Java [Javascript?] Using a function as an argument for another function

AI Thread Summary
The discussion revolves around correcting a JavaScript function implementation for calculating the cube of a number using a square function. The initial code incorrectly attempts to define the cube function by directly referencing the square function without calling it. The correct implementation involves calling the square function with an argument within the cube function. Participants clarify that the original title of the thread was misleading, as it suggested using a function as an argument rather than calling one function within another. A solution is proposed that correctly defines the cube function as returning the product of the square of a number and the number itself. Additionally, there is a discussion about the use of functions as parameters in JavaScript, with examples provided to illustrate how this can be done effectively. The conversation emphasizes the importance of experimenting with code to understand function interactions better.
mindauggas
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Code:
var square = function (x) {
  return x * x;
};

var cube = function (x) {
  return square * x;
};

This is my shot, its not right. Please indicate or suggest the correct answear. :)
 
Last edited by a moderator:
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mindauggas said:
Code:
var square = function (x) {
  return x * x;
};

var cube = function (x) {
  return square * x;
};

This is my shot, its not right. Please indicate or suggest the correct answear. :)

No, it's not correct.
Code:
var cube = function (x) {
  return x * square(x);
};
Your thread title is misleading - you aren't using a function as an argument of another function. One function is calling another function. There's a difference.
 
Last edited:


In C or C++, you can use pointers to functions as arguments:
Code:
#include <stdio.h>

int f1xf2(int (*pf1)(int), int (*pf2)(int), int v){return(pf1(v) * pf2(v));}
int returnv(int v){return(v);};
int square(int v){return(f1xf2(returnv, returnv, v));}
int cube(int v){return(f1xf2(returnv, square, v));}
int quad(int v){return(f1xf2(square, square, v));}

int main(int argc, char ** argv)
{
    printf("square 2 = %3d\n", square(2));
    printf("cube   3 = %3d\n", cube(3));
    printf("quad   4 = %3d\n", quad(4));
    return(0);
}
 
Last edited:


I am confused...it always bugs me when people do not state what language they are talking about...sure, if I knew, I would recognize it; but often times I don't and I can still help, if I knew what to google.

So, the last post is C, sure...what's the first one, too? or is it matlab?
 


Looks like JavaScript to me.
 
Sorry, it's JavaScript
 
Once again, sorry. This time for abandoning my own topic. Like I said the programming language is JavaScript. And yes, thanks for pointing out, that the title is misleading: the function is indeed calling another function, not using it as an argument.

Returning to the problem. "When we call the square function as a part of the cube function, we still have to define the parameter associated with the square function" - a hint provided says thus. But how to write this assignment? I have never assigned a parameter for a function in the body of another function.

Any hints?
Should I assign a parameter the same way as a variable?
 
These kind of simple questions is something you don't need a forum for...you can play around with the language and its compiler/interpreter and test stuff at will...it does not cost any money and you are not going to break anything...play around, experiment!

...if, on the other hand, you have a major logic problem or a bug that you cannot find...then, the forum could help.
 
mindauggas said:
Returning to the problem. "When we call the square function as a part of the cube function, we still have to define the parameter associated with the square function" - a hint provided says thus. But how to write this assignment? I have never assigned a parameter for a function in the body of another function.

Any hints?
Should I assign a parameter the same way as a variable?

Can you state the original problem/assignment as it is given to you?
 
  • #10
Using a function as an argument to another function would work like this:

Code:
var square = function(x) { return x * x; }
var cube = function(f,x) { return x * f(x); }

cube(square,2);

Though in this context I don't know why you'd want to pass in a function. I don't imagine you'd ever be swapping it out.
 
  • #11
DavidSnider said:
Using a function as an argument to another function would work like this:

Code:
var square = function(x) { return x * x; }
var cube = function(f,x) { return x * f(x); }

cube(square,2);

Though in this context I don't know why you'd want to pass in a function. I don't imagine you'd ever be swapping it out.

Thanks, this solved it.

Code:
var square = function (x) {
  return x * x;
};

var cube = function (x) {
  return square(x) * x;
};

The answer to your question: it just for the sake of knowing how to use functions in all/allmost alt the ways legitimate.
 
  • #12
I thought the idea was to use functions as parameters to other functions. I'm not sure if this syntax is correct for javascript, but it should look something like this:

Code:
var f1xf2 = function(f1, f2, x) {return f1(x) * f2(x);}
var returnx = function(x) {return x;}
var square = function(x) {return x * x;}
var cube = function(x) {return f1xf2(returnx, square, x);}
 
  • #13
rcgldr said:
I thought the idea was to use functions as parameters to other functions. I'm not sure if this syntax is correct for javascript, but it should look something like this:

Code:
var f1xf2 = function(f1, f2, x) {return f1(x) * f2(x);}
var returnx = function(x) {return x;}
var square = function(x) {return x * x;}
var cube = function(x) {return f1xf2(returnx, square, x);}

The original title probably mislead you, I apologize for this. the title should have stated that one function calls another function :)
 

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