Arrow operator for array length? No output when printed, How?

I have this code to iterate over an array, instead of using sizeof(array)/sizeof(*array) I used an arrow operator (I don't understand really well how it works).

I don't understand why when I print out array->length() it is just a blank output, and how can the for loop work and terminate properly with the set condition of i < array->length().

#include <iostream>

using namespace std;

int main() {

    string cars[] = {"Corvette", "Ford", "Mustang"};

    cars[1] = "Camaro", "Ford";
    for (int i = 0; i < cars->length(); i++)
    {
        cout << cars[i] << '\n';
    }

    cout << cars->length();
    return 0;
}

In cars->length(), the array std::string cars[3] decays into a pointer to the first element. Using -> on that pointer dereferences it, and calling length() on that first element will return the length of the element (8 for "Corvette") - not the size of the array.

Since your loop is equivalent to:

for (int i = 0; i < 8; i++)

and you dereference cars[i] inside the loop, you access the array out of bounds. The array only has 3 elements. This causes undefined behavior, and your program could do just about anything - or nothing, or crash.

Another problem in this program is this use of the comma operator:

cars[1] = "Camaro", "Ford";

This assigns "Camaro" to cars[1], but the result of the full expression is "Ford". Example:

auto x = (cars[1] = "Camaro", "Ford");

x is now "Ford" while cars[1] is "Camaro".

@TedLyngmo has already covered the problems here, and why the code doesn't work as well as it might appear to.

I'll try to cover some of the ways to make the code work. The preferred one is probably a range-based for loop, like this:

    for (auto const &car : cars)
    {
        cout << car << '\n';
    }

Another alternative to using sizeof(array)/sizeof(array[0]) is something like this:

template <class T, size_t N>
size_t length(T (&array)[N]) {
    return N;
}

With this, you can do something more like your original code:

for (int i=0; i<length(cars); i++) {
    std::cout << cars[i] << '\n';
}

The big advantage of this compared to sizeof(array)/sizeof(array[0]) is that if you have a pointer instead of an actual array, the sizeof code will produce incorrect results, but this simply won't compile at all. You generally don't want to actually define this yourself though--the standard library provides an std::size that will do the job for you, without having to remember the syntax for passing an array by reference.

I'd be remiss if I failed to mention that the generally preferred way would be to use an std::vector though:

std::vector<std::string> cars = {"Corvette", "Ford", "Mustang"};

With this, you can use the range-based for loop (the first one I showed above) or a cars.length()--which will now work correctly, because cars is a vector, which has a length() member that returns the number of elements in the vector.