You returned a pointer to a temporary. It's nasty but easy to spot.
Would you return `input.c_str()`? You might think twice about that, but it's the same thing. My favorite fix is to take a `std::string_view` as the parameter.
The result of the undefined behavior in your program is surprising, though. As it usually is.
quuxplusone 6 hours ago [-]
This is why I call string_view a "parameter-only type": it's fine to take as a parameter, but not fine to return or to store as a variable. (For sane definitions of "fine" and "not fine," of course. The blog post demonstrates one way it can go "not fine.")
but when using it as a parameter it IS stored as a variable, just only within that function, right?
dingaling911 5 hours ago [-]
Oh this is nasty. But the proposed solution of making admin_id a singleton is weird.
The real fix is don't pass in a temporary to the sanitize function.
stefanos82 8 hours ago [-]
With GCC and `-fsanitize=address,undefined` flag(s) I have catched it during runtime.
I've got a nice `==31604==ERROR: AddressSanitizer: stack-use-after-scope on address`
blastonico 5 hours ago [-]
std::string_view username = Sanitize("foobarbazboo");
^- temporary std::string created here
std::string_view Sanitize(const std::string& input) { ... }
^ ^- reference to that temporary string
+- returns a view to a temporary string
std::string_view username = // possibly data freed and dangling string
ranger_danger 5 hours ago [-]
> For example, this only works because the string foobarbazboo is the same length as admin_id
12 != 8
I don't get it...
> As you can guess, this is an example of undefined behavior causing unintuitive results.
No I couldn't... and what undefined behavior where? They don't explain anything.
dingaling911 5 hours ago [-]
He means the string value of the variable admin_id, not the variable name admin_id.
And the bug comes from the temporary string object created when he passes in the string to Sanitize.
Sanitize("foobarbaz") -> creates temporary string -> string_view from temporary string -> returns string_view of temporary string
temporary string destructed
operations on returned string_view -> oops, use after free.q
ranger_danger 4 hours ago [-]
But if Sanitize has a return type of string_view, then returning the temporary variable should make a copy of it (or move it), right? And the string "foobarbazboo" is static so that's not being freed either.
And the while loop shouldn't even run at all since front() never equals ' '.
dingaling911 4 hours ago [-]
string_view is literally a reference. Copying a string_view just copies the reference.
ranger_danger 3 hours ago [-]
But it's not actually taken as a reference (with &) anywhere I see... and even if it's just internally referencing the (static) string, there shouldn't be any issue with that memory either right?
magackame 5 hours ago [-]
The value of `admin_id` is "root_1231231", which is indeed the same length as "foobarbazboo". Otherwise, yeah, no idea where the bug comes from.
Rendered at 05:58:03 GMT+0000 (Coordinated Universal Time) with Vercel.
Would you return `input.c_str()`? You might think twice about that, but it's the same thing. My favorite fix is to take a `std::string_view` as the parameter.
The result of the undefined behavior in your program is surprising, though. As it usually is.https://quuxplusone.github.io/blog/2018/03/27/string-view-is...
The real fix is don't pass in a temporary to the sanitize function.
I've got a nice `==31604==ERROR: AddressSanitizer: stack-use-after-scope on address`
12 != 8
I don't get it...
> As you can guess, this is an example of undefined behavior causing unintuitive results.
No I couldn't... and what undefined behavior where? They don't explain anything.
And the bug comes from the temporary string object created when he passes in the string to Sanitize.
Sanitize("foobarbaz") -> creates temporary string -> string_view from temporary string -> returns string_view of temporary string
temporary string destructed
operations on returned string_view -> oops, use after free.q
And the while loop shouldn't even run at all since front() never equals ' '.