[jlebar]

> even without LTO, the compiler knows what malloc and free do at a semantic level.

  int *a = ...;
  int b = *a;
  malloc(42);
  b = *a;
  printf(b);

If malloc is an opaque function, can the compiler nonetheless eliminate the second `b = *a` load as dead because it "knows what malloc does"? Certainly not, right?

[jepler]

https://godbolt.org/z/z-92qR

Interestingly, the compiler emits no call to malloc, even. What!?

[steerablesafe]

With -fno-builtin. It's still "cheating" as b is local.

https://godbolt.org/z/U5VGxt

Without cheating:

https://godbolt.org/z/rzcXg2

It doesn't optimize the two stores to b away.

[gpderetta]

The compiler knows the semantics of malloc. Malloc as it appear in the source code doesn't even need to map 1:1 to calls to the library function, the compiler can remove and coalesce calls at will.