Why my string has diffrent length than i choosed? - c++

Question

I want to create a string with numbers. So i define length of my string array to 10, but when i start the program in console is 11 chars.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#define N 10
using namespace std;
int main()
{
srand(time(0));
int numArr[N];
for(int i = 0; i < N; i++)
numArr[i] = rand() % 26 + 97;
for(int i = 0; i < N; i++)
std::cout << numArr[i] << " ";
std::cout << std::endl;
char str[N] = "";
for(int i = 0; i < N; i++)
str[i] = numArr[i];
std::cout << str << endl;
std::cout << strlen(str);
return 0;
}

Answer 1

A string needs to be terminated with \0 to get its length with strlen, in the codestr isn't terminated with \0, when you add it as last character strlen gives correct answer
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#define N 10
using namespace std;
int main()
{
srand(time(0));
int numArr[N];
for(int i = 0; i < N; i++)
numArr[i] = rand() % 26 + 97;
for(int i = 0; i < N; i++)
std::cout << numArr[i] << " ";
std::cout << std::endl;
char str[N + 1] = "";
for(int i = 0; i < N; i++)
str[i] = numArr[i];
str[N] = '\0'; // terminate with \0
std::cout << str << endl;
std::cout << strlen(str);
return 0;
}

Answer 2

It is rather risky to use char[N] when you can use std::string.
Here is safer way to write your code (and you will now get the 10 characters lenght you expect).
int main()
{
srand(time(0));
char numArr[N]; // I changed your type from int to char because you can only pack chars in a string
for(char i = 0; i < N; i++)
numArr[i] = (char)(rand() % 26 + 97);
for(int i = 0; i < N; i++)
std::cout << numArr[i] << " ";
std::cout << std::endl;
std::string str;
str.reserve(N); // not strictly required: this is an optimization
for(int i = 0; i < N; i++)
str.push_back(numArr[i]);
std::cout << str << endl;
std::cout << str.length();
return 0;
}