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Accessing values of a ScalarField

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Old   December 20, 2020, 15:54
Default Accessing values of a ScalarField
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Saumava Dey
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If I define a volScalarField within my code, does it by default correspond to the mesh size I am working with?

The mesh I am working with is having 100 cells. I am asking this because if I am printing the 150th element value by the command:

Info << h.internalField()[150] << endl;

it is still printing some value, which in my knowledge is not supposed to happen.

Can anyone explain why is this happening?
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Old   December 20, 2020, 16:02
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Domenico Lahaye
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Boundary values, i.e., values of volScalarField in center of face on the boundary?
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Old   December 21, 2020, 01:35
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If it is printing the boundary values then it should have some limit. If I am trying to print the 1000th value for a 100 cell mesh, then also it is printing some value. So, the mystery still remains persistent.
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Old   December 21, 2020, 08:12
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Run in debugger?
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Old   January 1, 2021, 11:37
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A part of my code reads like this:

C = (T1-T2)/(H1-H2);

All variables, C, T1, T2, H1, H2 are volScalarFields corresponding to a particular 1D mesh.

This step is not working.

Individually when I am printing the values of the fields (T1-T2) and (H1-H2), the values are finite corresponding to the number of cells.

Can't figure out where is the error.
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Old   January 1, 2021, 13:43
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Domenico Lahaye
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The division of vector is mathematically not defined.

What are you trying to accomplish?
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Old   January 2, 2021, 13:52
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I require to find out the value of C for each cell which can be calculated as:
C[i] = (T1[i] - T2[i])/(H1[i] - H2[i]);

All variables, C, T1, T2, H1, H2 are volScalarFields corresponding to a particular 1D mesh.
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Old   January 2, 2021, 15:22
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Again, the division of vectors is mathematically not defined.
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Old   January 3, 2021, 07:32
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This is as expected, and is a feature of C++ (ie nothing to do with OpenFOAM). For example, look at the following code:
Code:
#include<iostream>
using namespace std;
int main()
{
		int a[3] = {40, 41, 42};
		cout << "a[0] = " << a[0] << endl;
		cout << "a[2] = " << a[2] << endl;
		cout << "a[5] = " << a[5] << endl;
		return 0;
}
I define an array with 3 entries .. but can still write the contents of the 6th - the output (try it) is:
Quote:
a[0] = 40
a[2] = 42
a[5] = 6294224
So what is going on? Well, have a read up on arrays in C++ ... you'll realise that arrays are addressed by pointers, in other words a[5] is just equivalent to *(a+5), and there is nothing stopping you calculating whatever pointer address you want and then going and checking the memory contents of that address (so long as the address exists). The contents may be junk, of course ... like in the above example.

So, long and short of it - you can't rely on the code spitting out its dummy if you go beyond the array dimensions; instead, you need to check yourself.
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