Output is 21212 but how? And what does m++ here means and even after declaring i as int 2 times why it's not showing any error?

Question

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

Answer 1

as m is a pointer to an array when you increase the value of m the effect is like you shift the 0 index by one, in this case:
m : 1 1 1 1 1 // entry point
^ // pointer is here
m : 2 1 1 1 1 // m[0] = 2
^
m : 2 1 1 1 1
^ // m++
m : 2 1 2 1 1 // m[1] = 2
^
m : 2 1 2 1 1
^ // m++
m : 2 1 2 1 2 // m[2] = 2

not sure if that is clear enough, but that's the best I can do in text 😃

Answer 2

m++ acts as an address skipping element here. If you have been noticing, f function's for loop sets to be repeated for 3 times and inside the for loop you see that after each multiplication, m shifts the address of the last indexed cell, 4 bytes further ( since it's an int type) and as a result, the multiplication has an alternate pattern.

Answer 3

Thank you Nikolay Nachev but why it's not showing any error even after declaring i as int 2 times?