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Question about exact time of destruction of temporaries in C++


Question about exact time of destruction of temporaries in C++

By : user3849244
Date : October 23 2020, 03:00 PM
will help you As other answers have stated you cannot take the address of a temporary. However, if you change the signature of
code :


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Destruction of string temporaries in thrown exceptions

Destruction of string temporaries in thrown exceptions


By : Rinor Nishevci
Date : March 29 2020, 07:55 AM
this one helps. As a temporary object (12.2), the result of the + will be destroyed as the last step in the evaluation of the full-expression (1.9/9) that contains it. In this case the full-expression is the throw-expression.
A throw-expression constructs a temporary object (the exception-object) (15.1) (std::runtime_error in this case). All the temporaries in the throw-expression will be destroyed after the exception-object has been constructed. The exception is thrown only once the evaluation of the throw-expression has completed, as the destruction of temporaries is part of this evaluation they will be destroyed before the destruction of automatic variables constructed since the try block was entered (15.2) and before the handler is entered.
Question about using string::swap() with temporaries

Question about using string::swap() with temporaries


By : truecool
Date : March 29 2020, 07:55 AM
To fix this issue After having used the swap-with-temporary idiom enough times, with lines like
code :
std::vector<int>().swap(v); // clear and minimize capacity
std::vector<int>(v).swap(v); // shrink to fit
Compiler policies on destruction of temporaries

Compiler policies on destruction of temporaries


By : user2946740
Date : March 29 2020, 07:55 AM
hop of those help? Likely, the string is still invalid, it just so happens that that section of memory hasn't yet been de-allocated at the operating system level and it "happens" to work. This program exhibits undefined behaviour- which always includes "may continue to work as if nothing went wrong". Visual Studio is completely correct here to crash your program or pretty much anything.
Why do MSVC optimizations break SSE code when function arguments are const refs to temporaries or temporaries copied by

Why do MSVC optimizations break SSE code when function arguments are const refs to temporaries or temporaries copied by


By : user3289075
Date : March 29 2020, 07:55 AM
should help you out You can add (non-virtual) member functions to a struct without really affecting the layout. So add destructor to print "I'm here %p" when the structure is destroyed, and print "I'm there" in your function. (Include the this address you you can make sense of other temporary copies being used).
Then you can observe the lifetime in the optimized code. See if that is your problem: I am suspicious that bad lifetime actually means anything because the place it was is still valid address in your stack frame.
Destruction order involving temporaries in Rust

Destruction order involving temporaries in Rust


By : Vish_mank
Date : March 29 2020, 07:55 AM
To fix the issue you can do Temporaries are dropped at the end of the statement, just like in C++. However, IIRC, the order of destruction in Rust is unspecified (we'll see the consequences of this below), though the current implementation seems to simply drop values in reverse order of construction.
There's a big difference between let _ = x; and let _b = x;. _ isn't an identifier in Rust: it's a wildcard pattern. Since this pattern doesn't find any variables, the final value is effectively dropped at the end of the statement.
code :
fn main() {
    let _ = B(&A as *const A);
    println!("End of main.");
}
Drop B.
Drop A.
End of main.
Drop A.
End of main.
Drop B.
struct A;
struct B<'a>(&'a A);

fn main() {
    let _ = B(&A);
}
struct A;
impl Drop for A { fn drop(&mut self) { println!("Drop A.") } }

struct B<'a>(&'a A);
impl<'a> Drop for B<'a> { fn drop(&mut self) { println!("Drop B.") } }

fn main() {
    let _ = B(&A);
}
error: borrowed value does not live long enough
 --> <anon>:8:16
  |
8 |     let _ = B(&A);
  |                ^ does not live long enough
  |
note: reference must be valid for the destruction scope surrounding statement at 8:4...
 --> <anon>:8:5
  |
8 |     let _ = B(&A);
  |     ^^^^^^^^^^^^^^
note: ...but borrowed value is only valid for the statement at 8:4
 --> <anon>:8:5
  |
8 |     let _ = B(&A);
  |     ^^^^^^^^^^^^^^
help: consider using a `let` binding to increase its lifetime
 --> <anon>:8:5
  |
8 |     let _ = B(&A);
  |     ^^^^^^^^^^^^^^
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