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Can the storage of trivially copyable objects be safely reallocated with realloc?


Can the storage of trivially copyable objects be safely reallocated with realloc?

By : Allison Barr
Date : November 20 2020, 03:01 PM
Does that help No, realloc cannot be used to safely move objects, even of trivially copyable types, because realloc cannot create new objects in uninitialized storage.
In particular, according to C++14 [basic.life]/1:
code :


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Is copying trivially copyable objects always defined in C++14?

Is copying trivially copyable objects always defined in C++14?


By : Muhammad Ramadan
Date : March 29 2020, 07:55 AM
hop of those help? There are a couple things at play here:
an expression evaluating to an indeterminate value causes undefined behavior, with certain exceptions (8.5p12) unsigned char (and possibly char, if unsigned) is the exception variables with automatic storage duration and whose types have trivial default initialization initially have indeterminate values (5.3.4p17)
Atomic objects of trivially copyable types in GCC

Atomic objects of trivially copyable types in GCC


By : NotADoctor
Date : March 29 2020, 07:55 AM
hope this fix your issue This answer is compiled from the comments.
You need to explicitly link the atomic operations library with your program by specifying -latomic on the command line.
Google Sparsehash uses realloc() on type which is not trivially copyable

Google Sparsehash uses realloc() on type which is not trivially copyable


By : user1817270
Date : March 29 2020, 07:55 AM
I hope this helps you . Consider this simple program: ,
Yes, it is undefined behavior.
code :
class bang {
    bang* p;
public:
    bang() : p(this) {}
    bang(bang const&) : bang() {}
    bang& operator=(bang const&) { return *this; }
    ~bang() { if (p != this) std::abort(); }
};
About aligned storage and trivially copyable/destructible types

About aligned storage and trivially copyable/destructible types


By : user2372424
Date : March 29 2020, 07:55 AM
Does that help This problem reduces to basically what LanguageLawyer suggested:
code :
alignas(int) unsigned char buff1[sizeof(int)];
alignas(int) unsigned char buff2[sizeof(int)];

new (buff1) int {42};
std::memcpy(buff2, buff1, sizeof(buff1));

assert(*std::launder(reinterpret_cast<int*>(buff2)) == 42); // is it ok?
assert(vec.size() == 1); // ok
assert(!vec[0].weak);    // ok
assert(vec[0].fn(vec[0].storage) == 42); // UB
Copy trivially copyable types using temporary storage areas: is it allowed?

Copy trivially copyable types using temporary storage areas: is it allowed?


By : Joe
Date : March 29 2020, 07:55 AM
hope this fix your issue It reads fine to me.
You've copied the underlying bytes of obj1 into obj2. Both are trivial and of the same type. The prose you quote permits this explicitly.
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