How Can I Optimize My ROOT Code for Array Initialization and Graphing?

[RJLiberator]

Homework Statement

The goal is to create two arrays that have the values
a = { 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 }
b = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

And then graph them so that a=mb
slope m thus being 2.

I am working on the first part before I graph them.

Homework Equations

The Attempt at a Solution

{
int a[10];
for(int n=0; n<10; n++)
{
   a[n] = 2+2*n;
}
cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<" "<<a[8]<<" "<<a[9]<<endl;
int b[10];
for(int r=0; r<10; r++)
{
  b[r] = r+1;
}
cout<<b[0]<<" "<<b[1]<<" "<<b[2]<<" "<<b[3]<<" "<<b[4]<<" "<<b[5]<<" "<<b[6]<<" "<<b[7]<<" "<<b[8]<<" "<<b[9]<<endl;
}

My question:
a) Is there anything I can do to make the code shorter?
b) My professor wants the output to be in the form a[10] = { 2, 4, 6, 8, 10, 12, 14, 16, 18, 20 } instead of vertical list. That's why I had to do the cout<<...<<endl. Is there anyway to make this output more easier than simply listing all the a[1]<<a[2]... etc.?
c) If I've done everything correctly here, how do I go from having these 2 arrays to inserting them into a simple xy graph? Any pointers to get me started there? I'm looking through tutorials now.

 

[RJLiberator]

As an update, I have had success creating the graph with the help of some pre-built code.

void CurrentGraph() {

  TCanvas *c1 = new TCanvas("c1","V=IR",200,10,700,500);

  c1->SetFillColor(42);
  c1->SetGrid();

  const Int_t n = 10;
  Double_t x[n], y[n];
  for (Int_t i=0;i<n;i++) {
  x[i] = i+1;
  y[i] = 2*x[i];
  printf(" i %i %f %f \n",i,x[i],y[i]);
  }
  TGraph *gr = new TGraph(n,x,y);
  gr->SetLineColor(2);
  gr->SetLineWidth(4);
  gr->SetMarkerColor(4);
  gr->SetMarkerStyle(21);
  gr->SetTitle("I = V/r");
  gr->GetXaxis()->SetTitle("Resistance");
  gr->GetYaxis()->SetTitle("Voltage");
  gr->Draw("ACP");

  c1->Update();
  c1->GetFrame()->SetFillColor(21);
  c1->GetFrame()->SetBorderSize(12);
  c1->Modified();
}

With that being stated, is there any better way to do what I am trying to do?

My professor suggested using the "Float" command some how... I am not sure how this one works.

 

[Mark44]

{
int a[10];
for(int n=0; n<10; n++)
{
   a[n] = 2+2*n;
}
count<<a[0]<<" "<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<" "<<a[8]<<" "<<a[9]<<endl;
int b[10];
for(int r=0; r<10; r++)
{
  b[r] = r+1;
}
count<<b[0]<<" "<<b[1]<<" "<<b[2]<<" "<<b[3]<<" "<<b[4]<<" "<<b[5]<<" "<<b[6]<<" "<<b[7]<<" "<<b[8]<<" "<<b[9]<<endl;
}

What is ROOT? Your code appears to be C++.

Here is your code from post 1, indented for better readability, and incorporating your output into the two for loops.

{
   int a[10];
   for(int n=0; n<10; n++)
   {
       a[n] = 2+2*n;
       count << a[n] << " ";
   }
   count << endl;
   // count<<a[0]<<" "<<a[1]<<" "<<a[2]<<" "<<a[3]<<" "<<a[4]<<" "<<a[5]<<" "<<a[6]<<" "<<a[7]<<" "<<a[8]<<" "<<a[9]<<endl;

   int b[10];
   for(int r=0; r<10; r++)
   {
      b[r] = r+1;
      count << b[r] << " ";
   }
   count << endl;
   // count<<b[0]<<" "<<b[1]<<" "<<b[2]<<" "<<b[3]<<" "<<b[4]<<" "<<b[5]<<" "<<b[6]<<" "<<b[7]<<" "<<b[8]<<" "<<b[9]<<endl;
}

There is a slight difference between what you did and what I have in the code just above: after the last array item is printed, my code will add one space and then the end-of-line character.In your code in post 2, why did you switch from int to Int_t? I don't seen any advantage in using Double_t instead of double, either.
Some advice: In your for loop, space things out -- it makes things easier for humans to read.
The loop should look like this:
for (Int_t i = 0; i < n; i++)
Notice the extra spaces I added.
Here's your code from post 2. My only change is the indentation of the for loop.

void CurrentGraph() {

  TCanvas *c1 = new TCanvas("c1","V=IR",200,10,700,500);

  c1->SetFillColor(42);
  c1->SetGrid();

  const Int_t n = 10;
  Double_t x[n], y[n];
  for (Int_t i=0;i<n;i++)
  {
      x[i] = i+1;
      y[i] = 2*x[i];
      printf(" i %i %f %f \n",i,x[i],y[i]);
  }
  TGraph *gr = new TGraph(n,x,y);
  gr->SetLineColor(2);
  gr->SetLineWidth(4);
  gr->SetMarkerColor(4);
  gr->SetMarkerStyle(21);
  gr->SetTitle("I = V/r");
  gr->GetXaxis()->SetTitle("Resistance");
  gr->GetYaxis()->SetTitle("Voltage");
  gr->Draw("ACP");

  c1->Update();
  c1->GetFrame()->SetFillColor(21);
  c1->GetFrame()->SetBorderSize(12);
  c1->Modified();
}

My professor suggested using the "Float" command some how... I am not sure how this one works.

float is a type, not a command. Your code from post 2 declares your arrays to be of type double, which is another floating point type. The float type is generally 4 bytes, while the double type is generally 8 bytes, so can store larger (or smaller) numbers with higher precision. Double_t, which you are using, is probably just an alias for double, so in your code, I don't see any reason to use float instead.

 

[RJLiberator]

What is ROOT? Your code appears to be C++.

Root is a modular scientific software framework. It provides all the functionalities needed to deal with big data processing, statistical analysis, visualisation and storage. It is mainly written in C++ but integrated with other languages such as Python and R.
https://root.cern.ch/
My course in Computational and Mathematical physics is working with ROOT. So, basically C++!

In your code in post 2, why did you switch from int to Int_t? I don't seen any advantage in using Double_t instead of double, either.

Ignorance on my part. So, essentially int = int_t and it would be simpler to use int. Not sure at all why the code was using int_t then. I will try it out with just int.

Some advice: In your for loop, space things out -- it makes things easier for humans to read.

Ah, that clears up some minor confusion I've been having.

float is a type, not a command. Your code from post 2 declares your arrays to be of type double, which is another floating point type. The float type is generally 4 bytes, while the double type is generally 8 bytes, so can store larger (or smaller) numbers with higher precision. Double_t, which you are using, is probably just an alias for double, so in your code, I don't see any reason to use float instead.

That makes it very, very clear to me and those words have helped me understand what I am doing a lot.
I thank you sir. I am going to try out some things, and perhaps report back.

As always, Mark44, you are truly greatness. :)

As an EDIT: I removed the "_t" part from int_t and double_t and the graph no longer showed when I ran it. :/

 

[Mark44]

As an EDIT: I removed the "_t" part from int_t and double_t and the graph no longer showed when I ran it. :/

Did you change all three of these lines in your CurrentGraph() function?

const Int_t n = 10;
Double_t x[n], y[n];
for (Int_t i=0;i<n;i++)

I don't have any idea why changing Double_T and Int_t to double and int would cause the problem you said.

 

[RJLiberator]

Yes, I change all three lines.
The program runs, but when it comes up, it appears that the background color that I set is covering over everything.
It is as if the graph part of it is hidden behind the background when I make the three changes.

EDIT: Got it.
The problem was that When doing Int_t the Int part was capitalized.
So I had to change Int_t to int instead of Int. Same thing with double.
:) Cheers.

 

[Mark44]

The problem was that When doing Int_t the Int part was capitalized.
So I had to change Int_t to int instead of Int. Same thing with double.

That would do it. All of the built-in types in C and C++ are lower-case: char, int, long, float, double, etc.

I'm surprised your code would even compile if you had Int instead of int.

 

[RJLiberator]

@Mark44 is there anything you can tell me about the following command:

printf(" i %i %f %f \n",i,x[i],y[i]);

I understand that this command prints the values out into the command terminal.

But why does %i go from 0 to 10 and then the first %f list the array from 1 to 10 and then the next %f list the array that goes 2, 4, 6, 8, etc. And what does \n signify?

Then why do we need another i after that before x[ i ], y[ i ]?

My only thought is that the are connected in the order they come. The first %f goes with x[ i ] and similarly the second %f goes with y[ i ] based on the way the code was written.

 

[Mark44]

@Mark44 is there anything you can tell me about the following command:

printf(" i %i %f %f \n",i,x[i],y[i]);

I understand that this command prints the values out into the command terminal.

But why does %i go from 0 to 10 and then the first %f list the array from 1 to 10 and then the next %f list the array that goes 2, 4, 6, 8, etc. And what does \n signify?

Here's the context of your question.

const Int_t n = 10;
  Double_t x[n], y[n];
  for (Int_t i=0;i<n;i++)
  {
      x[i] = i+1;
      y[i] = 2*x[i];
      printf(" i %i %f %f \n",i,x[i],y[i]);
  }

%i is not a variable, so it doesn't take on values. In the printf() call, the first argument is the control string. It consists of literals that will be printed (such as i and \n) and conversion specifiers (%i and %f). The 2nd, 3rd, and 4th arguments are expressions whose values will be displayed based on the conversion specifiers in the control string, in the same order.

So it's not %i that goes from 0 to 10 -- the loop control variable goes from 0 through 9. Each iteration of the loop causes another element of each array to be set, and causes the printf() function to be called (don't call it a "command" -- you'll seem like a newbie).

The first time through the loop, x[0] gets set to 1, y[0] gets set to 2, and printf() prints this:
i 0 1.0 2.0
and moves the cursor to the next line, as a result of the \n (newline) character.
The next iteration of the loop prints the next value of i (1) and the two array values, and so on, until the loop finishes.

Then why do we need another i after that before x[ i ], y[ i ]?

My only thought is that the are connected in the order they come. The first %f goes with x[ i ] and similarly the second %f goes with y[ i ] based on the way the code was written.

Yes.

 

[rootone]

what does \n signify?

\n means only 'start a new line' for the printout.