Why would you use a void pointer in this code?

Question

#include <iostream> using namespace std; int main() {     char *pc;      int *pi;     double *pd;      pc = (char *)10000;              pi = (int *)10000;               pd = (double *)10000;                // 1)     cout << "before pc = " << (void *)pc << " pi = " << pi << " pd = " << pd << endl;      pc++;     pi++;     pd++;     // 2)     cout << "after increase pc = " <<  (void *)pc << " pi = " << pi << " pd = " << pd << endl;      return 0; } 

In this code(1, 2), why is variable pc cast to a void pointer?

I am checking that a run-time error does not occur if you do not print the variable pc.

Answer 1

Because otherwise, the overloaded operator << (std::ostream&, const char*) would be called, which doesn't print an address, but a C-string.

For example:

std::cout << "Boo!";  

prints Boo!, whereas

std::cout << (void*)"Boo!"; 

prints the address that string literal is located at.

Answer 2

Because char* when printed using cout << something will try to print a string (cout << "Hello, World" << endl; uses char * [pedantically, in this example, a const char *] to represent the "Hello, World").

Since you don't want to print a string at address 10000 (it would MOST LIKELY crash), the code needs to do something to avoid the pointer being used as a string.

So by casting to void* you get the actual address printed, which is the default for pointer types in general, EXCEPT char *.