working of range for loop and C-style arrays in C++

Question

C-style Arrays are not self-describing as it doesn't has any information about it's size, so for iterating it's elements we need size information somehow. I'm confused as how range-for loop gets the size information. It is expected to fail. Consider 2 conflicting examples,

#include <iostream>
using namespace std;

int main() { 
    int a[] = { 1, 2, 3, 4, 5, 6 };
    for(auto x : a)
        cout<<x<<" ";
    cout<<endl;
    return 0;
}

It ran successfully(unexpected), but

#include <iostream>
using namespace std;

void print_a(int*);

void print_a(int a[])
{
    for(auto x : a)
        cout<<x<<" ";
    cout<<endl;
}
int main() { 
    int a[] = { 1, 2, 3, 4, 5, 6 };
    print_a(a);
    return 0;
}

This produced errors(expected).

Can someone explain as how range-for loop actually works ?

"C-style Arrays are not self-describing as it doesn't has any information about it's size" That's not right. It does contain this information (implicitly - it's not stored anywhere, but compiler knows it), but arrays decay to a pointer when passed to a function. — Yksisarvinen, Jun 02 '20 at 14:32
@Yksisarvinen but then how actually is range for getting this info about size — beta_me me_beta, Jun 02 '20 at 14:34
In the first example, the compiler has visibility that `a` is an array and of the size of that array, so can work out the range the `for` loop can iterate over. I assume you don't need an explanation of why the second example doesn't work. — Peter, Jun 02 '20 at 14:35
"Arrays are not self-describing because the number of elements of an array is not guaranteed to be stored with the array. This implies that to traverse an array that does not contain a terminator the way C-style strings do, we must somehow supply the number of elements" quoting from 'TC++PL' — beta_me me_beta, Jun 02 '20 at 14:36
[How does the range-based for work for plain arrays?](https://stackoverflow.com/questions/7939399/how-does-the-range-based-for-work-for-plain-arrays) and [cppreference for range-based loops](https://en.cppreference.com/w/cpp/language/range-for). In short, C-style arrays are handled separately when used in range-based for loops. Note that it only works with actual array type, not arrays that decayed to pointers. — Yksisarvinen, Jun 02 '20 at 14:37
The compiler can, however, examine the definition of an array to obtain its size. It can't do that if only given a pointer (which can point to a single value, or to the first of any number of values). — Peter, Jun 02 '20 at 14:37
`a` in main is of type `int[6]`, ie the size is part of its type. `a` in `print_a` is a `int*` which has lost all information on the array size — 463035818_is_not_an_ai, Jun 02 '20 at 14:41
@idclev463035818 even if we had int[] in declaration of print_a, it will eventually decay to int* — beta_me me_beta, Jun 02 '20 at 14:53
`void print_a(int*);` and `void print_a(int a[]);` are two declarations of the same function — Caleth, Jun 02 '20 at 15:12

score -1 · Answer 1 · answered Jun 02 '20 at 14:37

-1

range-for loop run from start to end that says it needs to know the length. of statically allocated arrays AKA int array[10] the compiler know the length on compile-time so he knows to terminate them, but on the dynamically allocated array, the compiler doesn't know the length of the array so he cants rang loop them

answered Jun 02 '20 at 14:37

yaodav

1,126
12
34

That's not entirely true, since statically allocated arrays can decay to pointers and thus lose information about their length. The code in this question is a prime example for this. – Lukas-T Jun 02 '20 at 14:38

working of range for loop and C-style arrays in C++

1 Answers1