Why can't I use a for each loop on array of int pointers in C++

Question

I'm trying to have an array of pointers (int *a[10]) and then use a range based for loop (as in the C++11 standard). However, the compiler complains D: - it says "error: invalid initialization of reference of type ‘int&’ from expression of type ‘int*’".

My code is basically:

main(){
    int *a[10];
    for(int& e : a){}
}

I had also tried: //... for(int* e : a){}

But having neither, inside the loop body, *e = 1 or e = 1 works.

I've read this question which links me to this page; by reading that, I think I understand that the container you're trying to iterate over must implement a begin() and an end() methods. I guess the primitive arrays don't implement these; is that the case? Also, is it the case that int& a = c; makes a have the same address as c, for some int c?

Is there another alternative?

Or for(int *&e : a) {} if you want to access the pointers themselves. — Quentin
– Quentin, Commented May 27, 2015 at 9:57
@AlexFarber I thought of that but it doesn't seem to be the behavior I want — someonewithpc
– someonewithpc, Commented May 27, 2015 at 9:58
@Quentin I do want to access them, but I don't understand how that works, then, because for(int *&e : a){*e = 1;} seems to do nothing, when I print it. — someonewithpc
– someonewithpc, Commented May 27, 2015 at 10:01

MSalters · Accepted Answer · 2015-05-27 11:59:08Z

3

*e = 1 suggests that you want to make 10 pointers to 10 integers, each of which has the value 1.

First, you need to have a loop that references all the 10 pointers:

int* a[10];
for(int*& e : a) // Reference so you can modify a.

Then, you need to create 10 pointers:

  e = new int;

and those have to be pointers to the value 1 so we complete the statement

 *e = 1;

or shorter

  e = new int (1);

When cleaning up, you'd have a similar loop:

 for (int& e : a) {
      delete e;
      e = nullptr;
 }

edited May 27, 2015 at 11:59

answered May 27, 2015 at 10:13

MSalters

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2 Comments

someonewithpc Over a year ago

It works if I use for(int *&e : a) but not if I do for(auto e : a). Why is that? With the latter I get "173680127\n-1161788088" as the output.

MSalters Over a year ago

That's because auto is deduced as int*. What happens then is that you modify a copy of the pointers in a. If you'd use auto& e: a, the deduced type of e would be int*& e. (In other words, the problem is not with auto but with leaving off the & reference.)

Jonathan Wakely · Accepted Answer · 2015-05-27 09:57:44Z

2

It's an array of int*, so of course you can't iterate over it using a variable of type int, that isn't the type of object in the array!

The objects in the array are int* so that's what your loop variable needs to be.

for (int* e : a) ...

Or

for (int*& e : a) ...

Or simply:

for (auto& e : a) ...

answered May 27, 2015 at 9:57

Jonathan Wakely

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1 Comment

someonewithpc Over a year ago

Right, I thought of the first one, of course, but it doesn't work as I want.

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