Re: DVD Cloner 2017 Crack Lifetime Activation Key Free Download
Odina Conkright
I think there is a subtle issue with what I'm trying to do here but I can't quite figure out why. I am attempting to clone a Box type (I know the type inside) but I think the lifetime is being propagated out somehow.
Boxing &T works as far as the types are concerned (because it's a Box), but it fails borrow checking. While T is guaranteed not to contain references to non-static data, the &T refers to the data inside &self which is not 'static, but has the anonymous lifetime of &self. This is actually pointed out by the compiler, but the error doesn't make sense without context.
The problem is that I'm trying to pass username to the default value method, which requires a str with a lifetime of 'r. I tried cloning but I can't figure how to tell it what the lifetime of the clone is going to be. I tried something along the following lines:
This says "for any lifetime that the caller of this function picks, I will return an Arg that contains references that will last that long". That's a pretty difficult promise to uphold, as the caller could request something that meets the 'static lifetime, a value that lasts longer than the call to main! In fact, the only way that you could meet the obligation of "any lifetime" is to return something that is 'static.
This is a very good sign that there is going to be a problem. See also Why can't I store a value and a reference to that value in the same struct?, which shows this case as a constructor. Many people jump to attempting to return the String along with the &str, so that answer might short-circuit that avenue as well. ^_^
username has a very specific lifetime and it's finite. If you look at a piece of code, it's generally straight-forward to find out the lifetime of an object: it's the scope of the block that the variable lives in without moving. In your example, username only lives during the block that's part of the match arm Ok(username) => // . As soon as that block exits, the value is destroyed.
env::var returns a Result, and you access the Ok variant, making username a String. The String implementation of clone takes a &String and returns a String. I'm not sure where the -> &'s str would come from.
This is a very common mistake. Check out Do Rust lifetimes influence the semantics of the compiled program? (and maybe Why are explicit lifetimes needed in Rust?) for some background information. You cannot change the lifetime of something other than by rewriting your code to make the referred-to value have a larger scope. The lifetime syntax reflects how long the variable lives, it does not control it. There's no (safe) way to "force" a lifetime.
If we dereference and re-reference a String, we end up with a &str. That &str will have a lifetime that corresponds to how long the String lives. This makes sense because the &str is just pointing to the String. When the String is deallocated, the &str would be pointing at memory that is no longer in a valid state.
Arg::default_value takes a &str as parameter, this means that the string is not stored in Arg, it is stored somewhere else. So the &str value must outlive the Arg that keeps the reference. If you use a &str obtained from a String value created in get_username_arg (that's the case for username), the Arg will outlive the &str (will live outside the get_username_arg while the &str lives only in the Ok block), so this generates a compiler error.
You do not show the Params definition, but it seems that it is just a "name space" for some functions. If this is the case, you can change these functions to receive &self as parameter (I know this is changing the signature, but the logic for creating args will stay in Params), and store username in Params:
Having issues with below code. The commented line is the issue. If I try and clone the config object inline I get the error that temporary value dropped while borrowed, which I take it to mean that I have to set config.clone() into a variable and then use it. However if I create a variable like this: let cfg = config.clone(), and then use the variable on the commented line, cfg.host.as_str(), the compiler complains that the new variable does not live long enough. And so I need a &'static str to pass in but I don't see how I can do that.
When the compiler tells you you need a 'static lifetime, usually it's a misleading error. It's trying to say that all temporary references are forbidden in this context. The only correct solution is not to use &str then.
In your case it's hard to say why it happens without seeing definition of SurrealWsConnection::new and SurrealDriver types. If new actually requires &'static str that's a poor design choice on their part, it should be taking String or Cow
Now I want to pass these bytes into a FlatBuffer function to create an object from them. Since FlatBuffers will hold on to the Vec I need to align the lifetime of the Redis returned value with the FlatBuffer object returned by my function.
Lifetimes are a way to associate input and output references which each other. In your example, there are no input references, so anything you output has to be valid for the entirety of the program. A reference to something allocated on your function's stack frame is not that.
Lifetimes don't do anything (they have absolutely zero influence on compiled code, and the compiler throws them away after initial check). Lifetimes only describe what the program is doing anyway, and work like assertions.
So ignoring lifetime syntax for a second, you're trying to take an address of a temporary variable temp_bytes, and then pass that address and return it (wrapped) from the function. That can't possibly work, because all variables are destroyed at the end of the function, so all references from variables become invalid. So logically there's no way for &lifetime_bytes to work, and there doesn't exist any valid syntax for it, because the intention of the operation is not valid.
If flatbuffers really really only supports references, then you can't use it in this function, because flatbuffers wants a reference to a permanently-stored value, and you don't have a place to store the temporary result somewhere where it will live longer than the function (sadly, you can't return both Vec and flatbuffers referencing it due to self-referential struct limitation in Rust).
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