bingo3131

7 ms·

1.
by bingo3131 1y ago
No idea how it works in other countries but in the UK there are fixed-rate mortgages and variable-rate mortgages. Fixed rate ones are where the interest rate is locked in at the time the mortgage is taken out so any changes to the national
2.
by bingo3131 1y ago
I would assume it would be because the poster would find themselves in negative equity (depending on how much was left to repay on the mortgage) with all the potential pitfalls that come with it, such as if you want to borrow further money
3.
by bingo3131 2y ago
! is commonly used as the unary not operator, so "a != b" makes sense as a shortcut for "!(a == b)". a not equals b.
4.
by bingo3131 2y ago
FYI: this is the latest draft of the proposal and it has not been voted into C++26 yet, but it is getting close.
5.
by bingo3131 2y ago
Rumour has it (not sure if this was ever confirmed) that one of the big reasons the second Xbox was called the Xbox 360 was to avoid unfavourable number comparisons with Sony. The Xbox launched vs the PS2, which meant the "Xbox 2"
6.
by bingo3131 2y ago
A lot of programmers refer to the compiler explorer as Godbolt due to the domain name, such as saying "Godbolt link", "try it on Godbolt", etc. They do not realise that the domain name is Godbolt.org not because "Go
7.
by bingo3131 2y ago
For those who missed it: each item on the iceberg is actually a link that explains the topic in more detail.
8.
by bingo3131 2y ago
Calling it a hint is very much over-simplifying things as there are also restrictions on what a constexpr function is and is not allowed to do, although those restrictions become fewer and fewer as time goes on as compilers are so advanced
9.
by bingo3131 2y ago
Regarding new/delete elision: https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2018/p09... You can also specify custom new/delete operations for your co-routine for control. I am n
10.
by bingo3131 2y ago
It is undefined behaviour to modify objects declared as const (except if those objects have mutable members, in which case the mutable members can be modified). const int a = 1; If you try to const_cast away the const of x to change its val
11.
by bingo3131 2y ago
A post about an obscure C++ feature on Hacker News? Comments are going to be a dumpster-fire as usual. https://en.cppreference.com/w/cpp/utility/launder "template <class T> [[nodiscard]] constexpr
12.
by bingo3131 3y ago
Misleading title? This is a Windows API bug with the slim reader/writer (SRW) locks. It's just that the bug was discovered via std::shared_mutex as that is implemented using SRW locks. SRW locks: https://learn.microsoft
13.
by bingo3131 3y ago
A new version of C++ comes out every 3 years, and every 3 years you will get loads of blog posts giving overviews of the changes as well as YouTube videos of conference talks on different things, which I watch in general as part of my on-go
14.
by bingo3131 3y ago
Yes. The main feature this is built upon is C++20's new string formatting features (very heavily inspired by the third party libfmt). C++23 just adds convinience functions to format and print at the same time. Type detection, formattin
15.
by bingo3131 3y ago
It's being actively worked on for C++2x. This video is a fairly recent update on how it's progressing: https://www.youtube.com/watch?v=AoLl_ZZqyOk Note for those who are not used to C++Now talks: they are intented
16.
by bingo3131 3y ago
https://en.wikipedia.org/wiki/GLib
17.
by bingo3131 4y ago
I just made a new empty C++ project in MSVC 2022 and by default it generates compile options for both 32-bit and 64-bit, with 64-bit selected by default. The preprocessor mode can be manually changed trivially. Project Properties -> C&#x
18.
by bingo3131 5y ago
The issue is that code outside of some_func does not know that f has been destroyed and a new object created in the same memory location, thus wouldn't know that the const member foo::x now has a new value. As for why the spec prohibit
19.
by bingo3131 5y ago
It is UB as the original object has non-static data members that are const-qualified.
20.
by bingo3131 5y ago
foo::x is declared const inside the struct in the example given. Because of this, regardles of how you access foo::x, it will always be const.