Re: how cast works?

On 09/08/2024 02:56, Bart wrote:

On 08/08/2024 23:32, David Brown wrote:

On 08/08/2024 21:09, Bart wrote:

 

Sorry but my function is perfectly valid. It's taking a Bool value and converting it to an int.

>
No, it is not.
>
Attempting to use the value of a non-static local variable that has not been initialised or assigned is undefined behaviour.  Your function is garbage.  No one can draw any conclusions about how meaningless code is compiled.

 FFS. You really think that makes a blind bit of difference?

Yes, I do.

A variable is not initialised, so the bool->int code shown must be just random rubbish generated by the compiler? You think I wouldn't spot if there was something amiss?

You wrote C code that had something amiss!
How often are you going to do this?  Write some piece of meaningless crap with undefined behaviour or - at best - no observable behaviour at all, compile with silly choices of flags, and then make nonsensical claims about what compilers do or how C is defined?  How hard is it for you to actually write a C function that has defined behaviour that does what you want it to do?  Get step 1 of these tests right, and we won't have to keep repeating this stuff.

 OK, let's initialise it and see what difference it actually makes. My code is now this:
    #include <stdbool.h>
    void BC(void) {
       _Bool b;
       int i;
       i=b;
   }
    void DB(void) {
       _Bool b=false;
       int i;
       i=b;
   }
 The output from gcc -O1 is this:
 BC:
         ret
DB:
         ret
 There is no difference.

There is a difference in the code.  One (BC) has no defined behaviour. The other (DB) has defined behaviour with no observable behaviour.  It's not a surprise that a compiler generates no code for either of them.

So, even initialised, it tells me nothing about what might be involved in bool->int conversion. It is useless.
 

Agreed.  Nobody suggested your code above as a good idea.

Now I'll try it with -O0 (line breaks added):
 BC:
         push    rbp
         mov     rbp, rsp
          movzx   eax, BYTE PTR [rbp-1]
         mov     DWORD PTR [rbp-8], eax
          nop
         pop     rbp
         ret
DB:
         push    rbp
         mov     rbp, rsp
         mov     BYTE PTR [rbp-1], 0
          movzx   eax, BYTE PTR [rbp-1]
         mov     DWORD PTR [rbp-8], eax
          nop
         pop     rbp
         ret
 Exactly the same code, except DB has an extra line to initialise that value.
 Are you surprised it is the same? I am 99% sure that you already knew this, but were pretending that the code was meaningless, for reasons that escape me.
 

Instead of trolling with what you know, without doubt, are pointless straw men, why not apply a little thought and write functions that make sense?  Or look at the functions that I wrote?
The stuff you write is either meaningless, or pointless, or perhaps both.  I have no doubt that you know this fine.
So the big question is, why are you writing it?  Is it just to provoke me?  Is it because you want to confuse the OP and other readers?  Do you like pretending to be a fool?
I'm giving up trying to help you - at least until you show some hint of trying to learn.  I will still make posts pointing out when you write nonsense that might confuse or mislead others, but I'll stop trying to explain things unless you specifically ask.