Should `--no-checks` imply `-DNDEBUG`

[jabraham17]

In C code with assert, compiling with -DNDEBUG turns off asserts.

For example

#include <assert.h>
int main() {
  assert(0);
}

If this program is compiled without -DNDEBUG, it will fail. But compiling with -DNDEBUG disables the assert.

This can effect Chapel programs in a few ways

For example, consider this extern block

extern {
  #include <assert.h>
  void foo(void);
  void foo(void) {
    assert(0);
  }
}
foo();

There is no way to turn off this assert in a release build (technically --ccflags -DNDEBUG would work), meaning there is always going to be extra code generated for assert

This also affects includes

// file.h
#include <assert.h>
static void foo(void) {
  assert(0);
}

require "file.h";
extern proc foo();
foo();

In this code, there is no way to remove the assert in a default release build (--fast)?

Should we have --no-checks imply -DNDEBUG to turn off C assertions? We already do this in the runtime, when make OPTIMIZE=1 is used. But this only affects C code in the runtime, not C code included in other ways (e.g., require's, passed via the command line, extern blocks)