why use (void *) to display char arrays' addr?

[q1670741824]

and (int *) works, too. How to understand this usage?

char arr[20] = "Hello!";
cout << (void *) arr;

i mean why void and int other than double or long something?

[q1670741824]

anybody?

[Garion]

It seems to be doing the same as this. That's the address of the array in both cases. Why it uses void. Idk.

#include <iostream>

int main()
{
    char arr[20] = "Hello!";
    std::cout << (void *)arr << " " << &arr;
    return 0;
}

I tried a google search and it sounds a void pointer has something to do with dynamic memory.

[Zhuge]

The purpose is so cout will output the address stored by the pointer; if you pass a char* it has special behavior that treats it as a C-string and outputs the characters at the memory address as though they were a C-string. This is not what you want if you want to see the address of the string, so you cast to a different pointer type to get the desired behavior.

[q1670741824]

oic thx :>