Why doesn't a pointer to an element in a c-string return just the element?

Question

I'm trying to understand how pointers work here. The findTheChar function searches through str for the character chr. If the chr is found, it returns a pointer into str where the character was first found, otherwise nullptr (not found). My question is why does the function print out "llo" instead of "l"? while the code I wrote in main return an "e" instead of "ello"?

#include <iostream>
using namespace std;

const char* findTheChar(const char* str, char chr)
{
    while (*str != 0)
    {
        if (*str == chr)
            return str;
        str++;
    }
    return nullptr;
}

int main()
{
    char x[6] = "hello";
    char* ptr = x;
    while (*ptr != 0)
    {
        if (*ptr == x[1])
            cout << *ptr << endl; //returns e
            ptr++;
    }
    cout << findTheChar("hello", 'l') << endl; // returns llo
}

Answer 1

cout << findTheChar("hello", 'l') << endl; //returns llo

Return type of findTheChar is const char *. When you try to print a const char * with std::cout it would print the character array pointed by the address upto terminating \0 character.

str -> +---+---+---+---+---+---+
       | h | e | l | l | o | \0|
       +---+---+---+---+---+---+
                 ^
                 |
         findTheChar(str, 'l')

If you want just one character, dereference the address (If it is not null). If you want to print the return address, you can typecast it to void *.

while the code i wrote in main return an "e" instead of "ello"

cout << *ptr << endl; //returns e

Here you are explicitly dereferencing the ptr as *ptr and thus you are printing a char and not const char *.

As you are using C++, you would be better with std::string and iterators.

// Returns offset of first occurance if found
// Returns -1 if not found
int findTheChar(const std::string& str, char chr );