爲什麼不使用「嚴格」（JavaScript）檢測未聲明的變量？

我試圖得到「嚴格使用」;指示工作，並有一些麻煩。在以下文件中，FireFox 9將（正確）檢測到第3行中沒有聲明someVar，但未能檢測到第19行中沒有聲明該變量。我很難理解爲什麼會出現這種情況。爲什麼不使用「嚴格」（JavaScript）檢測未聲明的變量？

"use strict"; // this will cause the browser to check for errors more aggresively 

someVar = 10; // this DOES get caught // LINE 3 

// debugger; // this will cause FireBug to open at the bottom of the page/window 
     // it will also cause the debugger to stop at this line 

    // Yep, using jQuery & anonymous functions 
$(document).ready(function(){ 
    alert("document is done loading, but not (necessarily) the images!"); 

    $("#btnToClick").click(function() { 

     alert("About to stop"); 
     var aVariable = 1; 
     debugger; // stop here! 
     alert("post stop " + aVariable); 

     // this lacks a "var" declaration: 
     theVar = 10; // LINE 19 // this is NOT getting caught 

     // needs a closing " 
     // alert("hi); 
     console.log("Program is printing information to help the developer debug a problem!"); 
    }); 

});

來源

2011-12-25 MikeTheTall