A global static array in C++

Question

I've got something like this:

static int n = 0; // global static int int *arr = new int[n]; // global int* int randToArray(int arr[], int min, int max) { srand(time(NULL)); for(int i = 0; i <= n; i++) { arr[i] = (rand() % max + min); } } void printArray() { if(n) { for(int i = 0; i < n; i++) cout << arr[i] << endl; } else cout << "The array hasn't been drawed yet."; }

And then in the main function I have a menu with a switch and options for getting random numbers and printing the array:

switch(option) { case 1: cout << "Set the size of the array: "; cin >> n; randToArray(arr, 1, 99); break; case 2: printArray(); wait(); break; }

I need my array to be available globally in order to use in several other functions.

Everything works except one thing: when I want to print the array I get working properly only 8 first elements. Then the terminal shows up some very large numbers.

n is zero? That won't work correctly. Also regardless of the size of n you want your loop to terminate on i < n as your array is index from 0 to n-1. — Jay Miller
– Jay Miller, Commented Nov 19, 2014 at 18:05

vsoftco · Accepted Answer · 2014-11-19 18:03:49Z

That's because you use

static int n = 0;

then allocate memory for zero elements.

Change the line static int n = 256; for example, to allocate memory for 256 elements.

Or, if you read n after, allocate the memory AFTER you have read n. That is, declare the array globally first (technically a pointer), as int *arr;, then

arr = new int[n];

after cin >> n;

Deduplicator · Accepted Answer · 2014-11-19 18:29:37Z

static is a much overloaded keyword.
The way you use it, it means translation-unit-local.

Also, you don't have any global array in your code, only a pointer initialized to point to the beginning of an allocation of 0 ints.
That allocation won't be changed if you later change n.

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