How do I base size of arrays on annother number?

Say I get the input from a user of how many arguements they want to enter: cin>>int num;
Then, I want to create an array with (num) slots in it. This is what I have, but it doesn't work. Everything else is the rest of my program, which, suprisingly, doesn't work (or generates more errors). I want to eventually build something which will calculate standard deviation for you, but I can't even get past the first part, finding the average.

Code:

---

#include <iostream.h>

double temp;
int main()
{
        double num;
        cin>>num;
        float arguments[num];
        for (int i=0; i<=num; i++)
        {
                cout<<"Enter data member number "<<i<<"-> ";
                cin>>arguements[i];
                double temp =+ arguements[i];
        }
        double mean = temp / num;
        return 0;
}

---

Gets:

--------------------Configuration: deviationcpp - Win32 Debug--------------------
Compiling...
deviationcpp.cpp
C:\My Documents\C++\Deviation\deviationcpp.cpp(9) : error C2057: expected constant expression
C:\My Documents\C++\Deviation\deviationcpp.cpp(9) : error C2466: cannot allocate an array of constant size 0
C:\My Documents\C++\Deviation\deviationcpp.cpp(9) : error C2133: 'arguments' : unknown size
C:\My Documents\C++\Deviation\deviationcpp.cpp(13) : error C2065: 'arguements' : undeclared identifier
C:\My Documents\C++\Deviation\deviationcpp.cpp(13) : error C2109: subscript requires array or pointer type
C:\My Documents\C++\Deviation\deviationcpp.cpp(13) : error C2679: binary '>>' : no operator defined which takes a right-hand operand of type 'int' (or there is no acceptable conversion)
C:\My Documents\C++\Deviation\deviationcpp.cpp(14) : error C2109: subscript requires array or pointer type
Error executing cl.exe.

deviationcpp.obj - 7 error(s), 0 warning(s)

> double temp;
Try to avoid using globals

> double num;
This needs to be an int - arrays can only have integer sizes.

> float arguments[num];
This should be
  float *arguments = new float[num];
This will create a dynamic array, as if you had declared
  float arguments[num];

> for (int i=0; i<=num; i++)
Like for all arrays, the loop should be
for (int i=0; i<num; i++)

> double temp =+ arguements[i];
1. temp is uninitialised
2. it goes out of scope before you have a chance to use it in calculating the average.

> If an array can only carry integer sizes, does that mean the user can't input numbers with decimals?
No - the size of an array (the bit in the [] ) and what each element contains (the type) are totally different.

float arr[1.234]; // is meaningless

but

float arr[2];
arr[0] = 1.234; // is fine

Quote:

---

double temp =+ arguements[i];
1. temp is uninitialised
2. it goes out of scope before you have a chance to use it in calculating the average.

---

Well, how would I keep temp 'in scope' long enough to calcualte the average? Sorry, Im still really bad on the syntax of c++.
EDIT: Yeah, I realized that I made a mistake about the size of an array to the data it can carry, so I deleted my post to cover up my stupidity :)

double temp = 0;
before you enter the loop in the first place.

Okay :) Everything compiles alright; exept when I output the mean it always equals 0.. why is this?

How many instances of data do you want to enter? ->6
Enter data member number 1-> 3
Enter data member number 2-> 5
Enter data member number 3-> 6
Enter data member number 4-> 2
Enter data member number 5-> 4
Enter data member number 6-> 3
The mean(average) of your data = 0
Press any key to continue

Code:

---

#include <iostream.h>

int main()
{
        double temp = 0;
        int num;
        cout<<"How many instances of data do you want to enter? ->";
        cin>>num;
        float *arguments = new float[num];
        for (int i=0; i<num; i++)
        {
                cout<<"Enter data member number "<<i+1<<"-> ";
                cin>>arguments[i];
                double temp =+ arguments[i];
        }
        double mean = temp / num;
        cout<<"The mean(average) of your data = "<<mean<<"\n";
        return 0;
}

---

> double temp =+ arguments[i];
You have two temp's in scope here - the one you've just declared, and the one at the start of main.

Unfortunately for you, you update this one, whilst the one at the start of main remains untouched.

temp =+ arguments[i];
is all you need.

Just an fyi whenever you use new always check for NULL pointer, most likely meaning insufficient memory (Dereferencing a null pointer is bad ju-ju). And always do a delete to put the allocated memory, from the new, back in the heap. They're just good programming practices.

I want to add error checking, but I don't know of any easy way to check if the input is numerical or not, if anyone could help me out, that'd be great; also, the 'float devmean = temp / num;' isn't returning the right value. I don't know if its the calculations before it, or whatnot.

Code:

---


#include <iostream.h>
#include <math.h>
#include <string>

// This program finds the standard deviation of a set of data.
// 1: Mean
// 2: (Data Member - mean) * 2.
// 3: Mean of data from 2.
// 4: Square root of numbers from 3.

int main()
{
        float temp = 0;
        int num;
        cout<<"How many instances of data do you want to enter? ->";
        cin>>num;
        cout<<"\n--------------------------\n";
        float *arguments = new float[num];
        for (int i=0; i<num; i++)
        {
                cout<<"Enter data member number "<<i+1<<"-> ";
                cin>>arguments[i];
                temp = temp + arguments[i];
        }
        float mean = temp / num;
        temp = 0;
        cout<<"\n\nThe mean(average) of your data = "<<mean<<"\n";
        cout<<"\n--------------------------\n";
        for (i=0; i<num; i++)
        {
                arguments[i] = arguments[i] - mean;
                arguments[i] = arguments[i] * 2;
                temp = temp + arguments[i];
                cout<<"Squared deviation of data member "<<i+1<<" = "<<arguments[i]<<"\n";
        }
        float devmean = temp / num;
        float squareroot = sqrt( devmean );
        cout<<"\n\n";
        cout<<"The mean of the squared deviations = "<<devmean;
        cout<<"\nStandard Deviation of your data = "<<squareroot<<"\n\n";
       
        return 0;
}

---

Don't know about how to make sure it's an int, but your float devmean = temp/num is probably running into typecasting problems. You're dividing a float (temp) by an int (num). Try a static cast.

If I knew what I static cast was, then maybe :)
On a side note, what is a static cast? Can you give me an example or modify my code so that it fits in?

Output:


The mean(average) of your data = 4.75

--------------------------
Squared deviation of data member 1 = -5.5
Squared deviation of data member 2 = -3.5
Squared deviation of data member 3 = -1.5
Squared deviation of data member 4 = 0.5
Squared deviation of data member 5 = 2.5
Squared deviation of data member 6 = 4.5
Squared deviation of data member 7 = 0.5
Squared deviation of data member 8 = -3.5
Squared deviation of data member 9 = 0.5
Squared deviation of data member 10 = 2.5
Squared deviation of data member 11 = 4.5
Squared deviation of data member 12 = -1.5


The mean of the squared deviations = 0
Standard Deviation of your data = 0

Press any key to continue

Based on my above code; can anyone see whats wrong?
Also, I might get output like this:

The mean(average) of your data = 5.17857

--------------------------
Squared deviation of data member 1 = -3.27714
Squared deviation of data member 2 = -3.75714
Squared deviation of data member 3 = 0.942858
Squared deviation of data member 4 = -3.45714
Squared deviation of data member 5 = -5.67714
Squared deviation of data member 6 = 8.50286
Squared deviation of data member 7 = 6.72286


The mean of the squared deviations = 5.44957e-007
Standard Deviation of your data = 0.000738212

The zero variance is because of a math error on your part.

You have each element set to its variation from the mean TIMES TWO, instead of its variation SQUARED. Because of how a mean works, if you sum the variations from the mean, you always get zero.

So, instead of:

arguments[i] = arguments[i] * 2;

you need:

arguments[i] = pow(arguments[i],2);

Oh, and, as mentioned, for good form you should have this line just before the return:

delete[] arguments;

This line frees the memory that was allocated to your array. Anytime you allocate memory with new, you should use delete somewhere to deallocate it.

There are a *few* exceptions to this, but as a rule of thumb, it's a really good idea to get yourself into the habit of matching every new with a delete.

BTW, you use delete[] instead of delete because you made an array with new.

One final comment:

If you're trying to calculate variance and standard deviation, you don't divide by num, you divide by (num - 1).

I forget why you do this, but variance and st.dev. are always over the number of samples minus one.