char a []和char * a之间的区别？

Question

#include <bits/stdc++.h>
using namespace std;

// Function to copy one string in to other
// using recursion
void myCopy(char str1[], char str2[], int index = 0)
{
    // copying each character from s1 to s2
    s2[index] = s1[index]; 

    // if string reach to end then stop 
    if (s1[index] == '\0')  
        return;

    // increase character index by one
    myCopy(s1, s2, index + 1); 
}

// Driver function
int main()
{
    char s1[100] = "GEEKSFORGEEKS";
    char s2[100] = "";
    myCopy(s1, s2);
    cout << s2;
    return 0;
}

I did not understand how the value of s2 is getting printed ....since we passed address of s1 and s2 to function mycopy(). mycopy() has two local array str1 and str2 as argument,so i was thinking two local array with values of s1 and s2 will be created.(call by values)

Shouldn't the function prototype be mycopy(char *s1,char *s2) for printing s2.(call by reference)

Answer 1

void myCopy（char＆str1 []，char＆str2 []，int索引= 0）

Answer 2

To be able to reinitialize your parameter (array s2 in this case) you need to pass it by reference .

Like this:

void myCopy(char &str1[], char &str2[], int index = 0)

This means that myCopy() will use arrays as they are, otherwise it will create local duplication of them.