why do i get a garbage value when i try to print array[size]?

Asked Jul 12 '21 at 15:08

Active Jul 12 '21 at 15:09

Viewed 65 times

#include <iostream>
using namespace std;
int main()
{
 
 int a[5]={1,2,3,4,5};
 cout<<a[5]<<endl; 
return 0; 
}

why do I get a garbage value instead of "out of range" error?

edited Jul 12 '21 at 15:09

HolyBlackCat

asked Jul 12 '21 at 15:08

Guru Harsha

4

valid indices are `0` till `4`. There is no 6th element in the array. What did you expect to get as output? – 463035818_is_not_an_ai Jul 12 '21 at 15:09
3

Does this answer your question? [Accessing an array out of bounds gives no error, why?](https://stackoverflow.com/questions/1239938/accessing-an-array-out-of-bounds-gives-no-error-why) – Algirdas Preidžius Jul 12 '21 at 15:09
1

Because you don't pay for what you don't use – Hatted Rooster Jul 12 '21 at 15:09
2

Because this is how the language works. Most compilers can check array bounds for you if you ask (look up "sanitizers"), but it's slower and hard to do in general, so usually it's disabled. – HolyBlackCat Jul 12 '21 at 15:09
read about [undefined behavior](https://en.cppreference.com/w/cpp/language/ub). You will encounter it more often and not always it is so obvious to spot as in this case – 463035818_is_not_an_ai Jul 12 '21 at 15:11
You would with `std::vector a{1, 2, 3, 4, 5};` followed by `a.at(5);` – Bathsheba Jul 12 '21 at 15:11
1

C++ have no range checking at all for arrays. To get an "out of range" exception you must use a `std::vector` or `std::array` and use its `at` function which have range-checking. – Some programmer dude Jul 12 '21 at 15:11
On the line before `cout< 4) throw std::logic_error("out of range");` – Eljay Jul 12 '21 at 15:21
In general C++ opts for speed over pretty much everything unless you explicitly ask for something other than speed. – user4581301 Jul 12 '21 at 15:35

0 Answers0