Re: how cast works?

On 07/08/2024 17:08, Keith Thompson wrote:

Thiago Adams <thiago.adams@gmail.com> writes:

How cast works?
Does it changes the memory?
For instance, from "unsigned int" to "signed char".
Is it just like discarding bytes or something else?

I also curious about how bool works.
Values converted to bool became 0 or 1.
When this conversion happens, at read or write? Both?
How much does it cost?
I also learned we cannot cast a pointer to nullptr_t.
But we can read a union with nullptr. In this case the value is 0
similar of what happens with bool.
Sample
----------------
#include <stdio.h>
struct value {
     int type;
     union{
         unsigned int ui;
         _Bool b;
         typeof(nullptr) p;
     } ;
};
int main(){
    struct value v;
    v.ui =  123;
    printf("%d / ", v.ui); //123
    printf("%d / ", v.b);  //1
    printf("%p", v.p);     //(nil)
}
--------
Using compiler explorer to understand bool.
void f(int i)
{
   bool b  = i;
   printf("%d / ", b);
}
cmp     dword ptr [rbp - 4], 0  //i think it checks 0 here
setne   al
and     al, 1
mov     byte ptr [rbp - 5], al
I think it convert at write in this case..but for union it converts
when we read. This is why
printf("%d / ", v.b);  //1
prints 1
void f()
{
    union {
     unsigned int ui;
     _Bool b;
    }  v;
    v.ui =  123;
    printf("%d / ", v.b);
}
Where b.b is converted to 1?
f:
         push    rbp
         mov     rbp, rsp
         sub     rsp, 16
         mov     dword ptr [rbp - 4], 123
         mov     al, byte ptr [rbp - 4]
         and     al, 1
         movzx   esi, al
         lea     rdi, [rip + .L.str]
         mov     al, 0
         call    printf@PLT
         add     rsp, 16
         pop     rbp
         ret
I think is at
al, 1
so the compiler uses 123 & 1 to cast number to bool.