Why is an exception being thrown in this dynamic array? - c++

Question

I am having trouble understanding why this exception is being thrown. I allocated an array to receive 100 int values and want to store all odd numbers under 200 into the array (which should be 100 integer values). I'm trying to understand why my code is not working.
I have called my function to allocate an array of 100 int values. After, I created a for-loop to iterate through and store integers into the array however I created an if statement to only store odd numbers. What I can't understand is if I put my counter to 200 and use the if statement an exception is thrown, but if I don't insert the if statement and only put my counter to 100 all numbers between 1-100 stored and an exception won't be thrown.
The only thing I can think of that's causing this is when my counter is at 200 and I have the if statement to catch all odd number, somehow all numbers under 200 are being stored in the array causing the exception to be thrown.
int *allocIntArray(int);
int main() {
int *a;
a = allocIntArray(100);
for (int count = 1; count < 200; count++) {
if (a[count] % 2 == 1) {
a[count] = count;
cout << a[count] << endl;
}
}
delete[] a;
return 0;
}
int *allocIntArray(int size) {
int *newarray = new int[size]();
return newarray;
}
When I look at the program output, it only displays the odd numbers yet the exception is being thrown. That tells me my if statement is working yet something is being muddied up.
What am I missing?
Thanks for your time and knowledge.

Answer 1

Cause of the error
If you have an array a that was created with n elements, it is undefined behavior when trying to access an array element out of bouds. So the index MUST always be between 0 and n-1.
So the behavior of your program is undefined as soon as count is 100, since evaluating the condition in the if-clause already tries to access out of bounds.
Adjustment that does what you want
Now in addition, there is a serious bug in your program logic: If you want to add numbers that satisfy some kind of condition, you need 2 counters: one for iterating on the numbers, and one for the last index used in the array:
for (int nextitem=0, count = 1; count < 200; count++) {
if (count % 2 == 1) { // not a[count], you need to test number itself
a[nextitem++] = count;
cout << count << endl;
if (nextitem == 100) { // attention: hard numbers should be avoided
cout << "Array full: " << nextitem << " items reached at " << count <<endl;
break; // exit the for loop
}
}
}
But, this solution requires you to keep track of the last item in the array, and the size of the array (it's hard-coded here).
Vectors
You are probably learning. But in C++ a better solution would be to use vector instead of an array, and use push_back(). Vectors manage the memory, so that you can focus on your algorithm. The full program would then look like:
vector<int> a;
for (int count = 1; count < 200; count++) {
if (count % 2 == 1) {
a.push_back(count);
cout << count << endl;
}
}
cout << "Added " << a.size() << " elements" <<endl;
cout << "10th element: "<< a[9] << endl;

Answer 2

The problem is not how many numbers you're storing but where you're storing them; you're storing 101 in a[101], which is obviously wrong.
If the i:th odd number is C, the correct index is i-1, not C.
The most readable change is probably to introduce a new counter variable.
int main() {
int a[100] = {0};
int count = 0;
for (int number = 1; number < 200; number++) {
if (number % 2 == 1) {
a[count] = number;
count += 1;
}
}
}
I think transforming this from a search problem to a generation problem makes it easier to get right.
If you happen to remember that every odd number C can be written on the form 2 * A + 1for some A, you' will see that the sequence you're looking for is
2*0+1, 2*1+1, 2*2+1, ..., 2*99+1
so
int main()
{
int numbers[100] = {0};
for (int i = 0; i < 100; i++)
{
numbers[i] = 2 * i + 1;
}
}
You can also go the other way around, looping over the odd numbers and storing them in the right place:
int main()
{
int numbers[100] = {0};
for (int i = 1; i < 200; i += 2) // This loops over the odd numbers.
{
numbers[i/2] = i; // Integer division makes this work.
}
}