C++ code

Question

Can someone explain this C++ code?

#include <iostream>
using namespace std;

int main() {
int* p=0; int x=1;
cout << p << endl;
cout << &p << endl; // cout << *p << endl; doesn't work?
p=&x;
cout << *p << endl;
cout << &x;
char *ptr = "hello";
printf("\n%s",ptr);
return 0;
}

Answer 1

Edward Finkelstein the pointer "p" is a null pointer at the time of initialization.
And when you do
cout<<*p;
You are trying to dereference a NULL pointer which leads to undefined behaviour.

Answer 2

Mr.Imperfect yes not just because it's read only but you also get compiler warnings if you do so.

Answer 3

Edward Finkelstein

This method ( if(p) or if (p != NULL) ) checks if the pointer is pointing to null or not.

As far as I know ,there is no way to check if a pointer is pointing to valid memory address or not.

That's why we either make a pointer point to a valid memory location or make it null pointer ( pointing to nothing ) making evaluation process possible.

Answer 4

Edward Finkelstein
Using "cout" should also give you output.

But using
cout<<ptr<<endl;
Will print output in previous line and then put a newline character, that's why you must have missed the output.

As for the warning is concerned then it is obviously because of the same reason it states, you are pointing that pointer to a string constant means you can only use it for read only purposes, any attempt to change it will result in error.

Answer 5

You get a warning in this:

https://code.sololearn.com/c2lATVt63B7F/?ref=app

Answer 6

Mr.Imperfect if it would have been any other memory address then it would have cause a problem, but 0 is a special case as it makes pointer point to NULL which is generally considers as a good practice as now you can easily test if the pointer is valid or not by doing if(p) { otherwise it would be containing a garbage value }

Answer 7

Edward Finkelstein what you said about it being C++ specific is correct. No idea why someone downvoted you.

Answer 8

Nishant Prateek who said his code is not working ?

Answer 9

In C++ don't assign a string constant to a char*

Answer 10

You don't get a warning in this:

https://code.sololearn.com/cj3d7vmbeEvC/?ref=app

Answer 11

Arsenic
So int* p=0; makes p point to NULL, which evaluates to false in the expression if(p). So null pointer is not a valid pointer? This I did not know.

Answer 12

First of all your code has an error because printf() is a function of standard liberary of c language

Answer 13

Nishant Prateek
Do cout << ptr << endl; and you get no output, with printf you get output. The compiler says the problem/warning is the initialization of char pointer to string constant.

Answer 14

it seems like this is a C++ thing, in c compiler i run code with char pointer and i get no such warning.

Answer 15

Better to use cout instead of printf in C++.

Answer 16

Mr. Imperfect, thanks. Why does monitor dump core when I replace cout << &p << endl; with cout << *p << endl; ?

Answer 17

And char *ptr != "hello"
You have written char *ptr ="hello"
It not a string it is a character
That's why your code is not working

Answer 18

You can declare other character variable and allocate the pointer to that variable using '&' operator and them print the output

Answer 19

I mean it's showing warnings in runtime and warnings doesn't means that the code will not working

Answer 20

It is for getting rid out of warnings