When I run this code firstly shows me 1 and then undefined. But I still couldn't understand it.
alert(alert(1) && alert(2));Here is some explanation :
The call to alert returns undefined (it just shows a message, so there's no meaningful return).
Because of that, && evaluates the left operand (outputs 1), and immediately stops, because undefined is a falsy value. And && looks for a falsy value and returns it, so it's done.
alert does not return anything
So alert(1) runs, since alert does not return anything it is undefined. As a result you have alert(undefined && alert(2)). The alert(2) will not execute because left part needs to be truthy to execute. Undefined is falsey. That evaluates to alert(undefined).
So you get alert of 1 for the first alert and you get undefined for the outside alert and the alert(2) is never executed.
Code written out in a less confusing manner
var action = alert(1) && alert(2) // only runs alert(1)
console.log(action) // shows variable is undefined
alert(action) // alerts undefined
&&or||operator is not necessarily of typeBoolean. The value produced will always be the value of one of the two operand expressions." (Source)