Modifying the value of a const protection

Look at the code first:

#include <stdio.h>

void Main ()

{

const int num = 10;int *a = (int *) (void *) &num; Give the address to A*a = 20; The address is assigned a value of 20printf ("*a=%d\t num=%d\n", *a,num);

}

Here, by forcing the type conversion (int *) (void *), &num can actually change the value in memory, but for NUM, the compiler has processed it as a constant, so you can see the assembly (the reply provided by the Great God).

By looking at the assembly of printf:

printf ("*a =%d\t num =%d\n", *a,num);
009717BE Push 0Ah
009717C0 mov Eax,dword ptr [a]
009717C3 mov ecx,dword ptr [eax]
009717C5 push ECX
009717C6 Push 976b30h
009717CB call _printf (0971320h)
009717d0 Add esp,0ch

As you can see, push 0Ah here is a constant in the press.

Modifying the value of a const protection