javascript - 消除就地定义为参数的函数？

Question

如何消除就地定义为参数的函数？例如，返回两个参数相加的函数应编码为命名函数，另一个函数应编码为分配给变量的匿名函数。

var outputAreaRef = document.getElementById("outputArea");
var output = "";

function flexible(fOperation, operand1, operand2)
{
var result  = fOperation(operand1, operand2);

return result;
}

output += flexible(function(num1, num2) {return num1 + num2}, 3, 5) + "<br/>";
output += flexible(function(num1, num2) {return num1 * num2}, 3, 5) + "<br/>"; 

outputAreaRef.innerHTML = output;

Answer 1

如果我正确理解了这个问题，您正在寻找类似的内容:

 var outputAreaRef = document.getElementById("outputArea");
 var output = "";

 function flexible(fOperation, operand1, operand2) {
     var result = fOperation(operand1, operand2);
     return result;
 }

 function sum(a, b) {
     return a + b;
 }

 output += flexible(sum, 3, 5) + "<br/>";
 output += flexible(function (num1, num2) {
     return num1 * num2
 }, 3, 5) + "<br/>";

 outputAreaRef.innerHTML = output;

参见https://jsfiddle.net/nwp1k93f/

此外，您还可以看看

• How to execute a method passed as parameter to function
• pass a function as a parameter and then execute it in a jquery function

注意:

• 在 JavaScript 中，此类函数称为回调或高阶函数
• 详细说明可以在这里找到:http://javascriptissexy.com/understand-javascript-callback-functions-and-use-them/