Getting_Very_Wierd_Output

Question

When I am compling the below program the expected output for s variable is "b". But it is getting mixed up with the above strings in the buffer.

#include <iostream>
#include<string.h>// for string methods.
#include<iomanip>
using namespace std;

int main()
{
char a[12] ="Hello World";
// cout<<a<<endl;
// cin>>setw(12)>>a;
// cout<<a<<endl;
// cout<<strlen(a)<<endl;
char s[]={'b'};
// cout<<strlen(s)<<endl;
cout<<s<<endl;
return 0;
}

Answer 1

Dayala Badrinadh Reddy a is initialized by the copy constructor. Any literal string is appended \0 at the end, when you write:

char a[12] = "Hello World";

The entire temporary string is copied to a, including the null terminator.

Using list initialization, no temporary is created, and you have to add the null terminator explicitly.

Answer 2

Yeah But the stack implementation what you said , Is it the way how compiler works on string I mean store strings !?
And when we are initialising a char array using {' ',...} we never mentioned '\0' specifically. Because it is said that the compiler takes cares of it and appends '\0' automatically at the end of string initialised that way. Then why are we specifically mentioning above char s[]={..,'\0'}

Answer 3

this is c++ but it's not Python