How to get the length of an array from a pointer? [duplicate]

Answer 1

C++ has proper string type:

std::string

which you may find helpful here. Even if you're passing it to function that accepts const char*, it has .c_str() method that allows you to pass it to function that accept a pointer. If the other function needs to modify the string, you can use &str[0] which is valid for many implementations of C++, and is required to work for C++11. Just make sure you resize() them to the correct size.

Some of the other containers in C++ are:

std::array (C++11) Array of constant size. Better than plain old C array, as it has size() method. std::vector Dynamic array (Java ArrayList equivalent)

As for your question - there is no way to find size of a pointed array. How could you even do that? It's just a stupid pointer.

Answer 2

If you want to keep simple and not use #include <array.h>, you can make a struct where you array sizes are. A little bit like:

struct
{
    int* int_Array;
    int array_Size;
}

That way, you can to whatever you want (like deleting your instances or something)

Answer 3

You can't, I'm afraid. You need to pass the length of the array to anyone who needs it. Or you can use a std::array or std::vector or similar, which keep track of the length themselves.

Answer 4

It is true that you cannot get the array size from a pointer to an element of the array.

If the reason you only have a pointer is because you are implementing a function that takes an array parameter as an argument like this:

void foo (T *p) {
    // p is a pointer to T
}

then you can use a template function instead to get the array size to the function.

template <unsigned N, typename T>
void foo (T (&p)[N]) {
    // p is a reference to an array[N] of T
    std::cout << "array has " << N << " elements" << std::endl;
    std::cout << "array has "
              << sizeof(p)/sizeof(p[0])
              << " elements"
              << std::endl;
}

int main ()
{
    int array[40];
    char array2[25];
    foo(array);
    foo(array2);
    return 0;
}

Answer 5

You can't. And the moral is, don't use pointers and arrays, use vectors. You can always get the size of a vector.