I am not understanding the output of this ,

#include <iostream>
using namespace std; 

// pointers and arrays 
char ch1 = 'a' ,  ch2= 'b';

char ch3[6] = {'c', 'd', 'e', 'f', 'g' , 'h'};

char *ptr[3]; 

int main ()
    ptr[0] = &ch1 ;

    ptr[2] = ch3;
    cout << &ch1 << endl;

The out put is abcdefgh .

isn't &ch1 supposed to give the address of ch1 ? I cannot make sense why the output should be abcdefgh.

  • 3
    it's called undefined behavior Commented Jun 11, 2014 at 15:42
  • @ratchetfreak i cannot understand what array pointer is , its confusing .
    – Theorem
    Commented Jun 11, 2014 at 15:45
  • 1
    then don't try to understand a faulty program you won't learn anything Commented Jun 11, 2014 at 15:50
  • The declaration and assignments to ptr do nothing and could be deleted without changing the behavior. Commented Jun 11, 2014 at 16:37
  • 1
    @kevincline Hard to say that for sure without seeing a dump of the executable. As @AustinMullins points out, one of the bytes of the address of ch1 may actually be providing the NUL that terminates the "string" (although it is possible - but not guaranteed - that ch3 would be followed by some zero bytes if ptr were not present).
    – nobody
    Commented Jun 11, 2014 at 16:47

2 Answers 2


The expression &ch1 does indeed produce the address of ch1. The reason you see abcdefgh printed is cout has a char* overload that prints the NUL-terminated string starting at the given address. In this case, it just so happens that ch2, ch3, and a NUL immediately follow ch1 in memory so you get abcdefgh printed.

As @rachetfreak mentions, this is undefined behavior. It is illegal to access more than one character through &ch1 because it is a pointer to a single character. On different platforms you may get different output, or a crash.

If you want to actually print the address of ch1 you can cast it to void* so that cout won't assume it's a pointer to a string:

cout << (void*)&ch1 << endl;

In the code shown, ch1 and ch2 are characters which are declared consecutively, and ch3 is actually an array of characters. The << operator interprets the address of ch1 as a C-style string, which is just a pointer to a null-terminated character array. So, the interesting thing to me is that storing the address of ch1 into ptr[0] results in a null character (the number 0) appearing in the byte after 'g`. As @AndrewMedico stated, this is compiler-specific behavior. I think the code was intended to show how C-strings are interpreted by IO streams, and how easy it is for bugs and security holes to creep in to otherwise good code.

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