How to correctly compose a logical expression in an asterisk?

Question

Asterisk

Good afternoon
, there is such a macro:

exten => s,n,GotoIf($[$["${ARG1}" != "exten"]|$["${ARG1}" != "out"]]?return)
exten => s,n,...................
exten => s,n,...................
exten => s,n,...................
exten => s,n(return),Return()

In theory, it should be: if arg1 is NOT exten OR arg1 is NOT OUT then go to return ...
IF ARG1 is EXTEN or OUT then continue executing the macro in normal mode ...
But in the end it turns out as if a logical operator was specified in the expression And ...
i.e. at the moment, the macro always goes into return.
Please tell me how to correctly compose a macro using logic: If ARG1 = EXTEN or ARG1 = OUT then
PS This macro will work fine:

exten => s,n,GotoIf($["${ARG1}" != "exten"]?return)
exten => s,n,...................
exten => s,n,...................
exten => s,n,...................
exten => s,n(return),Return()

Answer 1

As the official manual says, it is correct to do this:
But, if you can’t do this, then you can always do just two conditions:

exten => s,n,GotoIf($["${ARG1}" != "exten"]?return)
exten => s,n,GotoIf($["${ARG1}" != "out"]?return)
exten => s,n,...................
exten => s,n,...................
exten => s,n,...................
exten => s,n(return),Return()

Answer 2

exten => s,n,GotoIf($["${ARG1}" != "exten" | "${ARG1}" != "out"]?return)

Answer 3

У Вас оператор "ИЛИ". Поэтому, если результатом проверки первого условия будет "ИСТИНА", то до проверки второго условия не дойдет. Хотите, чтобы проверялись оба условия, используйте оператор "И".