Pointer of a character in C++

Question

Going by the books, the first cout line should print me the address of the location where the char variable b is stored, which seems to be the case for the int variable a too. But the first cout statement prints out an odd 'dh^#' while the second statement correctly prints a hex value ' ox23fd68'. Why is this happening?

 #include<iostream>
    using namespace std;

    int main()
    {
        char b='d';
        int a=10;
        char *c=new char[10];
        c=&b;
        int *e=&a;
        cout<<"c: "<<c<<endl;
        cout<<"e: "<<e;
    }

This is not a bad question, but probably answered elsewhere. The short answer is that C++ has inherited C's treatment of `char *` as "Probably a collection of printable things". — Drew Dormann, Jun 10 '16 at 14:45
If this is what your book is teaching you, please throw it out and try something off [The Book List](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) — The Forest And The Trees, Jun 10 '16 at 15:01
I'd be more inclined to think the OP was going by the book, and then extrapolated what they had learned, only to find out, that `char*` are handled differently from all other pointer types. As common as it may be to those that have seen C-style strings all of their (developer) lives, this can be justified as *surprising*. — IInspectable, Jun 10 '16 at 15:07
The author explains pointers with an example of an integer variable and says it works the same for every other data type. I guess it's a lack of thoroughness. Thank you for your suggestions anyway. :) — Pallavi Bagchi, Jun 10 '16 at 15:22

IInspectable · Answer 1 · 2016-06-10T14:55:32.533

There is a non-member overload operator<<(std::basic_ostream) for the const char* type, that doesn't write the address, but rather the (presumed) C-style string¹⁾. In your case, since you have assigned the address of a single character, there is no NUL terminator, and thus no valid C-style string. The code exhibits undefined behavior.

The behavior for int* is different, as there is no special handling for pointers to int, and the statement writes the address to the stream, as expected.

If you want to get the address of the character instead, use a static_cast:

std::cout << static_cast<void*>( c ) << std::endl;

¹⁾ A C-style string is a sequence of characters, terminated by a NUL character ('\0').

TrueY · Accepted Answer · 2016-06-10T14:53:15.620

Actually this program has problem. There is a memory leak.

char *c=new char[10];
c=&b;

This allocates 10 characters on heap, but then the pointer to heap is overwritten with the address of the variable b.

When a char* is written to cout with operator<< then it is considered as a null terminated C-string. As the address of b was initialized to a single character containing d op<< continues to search on the stack finding the first null character. It seems the it was found after a few characters, so dh^# is written (the d is the value of variable b the rest is just some random characters found on the stack before the 1st \0 char).

If you want to get the address try to use static_cast<void*>(c).

My example:

int main() {
  char *c;
  char b = 'd';
  c = &b; 
  cout << c << ", " << static_cast<void*>(c) << endl;
}

An the output:

dÌÿÿ, 0xffffcc07

See the strange characters after 'd'.

I hope this could help a bit!

Pointer of a character in C++

2 Answers2