Pointer Arithmetic Question (size of array) [C++] [duplicate]

Question

#include <bits/stdc++.h> 
using namespace std; 

int main() 
{ 
    int  arr[] = {1, 2, 3, 4, 5, 6}; 
    int size = *(&arr + 1) - arr; 

    return 0; 
}

How does int size = *(&arr + 1) - arr; find the size of the array exactly? I've read the explanation from geeksforgeeks and still a bit confused. I thought if you dereferenced (&arr + 1) then it would give you a nonexistent value since you're skipping ahead 6 integers, which can be anything random in that memory address? And also, if you're able to dereference (&arr + 1) to an int type, then how are you able to subtract that value from arr?

This may be useful to you. stackoverflow.com/questions/33523585/… — baseballr
– baseballr, Commented Apr 30, 2020 at 2:44
This will 100% help: stackoverflow.com/questions/2528318/… — abhiarora
– abhiarora, Commented Apr 30, 2020 at 3:50
there's another question exactly about this but the site doesn't correctly search for &x + 1 etc. ;( — M.M
– M.M, Commented Apr 30, 2020 at 3:56
see here or here for discussion of whether this is legal or not — M.M
– M.M, Commented Apr 30, 2020 at 4:02

Matt Timmermans · Accepted Answer · 2020-04-30 03:29:11Z

1

*(&arr + 1) - arr is a pretty complicated way to write 6.

&arr is a pointer to an int[6]

(&arr + 1) is a pointer to an int[6] that starts after the one at arr, i.e., it's 6 ints higher in memory.

*(&arr + 1) is that imaginary array after arr. There is no array there, but we're not going to use it, so it doesn't matter.

In *(&arr + 1) - arr, the two integer arrays are converted to int * pointers to their first elements before the difference operation is applied. Since these arrays are 6 ints apart in memory, there are 6 ints between those two pointers, and the result is 6.

answered Apr 30, 2020 at 3:29

Matt Timmermans

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2 Comments

abhiarora Over a year ago

Initially, I was also thinking that but it should be defined behavior!

M.M Over a year ago

"There is no array there, but we're not going to use it, so it doesn't matter." - the code is still technically undefined behaviour since the semantics of the * operator are only defined for the case where the pointer points to a valid object, making the behaviour undefined by omission. There is an open CWG issue about the subject

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