How do I store value into a result array?

Question

1.00/5 (2 votes)

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I have 4 arrays. a=6 indexes, b=20 indexes, x&t=10001 indexes. I have to calculate 10001 values and then sum it and then sum it all up. I have to operate the formula using unique combinations of (a,b) against every value of (x,t).

C++

long double temps[10001];
    long double result = 0;
    long double resids[120];
    for (int i = 0; i < 6; i++)
    {
        for (int j = 0; j < 20; j++)
        {
            for (int k = 0; k < 10001; k++)
            {
                temps[i] = pow((x[k] - (a[i] + (0.955 - a[i])*(exp(-(b[j]) * t[k])))), 2);
                result += temps[i];
            }
            for (int l = 0; l < 120; l++)
            {
                resids[l] = result;
            }
        }
    }

Posted 21-Aug-15 0:33am

09hadi

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CPallini 21-Aug-15 6:44am

Could you please provide more details? It is not clear what the actual question is.

KarstenK 21-Aug-15 6:50am

Please clearify your question. Use in the text the same variable names as in your code. It is best practice to use long or meaningful names.

I guess you need a two dimensional array. Like results[6][20];

09hadi 21-Aug-15 7:11am

double a[6], b[20], x[10001], t[10001];
these are the arrays

Sergey Alexandrovich Kryukov 21-Aug-15 11:12am

And the problem is..?
—SA

09hadi 21-Aug-15 14:28pm

I cant find the place where the fourth loop should be placed in order to get the values into the resid array. I know this one is stupid

Sergey Alexandrovich Kryukov 21-Aug-15 15:21pm

I don't know what to say. Programming is not done by finding a place for a loop, or something like that. This is strange way of thinking. You start with your goal and then decide what you need to achieve it. The loop is not a goal, but an instrument.
—SA

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bling · Accepted Answer · 2015-08-21T09:36:00

Solution 1

I see multiple problems with this code. To specifically answer your question about the result array ...

C++

long double temps[10001];
long double result = 0;
long double resids[120];
for (int i = 0; i < 6; i++)
{
    for (int j = 0; j < 20; j++)
    {
        result = 0; // unsure
        for (int k = 0; k < 10001; k++)
        {
            temps[i] = pow((x[k] - (a[i] + (0.955 - a[i])*(exp(-(b[j]) * t[k])))), 2);
            result += temps[i];
        }
        resids[i * 20 + j] = result; // <-- 120 results go here.
    }
}

I can't tell whether "result" should be reset to zero at the start of each "i" pass or each "j" pass.

Posted 21-Aug-15 9:36am

bling

Comments

09hadi 22-Aug-15 3:44am

Thank you very much.

09hadi 24-Aug-15 14:16pm

can I trouble you a bit more? the result of these calculations comes down to 20 decimal points. which double is unable to handle. Can you help me with that as well? How can I cater to that?

[no name] 24-Aug-15 14:47pm

If you need more than 15 digits (double precision) AND you are okay with fixed point math, a signed long will give you almost 20 digits. Your long values will have an implied decimal point. As long as all values have the same implied decimal point, you can use the + and - operators without shifting. Multiplication and division are a bit more complicated. You'd have to write your own code to print these fixed point values. If you need a full 20 digits or more, use "long long" as fixed point values.

09hadi 27-Aug-15 6:57am

can u give me a sample please?

[no name] 27-Aug-15 17:57pm

How many digits do you want to the left of the decimal point? How many digits do you want to the right of the decimal point?

09hadi 31-Aug-15 1:52am

I want 1 digit right of the decimal and I want 25 on the left

[no name] 31-Aug-15 12:06pm

Use arithmetic like this:
float x = 1.2f;
float y = 123456789012345678.1f;
long value1 = (long)(x * 10.f);
long value2 = (long)(y * 10.f);
long sum = value1 + value2;
// to print this in decimal form ...
printf("%ld.%1ld", sum / 10, sum % 10);

You'll have to go through your code and change it to the above form. The above code is good to over 19 digits. Use "long long" for more digits.