamic scope of the execution of that command (only).

Why function g create global variables x and a, but the x does not override outside global one?

g assigns to variables x and a. But in the context of the call x is effectively local to it, on account of a value having been specified for that variable as part of the call. The same is not true of a, so the assignment to it affects the shell environment normally.

Well, I got it. Function g does not overide global x but assign for the outside one line variable x. — kingkong, Jul 12 '22 at 06:21

how bash one line variable affect variable scope

1 Answers1