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

Question

its output is 5...how?? can somebody provide dry run for it??

Answer 1

as i tried some checks
for i<2 a[0]=6
for i<3 a[0]=3
for i<4 a[0]=2
for i<5 a[0]=5
upto 4 i got how values come but after 4 i couldn't evaluate it

Answer 2

it is can not be 5. when i = 4 it is out of the massive

Answer 3

@maria milovanova
run the code and check it

Answer 4

I checked it. There is an error

Answer 5

#include <iostream>
using namespace std;

int main() {

int a[]={3,5,2,1};
for(int i=1;i<5;i++)
{
a[0]=a[i]+1;
}
cout<<a[0];
return 0;
}

run this one

Answer 6

I know that there should be "return 0". As I said, there is an error in the conditions of the loop. Array starts with 0, the size of this massive is 4. And that is why you can not initialize element a[4], it is more then the arrays size. P.S. sorry for mistakes. I don't know english well

Answer 7

that's the problem in this code. Even though array size is 3 ,how can i goes upto 5??? 😎

Answer 8

even if conditions will changed on i<4, it cannot be 5, it should be 2

Answer 9

it is because, when you out of the massive, it adds 1 to num, thet lies beyond the array

Answer 10

actually the on mobile app this code produces answer as 5 but on pc it produces the garbage values. 😂