Reputation: 4685
I have a function that needs to return a pointer to an array:
int * count()
{
static int myInt[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
return &myInt[10];
}
inside my main function I want to display one of the ints from that array, like here at index 3
int main(int argc, const char * argv[])
{
int myInt2[10] = *count();
std::cout << myInt2[3] << "\n\n";
return 0;
}
this however gives me the error: "Array initializer must be an initializer list"
how do I create an array within my main function that uses the pointer to get the same elements as the array at the pointer?
Upvotes: 0
Views: 177
Reputation: 4939
A few problems in your code:
1) you need to return a pointer to the beginning of the array in count:
return &myInt[0];
or
return myInt; //should suffice.
Then when you initialize myInt2:
int* myInt2 = count();
You can also copy one array into the other:
int myInt2[10];
std::copy(count(), count()+10, myInt2);
Note copying will create a second array using separate memory than the first.
Upvotes: 4
Reputation: 210352
You don't need pointers, references are fine.
int (&count())[10]
{
static int myInt[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
return myInt;
}
int main(int argc, const char * argv[])
{
int (&myInt2)[10] = count();
std::cout << myInt2[3] << "\n\n";
return 0;
}
Upvotes: 1