JavaScript is an interpretive language, and the simple understanding is to load the sentence to which it runs (meaning this, but the expression is not entirely accurate). Your code is understood in four steps (the following explanation does not consider the problem of variable promotion for the time being):

var xyx = 1; //xyz1xyz1
function fx(){console.log(xyx)}//fx
var xyx = 2; //xyz2 xyz2
fx();//fxfxxyzxyz2

if you consider variable promotion, the above code is equivalent to:

var xyz; //xyzxyz
xyz = 1;
function fx(){console.log(xyx)}
xyz = 2;
fx();

execute the function body only when the function is called?
according to js execution mechanism

1. first xyz variable is defined and assigned to 1,
2. redefine the assignment xyz as 2, overwriting the previous value
3. call fx (),

so it is 2

  var xyx = 1;
  function fx(){console.log(xyx)  }   // 1
  fx();
  var xyx = 2;

I don't know whether to

 
 are you struggling with variable declaration promotion and function declaration promotion? 
 the effect of declaring promotion is not involved here 
The code before the 
 call is executed, which you can understand as removing the declaration promotion and executing all declarations related to this class once before the function call 
.