[英]Why does leaving off a semicolon break this code?
Or put another way, why does semicolon insertion fail, leaving the code below broken. 换句话说,为什么分号插入失败,以下代码被破坏。
function Foo() { }
Foo.prototype.bar = function () {
console.log("bar");
} // <------------ missing semicolon
(function () {
Foo.prototype.la = function () {
console.log("la");
};
})();
Why is the JavaScript parsing engine trying to combine Foo.prototype.bar = function () {
with what's in my closure? 为什么JavaScript解析引擎试图将
Foo.prototype.bar = function () {
与我的闭包中的内容结合起来? Is there anything I could put in that closure that would make this sensible? 有什么我可以放在那个关闭会使这个明智吗?
I'm not advocating leaving off semicolons with the expectation that semicolon insertion will save you; 我不是主张不用分号,期望分号插入可以挽救你; I'm just wondering why (a more useful version of) the above code broke when I accidentally left one off.
我只是想知道为什么(一个更有用的版本)上面的代码在我意外地离开一个时打破了。
因为它看到了(在下面的行上并且认为你想要调用上面的内容(使用下面的函数作为参数)。
Think of it like this... 想想这样......
Foo.prototype.bar = function () { // <-- 1. function
console.log("bar");
}(function () { // <-- 2. call the 1. function, passing a function argument
Foo.prototype.la = function () {
console.log("la");
};
})(); // <-- 3. tries to invoke the return value of the 1. function,
// but "undefined" was returned.
I don't like using ()
for IIFE. 我不喜欢IIFE使用
()
。 I prefer other operators. 我更喜欢其他运营商。
Foo.prototype.bar = function () {
console.log("bar");
}
void function () {
Foo.prototype.la = function () {
console.log("la");
};
}();
If we go back to the original, and have the first function return a function, you'll see that one invoked. 如果我们回到原来的,并让第一个函数返回一个函数,你会看到一个函数被调用。
Foo.prototype.bar = function () { // <-- 1. function
console.log("bar");
return function() { alert('INVOKED'); }; // 2. return a function
}(function () { // <-- 3. call the 1. function, passing a function argument
Foo.prototype.la = function () {
console.log("la");
};
})(); // <-- 4. tries to invoke the return value of the 1. function,
// which will now call the returned function with the "alert()"
Updated to use a unary operator as suggested by @Lasse Reichstein , as a binary operator will still evaluate its left and right operands, and return the result, which will be used for the assignment. 更新为使用@Lasse Reichstein建议的一元运算符,因为二元运算符仍将评估其左右操作数,并返回将用于赋值的结果。
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