JavaScript basic operation and principle learning

 /* Temporary Object Test JS in the temporary object after the reference will be destroyed */function OnClick () {var s= "Hello World"; var word=s.substring (S.indexof ("") +1,s.length); alert (word); var s= "Test"; s.len=4; Console.log ("s.len=:", S.len);//output: undefined var t=s.len;//alert (t); var t=s.len=4;console.log ("s.len=", t);// 4console.log ("s.len=", S.len);//undefinedvar s= "test", B=1,c=true;var s=new String (s), var b=new number (b); var c=new Boolean (b); Console.log ("s", s),//== s and s are the same, and when = = = All equals, the two are different because S is a string type and S is an object. Console.log ("IsTrue", S===s?true:false);//falseconsole.log ("IsTrue", S==s?true:false);//trueconsole.log ("s", typeof (S));//stringconsole.log ("s", typeof (S)), the original value in//object//js cannot be changed, not in any case, if other methods are called, he will regenerate A// The value of the object can be changed by var h= "Hello"; Console.log ("ToUpper", H.touppercase ());//"Hello" Console.log ("H", h);//"Hello" var o={x:1}; Console.log (typeof (O)); Console.log ("o.x", o.x); O.x=2;console.log ("o.x", o.x); o.y=3;//creates a new property for object o and assigns a value of Console.log ( "O.x", o.x); Console.log ("O.y", o.y); var a=[1,2,3];a[0]=0;a[3]=4;//object is a reference type, in any case assigning A to B, the object itself is not copied once, it is simply an assignment referenceValue, personal understanding: This assignment is the same base address, so they are congruent. var a=[];var b=a;b[0]=1;console.log (A[0]) console.log ("IsTrue", A===b?true:false); Typechange ();} function Typechange () {var s=10+ "objects";//10 converted to string var sum= "7" * "4";//two are converted to numbers; var n=1-"x";//=>nan, character "X" cannot be converted to a digital console.log (n); n+= "Objects";//=>objects cannot be converted to a digital console.log (s,sum,n);}     function Ondown () {/* Various types convert to String tostring () */var b=new Boolean (false);     Console.log ("B", b); Console.log ([1,2,3].tostring ());//1,2,3 Console.log ((function (x) {f (x);}).     ToString ());//function (x) {f (x);} Console.log (New Date (2017,0,1). toString ()),//sun Jan 00:00:00 gmt+0800//ValueOf () convert to actual     The image does not return the original value */var ob={x:1};     Console.log ("ob=", Ob.valueof ())//ob= [Object Object] var d=new Date (2017,1,9);     Console.log ("d==", d.valueof ());//1486569600000 converted to 1970 since JS's internal representation value in milliseconds represents the date var a=checkscope ();     Console.log ("Checkscope:", a);} Scope= "glocal"; function Checkscope () {//scope= "local";//myscope= "local"; Console.log ("Scope ", scope);//return [Scope,myscope];} /* Eval () has the ability to change global variables, but when it is called by the definition alias, Eval executes the string as the top-level global variable. That is, if an alias is used in a function to perform an eval on a global variable, the code executed will define a new global variable and global function, or assign a value to a global variable, but cannot use or modify the local variable in the keynote function, so it does not affect the optimization within the function. Personal understanding: When using aliases to manipulate global variables, the return of global variables within a function does not change, or is called a new global variable, and the global variable itself executes the method of manipulating the global variable in the function as the top-level global variable, so the function returns the global variable unchanged, but the global variable itself changes.   */var Geval=eval;   var x= "Global", y= "global"; function f () {var x= "local"; Eval ("x+= ' changed ';");   return x;} function g () {var y= "local"; Geval ("y+= ' changed ';");   return y;} Console.log (f (), X),//localchanged Global Console.log (g (), y);//local globalchanged</script>

JavaScript basic operation and principle learning