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# C++ Returning Arrays from Functions [ C++ Arrays](#) C++ does not allow returning a complete array as a function argument. However, you can return a pointer to an array by specifying the array name without an index. If you want to return a one-dimensional array from a function, you must declare a function that returns a pointer, as follows: int * myFunction(){//. . .} Here is an example: ## Example int* myFunction() { int myArray={1, 2, 3}; return myArray; } **Note:** You cannot simply return a pointer to a local array, because the local array will be destroyed when the function ends, and the pointer pointing to it will become invalid. C++ does not support returning the address of a local variable outside the function unless the local variable is defined as a **static** variable. To avoid the above situation, you can use a static array or dynamically allocate an array. Using a static array requires creating a static array inside the function and returning its address, for example: int* myFunction() { static int myArray={1, 2, 3}; return myArray; } Now, let's look at the following function, which generates 10 random numbers and uses an array to return them, as follows: ## Example #include#include#includeusing namespace std; // Function to generate and return random numbers int * getRandom(){static int r; // Set the seed srand((unsigned)time(NULL)); for(int i = 0; i<10; ++i){r = rand(); cout<<r<<endl; }return r; }// Main function to call the above defined function int main(){// A pointer to an integer int *p; p = getRandom(); for(int i = 0; i<10; i++ ){cout<<"*(p + "<<i<<") : "; cout<< *(p + i)<<endl; }return 0; } When the above code is compiled and executed, it produces the following result: 624723190146873569580711358597