Javascript - 向作为参数传递的函数添加参数

Question

Here is the behaviour I'm looking for:

function one(func){
   func(5);
}

function two(arg1, arg2){
   console.log(arg1);
   console.log(arg2);
}

one(two(3)) //prints 3, 5

Can this behaviour or something similar be accomplished in javascript?

Answer 1

You can always use the bind() function to pass some arguments to your function. It'll create a new function with the first argument - arg1 - equal to the value of 3 in this example:

function one(func){
   func(5);
}

function two(arg1, arg2){
   console.log(arg1);
   console.log(arg2);
}

one(two.bind(null, 3))

You can read more about the bind() function here: MDN - Bind

Answer 2

Some workaround is possible

 function one() { var args = Array.prototype.slice.call(arguments); var func = args[0]; args.splice(0, 1); args.push(5); func.apply(this, args); } function two(arg1, arg2) { console.log(arg1); console.log(arg2); } one(two, 3)

Answer 3

There's a problem with your syntax: function one is expecting its single argument to be a function. Then, below, when you invoke it, you are not passing the function two , but whatever two returns when it's passed a single argument, probably undefined . I don't know what specifically you're trying to accomplish but I'd recommend a little research into closures .

Answer 4

as soon as one expects function as argument two(3) should return function.
this condition is required so in order to achieve it your two function should be

function two(arg1){
    console.log(arg1);
        return function(arg2) {
        console.log(arg2);
    };
}

so two(3) function call gets passed as argument to one
so before assigning value to variable engine executes it. And execution of two(3) call logs 3 to console and returns function

   function(arg2) {
      console.log(arg2);
    };

and then engine assigns executed value(returned function) to func variable. so func parameter of one function now looks like

func = function(arg2) {
    console.log(arg2);
 };

one calls func with 5 passed in as argument. so 5 gets logged to console.

Answer 5

function one(arg){
   two(arg, 5); // func here is two so it requires two params...
}

function two(arg1, arg2){
   console.log(arg1);
   console.log(arg2);
}

one(3)// one expect function so can't execute function here!

Answer 6

Basically you can't specify the parameter in the function or it'll run. You need to specify the function aka one(two), but that obviously wouldn't work.

However if you dynamically create a function you should be able to accomplish the task like so :

function one(func){
   func(5);
}

function two(arg1, arg2){
   console.log(arg1);
   console.log(arg2);
}

one(function(val) { two(3, val) }) //prints 3, 5

Answer 7

well somehow this things work for me

 function one(func){ console.log("one is running"); func; } function two(args1, args2){ console.log("two is running"); console.log("args1 -> " + args1); console.log("args2 -> " + args2); } //to call it args1 = 6; args2 = 12 one(two(args1,args2));