为什么在c ++中分配char数组元素时，分配的字符会被破坏？

Question

I wrote a function in C++ which removes two characters from a char array. I think when I assign str[o+2] to str[o] , the str[o+2] shoud not be changed. But when I use cout to print it, I see that str[o+2] is altered with null.

#include<iostream>
#include<string.h>
using namespace std;
void shiftLeft(char*,int, int);
int main(){
    char str[100];
    cout<<"enter the string: ";
    cin>>str;
    cout<<"removes the letter with index i and i+1\nenter i:";
    int i;
    cin>>i;
    int n=strlen(str);
    shiftLeft(str,i,n);
    cout<<str;
    return 0;
}
void shiftLeft(char*str,int i, int n){
    for(int o=i-1; o<n; o++){
        str[o]=str[o+2];
    }
}

For example with input "abcdef" and i=3 , I expect output "abefef" but I get "abef" . where are the last "ef" ? Why are they ignored?

Answer 1

abcdef0???... <- the contents of array (0 means NUL, ? means indeterminate)
  ef0?        <- what is written to the position (shifted 2 characters)

Assigning str[o+2] to str[o] itself won't change str[o+2] , but later o will come to o+2 thanks to the for statement and then assigning str[o+2] to str[o] means assigning str[o+4] to str[o+2] with original o value.

Then, terminating null character is written and the string ends.