Interesting. Out of curiosity, what's a case where tmp = a.b(); tmp.c() works but a.b().c() doesn't? I'm still learning rust and the online book's Methods chapter doesn't say anything about such restrictions.
a.b().c()
b here is temporary, and so is freed at the end of the line, making the return of c dangling let temp = a.b();
temp.c();
Now that the return of b() is not temporary, it will last to the end of the scope and so c is fine.