我不明白范围

Question

1 - why when I run the below code I got undefind instead "a=1" ?

function f1(){a=1; f2();}
function f2(){return a;}
var a= 5;
a = f1();
​alert(a);​

like this example the resualt is "a=1" .

function f1(){a=1; f2();}
function f2(){alert(a);}
var a= 5;
f1();

Answer 1

With

a = f1();

you are assigning the result of calling f1 to a . Yet, f1 does not return anything, it evaluates to undefined . You'd need to use a return statement:

function f1(){a=1; return f2(); }

Btw, this is not a scope problem. You don't have any variables that are local to your functions, everything accesses the same a .

Answer 2

You probably forget a return statement to get your a value

function f1(){a=1; return f2();}
function f2(){return a;}
var a= 5;
a = f1();
​alert(a);​

Answer 3

f1 does not return anything that's why try the below

function f1(){a=1; return f2();}
function f2(){return a;}
var a= 5;
a = f1();
​alert(a);​

even if does not make lots of sense

Answer 4

您需要从f1显式返回。

Answer 5

第一个例子中的函数f1没有返回任何值，所以这就是原因

Answer 6

During the line a = f1(); the f1 function isn't returning anything so a is getting set to undefined.

I'm not positive what you are trying to do; if you add more I could make a suggestion for how to make it do what you want.

Answer 7

f1() doesn't return any value. Returning nothing is the same as returning undefined.