m3ph1st0s's programming puzzle 2 (C++): "Call hard!"

Question

I am back with a brand new puzzle for C/C++ addicts. Following the feedback received on the 1st puzzle, this time I will try to make the requirements crystal clear and to provide a perfectly valid code.

Here is the problem. Given the following code:

#include <iostream>

char message[6]="hello";

void print_hello_message() {
   std::cout << message;
}

int main() {
   char* q = &message[0];

   // insert your code here
}

You are required to insert two more lines where the comment specifies in order to make the program print "hello" and nothing more, with the following restrictions:

1. the first line will NOT call any function, including cout
2. the second line will contain AT MOST 5 letters (and as many other symbols as you want)
3. the first line may contain new variable declarations but may not initialize them
4. the symbols # and / will not be used at all.

5. no logical operators will be used

   The shortest code wins!

Answer 1

In thinking about it some more, the problem states "insert two more lines where the comment specifies" not "statements". Taking that into consideration, here is another shorter answer (though it depends on how you define "the first line will NOT call any function, including cout"):

std::
cout<<q;

The first line qualifies which cout will be used, and the second line includes the actual operator that results in calling operator<< for the std::ostream & const char* operands.

Answer 2

If I understand what you're asking for, this is what I'd add:

using std::cout;
cout<<q;

The first line brings std::cout into the current name space so that we can use it without qualification. using namespace std; would have worked as well but is longer so I didn't use it.

The second line uses the limit of five letters (plus three symbols and no necessary white space beyond whatever the platform uses to denote end of line).

There may be something shorter, but I can't think of it at the moment.

Answer 3

Use q to store a pointer to print_hello_message, then cast it back to a function pointer and invoke it.

q = (char *) &print_hello_message;
(*(void(*)())q)();

Short version:

q=(char*)print_hello_message;
(*(void(*)())q)();

Answer 4

#include <iostream>

char message[6]="hello";

void print_hello_message() {
   std::cout << message;
}

int main() {
   char* q = &message[0];

   while(*q!=' ')
       cout<<*q++;

}