Segmentation 11 in Linux issue - c++

Question

I am doing a buffer overflow problem and I trying to print hello world. Below is my code to
but I am getting a segmentation 11 issue when I run this file with another one. "./executable < input.cpp(This is the file below). I am doing something wrong to solve a buffer overflow issue?
#include<stdio.h>
using namespace std;
main()
{
printf("A");
//00000b00
for (int i = 0; i < 4; i++)
printf("%c%c%c%c",0x00,0x0b,0x00,0x00);
}
Below is the actual code that I am trying to print hello world. Above is my input string.
#include <iostream>
using namespace std;
int i;
unsigned int* p;
void f1() {
int a=10;
char str[4];
cout << "Please enter a string:";
while (!cin.eof()) {
cin.get(str[i]);
i++;
}
printf("address of str is:%x\n",str);
cout << "The string you entered is:";
printf("address of a is:%x\n",&a);
cout << str << endl;
}
void f2()
{
cout << "Hello World!\n";
}
main()
{
printf("The address of function f2:%08x\n",f2);
f1();
}

Answer 1

I am getting a segmentation 11 issue when I run this file with another one.
./executable < input.cpp
I am doing something wrong to solve a buffer overflow issue?
Yes. Buffer overflow attacks don't work like that - dumping a bunch of C source code into memory does not magically make the machine compile and run it. To generalize wildly, the data you dump into memory must contain:
Padding to force the following data to lie in the right place in the stack
A replacement address in the location of the old return address, pointing to the following executable code
Some more padding, usually a "NOP slide"
Some executable code
Please read the classic "Smashing the stack for fun and profit", and keep in mind that you may have to disable some protections (non-executable stack, ASLR, stack canary) to get these exploits to work on a modern system.

Answer 2

Use the %x modifier to print hexadecimal values
If this is a C program, then the use of namespace std makes no sense
#include<stdio.h>
int main(void)
{
puts("A");
for (int i = 0x0; i < 4; i++)
printf("%x\n", i);
return 0;
}
Op's post was updated:
#include<stdio.h>
#include <iostream>
using std::cout;
using std::cin;
using std::endl;
int i = 0; //Initialise i
void f1() {
int a=10;
char str[4];
cout << "Please enter a string: ";
while (!cin.eof() && i < 4 ) { //Have a condition on length of string
cin.get(str[i]);
i++;
}
str[i] = '\0'; //Set the eof character at end of the string
printf("address of str is: %p\n", str);
printf("address of a is: %p\n", &a);
cout << "The string you entered is: " << str << endl;
}
void f2() {
cout << "Hello World!\n";
}
int main()
{
printf("The address of function f2: %p\n", f2); //To print address use the %p option
f1();
return 0;
}