Incorrect extraction of the number of values.

[zusage]

When getting values, the following occurs.
Perhaps I'm using the syntax incorrectly.

$ echo 'foo(var1, var2, var3, var4, var5)' | rg '^foo((([^,]+,?){5}))' -r '$1'
var1, var2, var3, var4, var5 <--- Correct, since the number of required matches is {5}
Expected result: var1, var2, var3, var4, var5

$ echo 'foo(var1, var2, var3, var4, var5)' | rg '^foo((([^,]+,?){4}))' -r '$1'
Nothing <--- Correct, since the number of required matches is {4}
Expected result: Nothing

$ echo 'foo(var1, var2, var3, var4, var5)' | rg '^foo((([^,]+,?){6}))' -r '$1'
var1, var2, var3, var4, var5 <--- Incorrect, since the number of required matches is {6}
Expected result: Nothing

[BurntSushi]

It's correct. I believe you're misunderstand regex semantics here. A simpler example:

$ echo 'aaaaa' | rg '(a+){2}'
aaaaa
$ echo 'aaaaa' | rg '(a+){3}'
aaaaa
$ echo 'aaaaa' | rg '(a+){4}'
aaaaa
$ echo 'aaaaa' | rg '(a+){5}'
aaaaa
$ echo 'aaaaa' | rg '(a+){6}'
$

That is, ripgrep will prioritize a match over greediness. So that means a (re)+ can result in matching fewer repetitions of re if it means an overall match could be found. In your regex, ^foo((([^,]+,?){9})), it might be helpful to focus on the thing being repeated: [^,]+,?. Notice that that regex can match any single letter except for , since the , at the end is made optional by ?.

I believe what you want is to make sure you match a terminator:

$ echo 'foo(var1, var2, var3, var4, var5)' | rg '^foo((([^,]+(,|\))){5}))' -r '$1'
(var1, var2, var3, var4, var5)
$ echo 'foo(var1, var2, var3, var4, var5)' | rg '^foo((([^,]+(,|\))){6}))' -r '$1'

[zusage]

Could you please check the results? Perhaps I missed something important.
$ echo 'foo(var1, var2, var3, var4, var5)' | rg '^foo((([^,]+(,|))){5}))' -r '$1'
Expected result: var1, var2, var3, var4, var5

$ echo 'foo(var1, var2, var3, var4, var5)' | rg '^foo((([^,]+(,|))){4}))' -r '$1'
Expected result: Nothing

$ echo 'foo(var1, var2, var3, var4, var5)' | rg '^foo((([^,]+(,|))){6}))' -r '$1'
Expected result: Nothing

[BurntSushi]

,| is effectively equivalent to ,?. So you have the same problem as your original regex.

[BurntSushi]

Okay special attention to the fact that \( matches a literal (. That is, it escapes the parenthesis. I assume you meant to escape some of the parentheses, but I don't see any escapes in your regex.