为什么不使用“严格”（JavaScript）检测未声明的变量？

我试图得到“严格使用”;指示工作，并有一些麻烦。在以下文件中，FireFox 9将（正确）检测到第3行中没有声明someVar，但未能检测到第19行中没有声明该变量。我很难理解为什么会出现这种情况。为什么不使用“严格”（JavaScript）检测未声明的变量？

"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!"); 
    }); 

});

来源

2011-12-25 MikeTheTall