Thread: dynamic memory:

No, you haven't allocated any space for the array. Also, you should print the values after breaking out of the loop. Try this:

Code:

#include <iostream>
using namespace std;
int main(void)
{
    int max_size = 1024;
    int n=0;
    int num=0;
    int *number=new int[max_size];
    cout<<"Enter a maximum of " << max_size << " values and enter -1 to stop the program \n";
    for(;;) {
        if(n == max_size)
        {
            cout<<"Sorry, no more room!\n";
            break;
        }
        cin>>num;
        if(num == -1)
            break;        
        number[n++]=num;
    }
    cout<<"Numbers entered:\n";
    for (int j=0; j<n; j++) {
        cout<<number[j]<<"\n";
    }    
    delete [] number;
    return 0;
}

Originally Posted by jocdrew21

Is this correct?

No since you allocate zero-length array - you have no place to store values.

You should implement the above using std::vector first. Using raw dynamic memory would be next step

Also, in your loop you're checking for -1 before the user input is even read.

Originally Posted by Sebastiani

Also, in your loop you're checking for -1 before the user input is even read.

It i ok since initially num is initialized to 0...
Problem lies later - when num is inserted into the array even if stop value of -1 was entered.

Originally Posted by jocdrew21

How did I not allocate it any space?

Since 'n' was initialized to zero, you were essentially doing this:

Code:

int *number=new int[0];

So basically asking the system for zero ints. Indexing the array with a higher number does not somehow automagically allocate more space, it just equates to an out-of-bounds access (OOBA). That said, you could certainly design a custom class that would treat OOBA's as append requests where the internal memory was resized before returning. But we'll leave that for a later exercise...

Originally Posted by jocdrew21

Now before some says it, I know I am treating this kinda like a vector and it would be smarter to use a vector. I just am practicing with array's.

In that case, design and implement your own simplified dynamic array class, using std::vector for reference.

This part is certainly wrong:

Code:

for (int i=0; i<n; i++) {
    cout<<easier[i]<<endl;
    easier.erase(easier.begin(),easier.end());
}

On the first iteration, you print easier[0], then you erase the entire range of the vector. Therefore, even if n was greater than 1, easier[1] ceases to exist, hence you access the vector out of bounds on the next iteration. You don't need to call erase at all.

Furthermore, you don't need n: you can just call easier.size().

Originally Posted by jocdrew21

why is that I have it working and outputting the way I would like the way I wrote it

Undefined behaviour.

Originally Posted by jocdrew21

but it but when I use easier.size() I only output the last number. Granted I left the erase method too.

You probably only output the first number. Since you left the erase call in, on the next iteration, the size() member function returns 0, hence the loop ends.

Originally Posted by jocdrew21

By the way, you have given me some of the best advice next to vart and white flags. You guys are really amazing.

You're welcome

Originally Posted by jocdrew21

How long have you been programing?

13 years or so.