basic arrays question

[adam2016]

so I'm going to have to ask a very basic question which I am kind of shy to ask since I have been learning c++ for 2 years on and off

anyway my question is as follows,here I get some errors

how come this isn't allowed we are returning a pointer to an array and I know this is allowed because arrays decay down to pointers to their first element

but how come this is giving me an error?

also when I return the pointer from the array function isn't this pointer pointing to memory that has gone out of scope?

so number[5] = out of scope memory?

thanks

  #include <iostream>
using namespace std;

int* array(){

	int numbers[] = {1,23,4,45,6};
	return numbers;
}

int main() {

	int number[5] = array();
}

ALSO

#include <iostream>
using namespace std;

int* array(){

	int numbers[] = {1,23,4,45,6};
	return numbers;
}

int main() {

	const int* arr = array();

	for(int i = 0; i < 5; i++){

		cout << arr[i];

	}
}

I'm not surprised this code compiles but I am quite surprised the code above crashes,I thought by making the int pointer const I am keeping the array which was returned in scope?

thanks

[AbstractionAnon]

so number[5] = out of scope memory?

Yes. You allocated 5 elements [0]-[4]. So [5] is not a valid reference.

also when I return the pointer from the array function isn't this pointer pointing to memory that has gone out of scope?

Yes, numbers goes out of scope when array() exits.

I am quite surprised the code above crashes,I thought by making the int pointer const I am keeping the array which was returned in scope?

In the second snippet, numbers goes out of scope at line 8. The fact that you declared arr as const only means that you have told the compiler you're not going to modify arr within main(). The const declaration has nothing to do with whether the pointer returned by array() is in scope or not. It's not.

[jonnin]

the word you want is static. If you do this

static const int blah[5];

it will exist until program ends, even if its scope ends.

however if it is out of scope, you need a way to access it...

[adam2016]

thanks guys BUT

isn't using const with a reference prolonging the life of a object returned from a function

this isn't the same for pointers?

[adam2016]

also making a variable static how will this prolong the life of a variable?

I thought static just means the variable belongs to the class instead of instances of it

thanks

[AbstractionAnon]

this isn't the same for pointers?

No. As I explained before, you're returning a pointer to something that has gone out of scope. Therefore it no longer exists.

also making a variable static how will this prolong the life of a variable?

Making a variable static will make the variable exist for the life of the program. Thereby making it legal to return a pointer to it since it will not go out of scope.

I thought static just means the variable belongs to the class instead of instances of it

Using static on a class member variable makes the member variable exist for the life of the program. Since the member variable is scoped within the class name, you can think of it as belonging to the class, but the lifetime of a static member variable is still for the lifetime of the program.

[adam2016]

thanks Anon =)