Where is the array getting that value from?

Question

So here's a simple program that just search for two numbers in an array that sum up to a certain value k

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

int main()
{

unordered_set<int> hashtable;
int k =7;
int arr[5] = {1, 2, 3, 4, 5};
int s = sizeof(arr);
for (int i =0; i<s; i++){
    if( hashtable.find(k - arr[i])!= hashtable.end() )
    {
       cout << arr[i] << endl;
       cout<< "found one " << arr[i] << " and "<< k-arr[i]<< endl;
    } else {
        hashtable.insert(arr[i]);
    }
}

return 0;
}

And here's the out put, I am getting

4
found one 4 and 3
5
found one 5 and 2
7
found one 7 and 0
7
found one 7 and 0

Am I missing something?

Print the value of `s` and observe that it is not what you are assuming. Then read about what `sizeof` means in your favourite C++ book. — molbdnilo, Mar 01 '22 at 12:00
you miss to ask a question. Whats wrong with the output ? We only see the code and its output — 463035818_is_not_an_ai, Mar 01 '22 at 12:01
the output prints `7` which doesn't exist in the arr variable — YGouddi, Mar 01 '22 at 12:03
@YGouddi Please note that [`#include using namespace std;`](https://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h) is one of the worst things you could do. — πάντα ῥεῖ, Mar 01 '22 at 12:30
No, i was playing around with unordered_set and i forgot to remove it afterwards — YGouddi, Mar 01 '22 at 12:37

score 1 · Answer 1 · answered Mar 01 '22 at 12:04

1

You access the array outside of its bounds. The behaviour of the program is undefined.

sizeof does not yield the number of elements in an array. It yields the size of an object in bytes. When the size of the element is more than one byte - and int is more than one byte on most systems - then the number of bytes in the array is more than the number of elements.

A correct way to get the number of elements in an array is to use std::size:

int s = std::size(arr);

answered Mar 01 '22 at 12:04

eerorika

232,697
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I'm getting an error that size is not a member of `std` – YGouddi Mar 01 '22 at 12:08
@YGouddi Did you include the header that defines `std::size`? – eerorika Mar 01 '22 at 12:09
@YGouddi `std::size` is only available with C++17 and higher. Make sure you're using `C++17` so that you can use `std::size`. Also, don't forget to include the header for using `std::size`. For that you can write `#include ` etc. – Jason Mar 01 '22 at 12:10
which is what exactly? – YGouddi Mar 01 '22 at 12:10
@YGouddi Quite a few standard headers in fact. Check the documentation. – eerorika Mar 01 '22 at 12:15

score 1 · Answer 2 · answered Mar 01 '22 at 12:09

1

Since you use only arr[i] and not i itself, you can write for (auto a : arr). This will respect the array bounds, you don't need to calculate the maximum index. Hence, it avoids the wrong calculation (which the other answers fix)

answered Mar 01 '22 at 12:09

MSalters

173,980
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score 0 · Answer 3 · answered Mar 01 '22 at 12:03

0

Maybe there are other ways to get the size of an array but for now this will do :

int s = sizeof(arr)/sizeof(arr[0]);

answered Mar 01 '22 at 12:03

YGouddi

341
2
14

2

Prefer `std::size`, it's actually type safe (as in, won't compile if you got the types wrong, as opposed to this). – StoryTeller - Unslander Monica Mar 01 '22 at 12:05

Where is the array getting that value from?

3 Answers3