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 stackoverflow.com/questions/48165818/… and stackoverflow.com/questions/13998075/… — pynexj
– pynexj, Commented Jul 12, 2022 at 4:51
While we are on the topic, do also check the local keyword. — anishsane
– anishsane, Commented Jul 12, 2022 at 6:25
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
– kingkong, Commented Jul 12, 2022 at 6:48

John Bollinger · Accepted Answer · 2022-07-12 04:54:37Z

3

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.

answered Jul 12, 2022 at 4:54

John Bollinger

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1 Comment

kingkong Over a year ago

Well, I got it. Function g does not overide global x but assign for the outside one line variable x.

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