「x = x || 4」生成`ReferenceError：x未定義'錯誤

Question

爲什麼x = x || 4甚至x=(x||5)生成ReferenceError: x is not defined錯誤，但var x=x || 4按預期工作？

Answer 1

您正在嘗試使用之前未聲明的變量。這會導致參考錯誤。

這麼多錯誤的單詞。

[退出] @grc說得對。

Answer 2

x = x || 4表示在變量x中分配x或4。 如果x爲空4分配給變量x。

可能是你沒有聲明x變量。這就是爲什麼你得到x is not defined

如果你試圖在它下面的工作：

var x; 
x=x||4; 
alert(x); 

這也將工作：

x=x||4; 
var x; 
alert(x); 

Answer 3

這是因爲變量聲明首先處理（提升）。該MDN page on var解釋說得好：

Because variable declarations (and declarations in general) are processed before any code is executed, declaring a variable anywhere in the code is equivalent to declaring it at the top. This also means that a variable can appear to be used before it's declared. This behaviour is called "hoisting", as it appears that the variable declaration is moved to the top of the function or global code.

所以下面也將工作：

x = x || 4; 
var x;