This is false.
Here’s an example.
Example assembly output
With GCC 16.2 with -O1.
"serve_hot_chocolate()":
mov DWORD PTR "device_command"[rip], 42
"explode()":
mov DWORD PTR "device_command"[rip], 3
ret
"device_command":
.zero 4
Ruh roh. Exploding chocolate! The function explode() will be executed, even
though it was never called!
godbolt link
Another example: dereferencing a null pointer
Some people insisted that dereferencing a null pointer is fine. It’ll just
try to access address zero! And these people supposedly know what happens when
you try: if nothing is mapped there, you’ll get a segfault.
Again, not true.
Let’s say we change std::unreachable() to int* ptr = nullptr; *ptr = 99; in
the previous example. Now we get this:
"serve_hot_chocolate()":
mov DWORD PTR "device_command"[rip], 42
mov DWORD PTR ds:0, 0
ud2
Curious, eh? First it writes zero, not 99, to memory address zero. Then it
crashes the program using an invalid instruction. The number 99 never appears
in the generated program. But the null pointer write does. Presumably the null
pointer write is done in order to improve debugger error messages.
Now let’s check clang with godbolt:
"serve_hot_chocolate()":
mov dword ptr [rip + "device_command"], 42
"explode()":
mov dword ptr [rip + "device_command"], 3
ret
"device_command":
.long 0
Again we get exploding chocolate.
Let’s do some pointer arithmetic on the null pointer
Ok, so what happens if we add ptr += 10. That’s UB for three reasons:
- using a null pointer for anything other than comparison against other pointers,
even if an object would be valid at the null pointer (which it’s not),
creating a pointer that points outside that object, or if an array, beyond
one-past the array, is UB,
- even if that were valid (which it’s not), we then dereference the address.
Let’s try it.
"serve_hot_chocolate":
mov DWORD PTR "device_command"[rip], 42
mov DWORD PTR ds:40, 99
ret
Wow. Now it did what someone naively would expect, taking the code at face
value. Lucky.
Why did GCC (and clang, by the way) not detect that *(ptr + 10) = 99 is
equally UB as *ptr = 99, and do the same thing?
Wrong question. No compiler is obligated to “detect” UB at all. Looks to me
like GCC and clang, if we want to anthropomorphize them, saw “oh, null pointer
dereference? Well that can’t happen, so I’ll stop emitting instructions for
this function, to save on the binary size”. Because of the halting
problem, it cannot in the
general case know if the code will ever be executed.
So what does UB do?
Anything. I have now demonstrated three things UB can do, when UB was triggered
in the same place:
- Run whatever happens to be next in memory (exploding chocolate).
Store a (for all intents and purposes) random value at address zero (wrote
zero, not 99), and then deliberately crash.
Do what one would naively expect: write the value to the address seen in the
code at face value.
And to think that some people don’t believe in nasal
demons.