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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 chapt
by general_pizza 10y ago
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.
- steveklabnik 10y agoIt happens if b() returns an owned value that c() returns a reference to. In that case, 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.