有和没有返回语句的函数有区别吗？

Question

假设你有没有返回值

function a() { 
    // do some interesting things 
} 

function b() { 
    // do the same interesting things 
    return; 
} 

功能b显然更为冗长，但有它们之间的任何功能上的区别2层相同的功能呢？

Answer 1

没有真正的区别;两者都将返回undefined。

没有返回语句的函数将返回undefined，因为将使用空的return语句。

---

为了证实自己这一点，你可以运行这段代码 - FIDDLE：

​function a() { 
} 

function b() { 
    return; 
} 

var aResult = a(); 
var bResult = b(); 

alert(aResult === bResult); //alerts true 

Answer 2

一般要返回的值。因此，例如，

function b() { 
    return 'hello'; 
} 

a = b(); 

console.log(a); 

将输出“hello”到您的控制台。

Answer 3

亚当是正确的;这两个函数都会返回未定义的值，如果您不关心返回值（或者希望该值未定义），那么两种方法都绝对没问题。但是，在更复杂的程序中显式返回函数通常更好，特别是因为Javascript程序通常具有复杂的回调机制。例如，在这段代码（只是多一点比你复杂的）我相信return语句确实有助于澄清代码：

function someAsyncFunction(someVar, callback) { 
    // do something, and then... 
    callback(someVar); 
    // will return undefined 
    return; 
} 

function c(){ 
    var someState = null; 
    if (some condition) { 
     return someAsyncFunction(some variable, function() { 
      return "from the callback but not the function"; 
     }); 
     // we've passed the thread of execution to someAsyncFunction 
     // and explicitly returned from function c. If this line 
     // contained code, it would not be executed. 
    } else if (some other condition) { 
     someState = "some other condition satisfied"; 
    } else { 
     someState = "no condition satisfied"; 
    } 
    // Note that if you didn't return explicitly after calling 
    // someAsyncFunction you would end up calling doSomethingWith(null) 
    // here. There are obviously ways you could avoid this problem by 
    // structuring your code differently, but explicit returns really help 
    // avoid silly mistakes like this which can creep into complex programs. 
    doSomethingWith(someState); 
    return; 
} 

// Note that we don't care about the return value. 
c();