Why there isn't an error in this statement ( a[0-1]) ?

Question

Here is the code ,assume all necessary libraries included:
int main()
{
int a[5] = {1,2,3,4,5};
for (int i=0; i<5; i++)
{
a[i]= 2* a[i-1];
cout<<i;
}
return 0;
}

Answer 1

compiler ignored error because it is a runtime exception

Answer 2

Start the for loop at 1 to fix. Then change a[0] however you wanted it to be.

Answer 3

i=0
a [i-1] = a [ -1] = error; a>=0

Answer 4

There is an error i.e. a [-1]

Answer 5

Because there is no a[-1]. Arrays starts at index 0.

Try this instead :
...
for (int i=1; i<5; i++) {
a[i] = 2 * a[i-1];
cout << i;
}
...

Answer 6

Thank you for your help , I appreciate it.

Answer 7

Sorry for disturbing you , but why this error is ignored by compiler ?

Answer 8

no problem ;)

Answer 9

I thought the same thing , but when i run this program the output is 01234.

Answer 10

@mohammad, the output of this code (if you ignore the a[-1]) is 01234 because you wrote :
...
for (int i = 0; i < 5; i++) {
...
cout << i;
}
...

Answer 11

good question... :/

Answer 12

@NimWing Yuan , thanks :)