Question

What does this line do exactly?

(char*) (&input)

I know that it converts the int(input) to char array..or am I wrong? and could you guide me how does it compute?

update:
I think I now understand a little, then I created c++ code from your comments

#include <cstdlib>
#include <iostream>

using namespace std;
int main(){
 int input = 123456;
 int *p = &input;

 char *cp = (char*)p; 

for(int counter = 0;counter <sizeof(input); counter++){        
        cout << *(cp+counter) << endl;
        }


 system("pause");
 return 0;
}

but i think my code is wrong. because it only shows ‘@’ , big ‘r’? , and something like cross sign …

Answer 1

It’s just casting the address of input as a pointer to a char. You might see something slightly different like:

unsigned char *bytes = (unsigned char*)&input;

Which is useful for reading single bytes at the address of input, since char is a byte in size. Note that char is signed and ranges from -128 to 128 while unsigned char is (obviously) unsigned and more appropriate for reading bytes as it’s unsigned and ranges from 0 to 255.

You could then read specific bytes like:

first_byte = bytes[0];

Or perhaps you might want to set specific bytes such as:

bytes[0] = 0x7E;