javascript "use strict"; block scope not behaving correctly

Question

When using strict mode, exceptions are thrown when accessing undefined variables.

Consider the following:

"use strict";
alert(typeof mistypedVaraible + " one");
alert(mistypedVaraible + " two");

only the first alert fires because an exception is thrown when trying to access the undefined variable in the second alert. So define it before the second alert:

"use strict";
alert(typeof mistypedVaraible + " one");
if(typeof mistypedVaraible == "undefined") { var mistypedVaraible; }
alert(mistypedVaraible + " two");

Both alerts fire, because the test for undefined has declared the variable.

But if the test fails and the variable DOES NOT get defined, the 2nd alert fires anyways, as if the "if" block got executed to define the variable:

"use strict";
alert(typeof mistypedVaraible + " one");
if(typeof mistypedVaraible != "undefined") { var mistypedVaraible; }
alert(mistypedVaraible + " two");

What's going on here?

Both alerts will fire anyway - there's no condition that governs the second. — , May 25 '15 at 21:31
The variable declaration gets hoisted in both, what were you expecting to occur? — Jesse Kernaghan, May 25 '15 at 21:32

score 2 · Answer 1 · answered May 25 '15 at 21:32

Variable declarations are hoisted to the top.

Therefore, the examples become something like

"use strict";
var mistypedVaraible;
alert(typeof mistypedVaraible + " one");
if(typeof mistypedVaraible == "undefined") { mistypedVaraible; }
alert(mistypedVaraible + " two");

"use strict";
var mistypedVaraible;
alert(typeof mistypedVaraible + " one");
if(typeof mistypedVaraible != "undefined") { mistypedVaraible; }
alert(mistypedVaraible + " two");

So if doesn't matter whether the condition is true or false, the variable is declared anyways.

javascript "use strict"; block scope not behaving correctly

1 Answers1