function fn1(){
var a = 6;
function fn2(){
alert(a) ;
}
return fn2;
}
// var f = fn1();
// f() alert 6
fn1() //as in the above code, the variable f can be assigned to alert 6, but there is no response when calling fn1 directly. What is the principle of this, please?
because return is a function
, fn1 () gets a function
fn1 () () calls the function
because you fn1 () only once.
and your f executes the function once during the assignment, and then f itself executes the function again, which is equivalent to twice.
if you take a closer look at fn1, direct fn1 has a result. The result is that the internal function fn2:
is returned.var f = fn1(); // f fn2
f() // fn2() alert 6
// alert 6
fn1()() you just call the function fn1 and return a function reference
execute fn1 () ()
fn1 () ()
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