在JavaScript中使用setTimeout（）和setInterval（）时调用函数 [英] Calling a function when using setTimeout() and setInterval() in JavaScript

问题描述

如果我使用setTimeout（）和setInterval（）调用命名函数而没有括号，则它按预期工作。当我用括号调用相同的函数时，它会立即执行它或者给出错误。

If I call a named function using setTimeout() and setInterval() without the parentheses it works as expected. When I call the same function with the parentheses it either executes it immediately or gives an error.

我正在寻找一个更深入的了解这个问题然后我发现了什么网络。
你能告诉我为什么这是真的吗？

I am looking for a deeper understanding in this matter then what I have found on the web. Would you guys please explain to me why this is true?

var func = function(){
    console.log("Bowties are cool.");
}

setTimeout(func(), 1500);
// Prints "Bowties are cool." immediately

setInterval(func(), 1500);
// Throws an error

setInterval(func, 1500);
// Works as expected

setTimeout(console.log("Bowties are cool."),1500);
// This method has the same result as "setTimeout(func(), 1500)".

推荐答案

您必须将函数引用传递给 setTimeout（）和 setInterval（）。这意味着你传递一个没有（）的函数名，或者你传递一个匿名函数。

You must pass a function reference to both setTimeout() and setInterval(). That means you pass a function name without the () after it or you pass an anonymous function.

当你在函数名之后包含（），如 func（），你正在执行该函数，然后传递将结果返回 setInterval（）或返回 setTimeout（）。除非函数本身返回另一个函数引用，否则它将永远不会执行您想要的操作。对于Javascript程序员来说这是一个非常常见的错误（我在学习语言时犯了同样的错误）。

When you include the () after the function name as in func(), you are executing the function immediately and then passing the return result to setInterval() or to setTimeout(). Unless the function itself returns another function reference, this will never do what you want. This is a very common mistake for Javascript programmers (I made the same mistake myself when learning the language).

所以，正确的代码是：

setTimeout(func, 1500);
setInterval(func, 1500);

如果你知道其他语言使用指针，你可以想到一个函数的名称与（）之后它就像一个指向函数的指针，这就是Javascript中的函数引用，当你希望它能够执行某些函数时，你传递给函数的是什么功能稍后，而不是立即。

If you know other languages that use pointers, you can think of the name of a function with the () after it as like a pointer to the function and that's what a function reference is in Javascript and that's what you pass to a function when you want it to be able to execute some function later, not immediately.

当你错误地这样做时：

setTimeout(func(), 1500);
setInterval(func(), 1500);

它正在立即执行你的 func（）然后将返回结果传递给 setTimeout（）和 setInterval（）。由于你的 func（）没有返回任何内容，你基本上是这样做的：

It is executing your func() immediately and then passing the return result to setTimeout() and setInterval(). Since your func() doesn't return anything, you are essentially doing this:

func();
setTimeout(undefined, 1500);

您观察到的是什么，但不是您想要的。

Which is what you observe happening, but not what you want.

此外，如果你想执行特定的函数调用，例如 console.log（Bowties很酷。），然后你可以将它包装在另一个函数中：

Further, if you want to execute a specific function call such as console.log("Bowties are cool."), then you can wrap it in another function like this:

setTimeout(function() {
    console.log("Bowties are cool.")
}, 1500);

因此，您再次将函数引用传递给 setTimeout（）可以执行，而不是立即执行，这就是你正在做的事情。

So, again you are passing a function reference to setTimeout() that can be executed LATER rather than executing it immediately which is what you were doing.

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