Keep in mind, this is part of a homework assignment - so please, no direct answer. I just need some help in finding out the answer, so a link to a tutorial to help me understand the material would be great.
SML code:
datatype 'ingredient pizza =
Bottom
| Topping of ('ingredient * ('ingredient pizza));
datatype fish =
Anchovy
| Shark
| Tuna;
(* Testing Pizza Objects *)
val my_pizza1 = Topping(Tuna, Topping(Shark, Topping(Anchovy, Bottom)));
val my_pizza2 = Topping(Shark, Topping(Tuna, Topping(Anchovy, Bottom)));
val my_pizza3 = Topping(Anchovy, Topping(Shark, Topping(Tuna, Bottom)));
(* My Function Start *)
fun rem_ingredient Bottom = Bottom
| rem_ingredient(t) = fn(Topping(p)) => Topping(t, rem_ingredient(p))
| rem_ingredient(Topping(t,p)) = Topping(t, rem_ingredient(p));
(* My Function End *)
If I call the function rem_ingredient with 1 parameter
val rem_tuna = rem_ingredient Tuna;"
I should get a function that can then call
rem_tuna my_pizza3;
to remove Tuna from pizza3
If I call the same function with 2 parameters
rem_ingredient Tuna my_pizza2;
I should directly remove Tuna from the pizza2 object with the 2 parameters.
The Problem:
I keep getting the error: syntax error: replacing EQUALOP with DARROW on the 3rd constructor of rem_ingredient, I know I'm missing something that is probably obvious. We just start learning SML last week in Programming Languages and I'm still trying to understand it. Anyone pointing me in direction would be appreciated.
Again, no direct answer please, I want to learn the material, but I'm not sure what I'm trying to fix.
To get rid of the syntax error you need to put parentheses around the fn expression (since the following | pattern is otherwise taken to be part of the fn).
However, that is not your real problem here. The function as written does not have a consistent type, because the second case returns a function while the others do not.
Related
Please excuse the beginner question. I couldn't find an appropriate answer in any Mathematica tutorial.
I am confused why a definition as a function or a definition in terms of a simple replacement produce different results. Consider this example (Mathematica 9 code):
In[397]:= ClearAll["Global`*"]
In[398]:= Test := 3 c^2 + d^4
In[399]:= v[f_] := D[f, c]
In[400]:= v[Test]
Out[400]= 6 c
The first definition of this simple derivative function "v" acting on a variable is fine. Defining a replacement Test = ... to replace the variable produces the expected result (It derives 3c^2+d^4 with respect to c and answers 6c).
However if I define a function instead of a simple replacement this does not work:
In[401]:= TestFunction[a_, b_] := 3 a^2 + b^4
In[403]:= vFunction[f_[a_, b_]] := D[f[a, b], a]
In[405]:= vFunction[TestFunction[a, b]]
Out[405]= \!\(
\*SubscriptBox[\(\[PartialD]\), \(3\
\*SuperscriptBox[\(a\), \(2\)]\)]\((3\
\*SuperscriptBox[\(a\), \(2\)] +
\*SuperscriptBox[\(b\), \(4\)])\)\)
Why is that? I am risking to look like a moron here, but please enlighten me!
For your convenience, I uploaded a copy of my workbook here
Thanks a lot,
Michael
Do this instead
vFunction[f_,a_,b_]:=D[f[a,b],a];
and when you need derivatives simply use vFunction[TestFunction,a,b] to get it.
When you write down f[x], it means the evaluated value of f with argument value x. So, f[x] is technically not a function anymore. What you want as the argument of vFunction[] is the function TestFunction, not the evaluated value.
As a homework assignment, I'm writing a code that uses the bisection method to calculate the root of a function with one variable within a range. I created a user function that does the calculations, but one of the inputs of the function is supposed to be "fun" which is supposed to be set equal to the function.
Here is my code, before I go on:
function [ Ts ] = BisectionRoot( fun,a,b,TolMax )
%This function finds the value of Ts by finding the root of a given function within a given range to a given
%tolerance, using the Bisection Method.
Fa = fun(a);
Fb = fun(b);
if Fa * Fb > 0
disp('Error: The function has no roots in between the given bounds')
else
xNS = (a + b)/2;
toli = abs((b-a)/2);
FxNS = fun(xns);
if FxNS == 0
Ts = xNS;
break
end
if toli , TolMax
Ts = xNS;
break
end
if fun(a) * FxNS < 0
b = xNS;
else
a = xNS;
end
end
Ts
end
The input arguments are defined by our teacher, so I can't mess with them. We're supposed to set those variables in the command window before running the function. That way, we can use the program later on for other things. (Even though I think fzero() can be used to do this)
My problem is that I'm not sure how to set fun to something, and then use that in a way that I can do fun(a) or fun(b). In our book they do something they call defining f(x) as an anonymous function. They do this for an example problem:
F = # (x) 8-4.5*(x-sin(x))
But when I try doing that, I get the error, Error: Unexpected MATLAB operator.
If you guys want to try running the program to test your solutions before posting (hopefully my program works!) you can use these variables from an example in the book:
fun = 8 - 4.5*(x - sin(x))
a = 2
b = 3
TolMax = .001
The answer the get in the book for using those is 2.430664.
I'm sure the answer to this is incredibly easy and straightforward, but for some reason, I can't find a way to do it! Thank you for your help.
To get you going, it looks like your example is missing some syntax. Instead of either of these (from your question):
fun = 8 - 4.5*(x - sin(x)) % Missing function handle declaration symbol "#"
F = # (x) 8-4.5*(x-sin9(x)) %Unless you have defined it, there is no function "sin9"
Use
fun = #(x) 8 - 4.5*(x - sin(x))
Then you would call your function like this:
fun = #(x) 8 - 4.5*(x - sin(x));
a = 2;
b = 3;
TolMax = .001;
root = BisectionRoot( fun,a,b,TolMax );
To debug (which you will need to do), use the debugger.
The command dbstop if error stops execution and opens the file at the point of the problem, letting you examine the variable values and function stack.
Clicking on the "-" marks in the editor creates a break point, forcing the function to pause execution at that point, again so that you can examine the contents. Note that you can step through the code line by line using the debug buttons at the top of the editor.
dbquit quits debug mode
dbclear all clears all break points
I'm having some trouble with the error "1050: Cannot assign to a non-reference value." I'm still fairly new to coding, and so being unable to fix this error is frustrating, any help will be greatly appreciated.
var PracticeDummyHealth:int=50
var PlayerAttack:int=20;
public function PlayerAttackFunction(){
if(PracticeDummyHealth>0){
PracticeDummyHealth-PlayerAttack=PracticeDummyHealth;
}
}
An grammar construct which is not a Property/Variable name is on the left of the = assignment operator:
// expression = expression
PracticeDummyHealth-PlayerAttack=PracticeDummyHealth;
// which makes as much sense to ActionScript as .. it's not an equation solver :)
// 100 - 50 = 100
Compare with this valid code:
// variable = new_value
PracticeDummyHealth = PracticeDummyHealth - PlayerAttack;
// or
PracticeDummyHealth -= PlayerAttack;
Note that a "reference" (read: Property/Variable name) appears on the left of the = (or compound -=) in both of these cases. This terminology comes from the specification which deals with l-values and it is slightly unfortunate it doesn't yield a nicer error message here.
I've been looking around and I have not been able to find anything that has worked for me. I'm starting to learn more Lua and to start off I'm making a simple calculator. I was able to get each individual operation onto separate programs, but when I try to combine them I just can't get it to work. My script as it is now is
require "io"
require "operations.lua"
do
print ("Please enter the first number in your problem.")
x = io.read()
print ("Please enter the second number in your problem.")
y = io.read()
print ("Please choose the operation you wish to perform.")
print ("Use 1 for addition, 2 for subtraction, 3 for multiplication, and 4 for division.")
op = io.read()
op = 1 then
function addition
op = 2 then
function subtraction
op = 3 then
function multiplication
op = 4 then
function division
print (answer)
io.read()
end
and my operations.lua script is
function addition
return answer = x+y
end
function subtraction
return answer = x-y
end
function multiplication
return answer = x*y
end
function division
return answer = x/y
end
I've tried using
if op = 1 then
answer = x+y
print(answer)
if op = 2 then
answer = x-y
print(answer)
and I did that completing each operation. But it doesn't work. I can't even get the error code that it's returning because it closes so fast. What should I do?
In your example, make these changes: You require operations.lua without the extension. Include parameters in your operations function definitions. Return the operation expression directly versus returning a statement like answer = x+y.
All together:
Code for operations.lua
function addition(x,y)
return x + y
end
--more functions go here...
function division(x,y)
return x / y
end
Code for your hosting Lua script:
require "operations"
result = addition(5,7)
print(result)
result = division(9,3)
print(result)
Once you get that working, try re-adding your io logic.
Keep in mind that as it's coded, your functions will be defined globally. To avoid polluting the global table, consider defining operations.lua as a module. Take a look at the lua-users.org Modules Tutorial.
The right if-then-else syntax:
if op==1 then
answer = a+b
elseif op==2 then
answer = a*b
end
print(answer)
After: please check the correct function-declaration syntax.
After: return answer=x+y is incorrect. If you want set answer's value, set without return. If you want return the sum, please use return x+y.
And I think you should check Programming in Lua.
First of all, learn to use the command line so you can see the errors (on Windows that would be cmd.exe).
Second, change the second line to require("operations"). The way you did it the interpreter expects a directory operations with an underlying script lua.lua.
I'm new to Lua, so (naturally) I got stuck at the first thing I tried to program. I'm working with an example script provided with the Corona Developer package. Here's a simplified version of the function (irrelevant material removed) I'm trying to call:
function new( imageSet, slideBackground, top, bottom )
function g:jumpToImage(num)
print(num)
local i = 0
print("jumpToImage")
print("#images", #images)
for i = 1, #images do
if i < num then
images[i].x = -screenW*.5;
elseif i > num then
images[i].x = screenW*1.5 + pad
else
images[i].x = screenW*.5 - pad
end
end
imgNum = num
initImage(imgNum)
end
end
If I try to call that function like this:
local test = slideView.new( myImages )
test.jumpToImage(2)
I get this error:
attempt to compare number with nil
at line 225. It would seem that "num" is not getting passed into the function. Why is this?
Where are you declaring g? You're adding a method to g, which doesn't exist (as a local). Then you're never returning g either. But most likely those were just copying errors or something. The real error is probably the notation that you're using to call test:jumpToImage.
You declare g:jumpToImage(num). That colon there means that the first argument should be treated as self. So really, your function is g.jumpToImage(self, num)
Later, you call it as test.jumpToImage(2). That makes the actual arguments of self be 2 and num be nil. What you want to do is test:jumpToImage(2). The colon there makes the expression expand to test.jumpToImage(test, 2)
Take a look at this page for an explanation of Lua's : syntax.