Jeffrey Bosboom wrote:
Robert Ramey wrote:
The problem only occurs when program 1 contains only
ar << T
and program 2 contains
ar >> T and ar >> Tptr
But if only program 2 contains ar >> Tptr - how does the archive it's supposed to read get created? It can be only created by a program which also contains ar << Tptr. That is in order for serialization to make any sense at all both programs have to contain exactly the same sequence of loads and saves. Hence. they will both either serializations for the same set of types at compiler time.
I've been following this discussion, but now I'm a bit confused.
If program A contains ar << T and program B contains ar << Tptr, can program C read from files produced by both programs A and B? (C would thus require ar >> T in the code that reads files produced by A and ar
Tptr when reading files produced by B.)
The implementation of ar >> Tptr requires the usage of ar >> T by the library. So ar >> T is always there. Program A contains only ar << T Proram B contains only ar << Tptr (which implies ar << T as well) Program C is designed to read both input from program A as well as program B. So it must contain both: ar >> T and ar >> Tptr
I certainly think it "ought to", but as I understand it from the discussion, only the files produced by program B would be readable by program C.
LOL - dammit- looks like your correct. All this time to get a good example. So your example would look like for T = vector<char> or some other collection of primitives struct X { ... ar << T } struct Y { .. ar << Tptr } Program A ar << x; // only x serialized - struct Y not included in program Program B ar << y; Program C ar >> y; // OK ar >> x ;// error ! - can't read archive produced by Program A. // if x was produced by program B - then no problem So a good argument for at least including the compile time warning. Very astute. Robert Ramey