790 karma · joined November 7, 2014
There are some expensive hardware solutions to capture all network traffic and timestamp it at nanosecond level. E.g.,
https://www.endace.com/endace-high-speed-network-recorders.h...
I am little peeved by the maneuver at the beginning of the story though. You cannot use a slingshot maneuver around the sun to gain speed relative to the sun - that's simple conservation of energy. I know Geoffrey Landis is NASA scientist, so hard to understand how he could get that wrong, but it seems he did.
In an ideal world we'd have leaks of every group involved in the election: DNC, GOP, Hillary campaign, Bernie campaign, Trump campaign, Trump's tax records, Clinton foundation donor list, etc. However, having a subset, which was selected by some actor with ulterior motives, leaked is IMHO strictly worse than no leaks at all.
Also, there's been some claims that NSA and other agencies are running TOR exit nodes. Not sure if there's anything to it, but the point is state level actors have resources that nobody else does.
Besides, I am not sure what is your point. "Russian hackers would have covered their tracks, therefore it must not have been Russian hackers"?
https://johncarlosbaez.wordpress.com/2016/07/04/large-counta...
https://johncarlosbaez.wordpress.com/2016/07/07/large-counta...
[1] http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p029...
if constexpr ( COND) { CODE };
would compile even if CODE only compiles when COND is true. There a lot of nice use cases for that, and that's how the static_if in D works.You can still do this sort of stuff with template overloads, but it's much more cumbersome.
template<typename T>
void func(T x) {
if constexpr ( std::is_same< T, ClassWithMemberF>::value ) {
x.f();
}
}
will not work if you pass in a type for x that does not have a member function f. Is my understanding right? If so, constexpr if is not nearly as nice as I thought... if constexpr (false) { static_assert(false); }
Will fail to compile because the static_assert will trigger. This seems strange to me, anybody have an idea what the reasoning here is?For a cash dividend that's not true. Something real happens: the company has less cash afterwards (and the shareholder more).
There's a school of thought that says that the value of share should be the discounted sum of all future _cash_ dividends. According to it a company that only does share dividends should be worth nothing.
let result = flatuseStruct(outputData)
for _ in 0..<iterations {
assert(result == 8644311667)
total = total + UInt64(result)
}
The counter will still be correct, but you are not measuring what you think you are measuring.If ZF is inconsistent... I think that would make a lot of mathematicians panic.
To see a better breakdown look at the 2016 rich list at
http://www.institutionalinvestorsalpha.com/Article/3552805/T...
Note the following paragraph:
In a year when roughly half of all hedge funds lost money, a
number of frequent Rich List members are conspicuously missing
from the ranking because their funds finished in the red.
So for guys previously in this list their "salary" for 2016 was negative. template<typename T>
struct OnScopeExit {
OnScopeExit(T f) : f_(f) {};
~OnScopeExit() { f_();};
T f_;
};
template<typename T>
OnScopeExit<T> make_scope_exit(T f) { return OnScopeExit<T>(f); }
And then you use it as auto x = make_scope_exit([]() { std::cout << "Exitced Scope!" << std::endl; }); auto f = []() { std::cout << "Exited Scope!" << std::endl; };
OnScopeExit<decltype(f)> x(f);
The way unique_ptr does it is by using type erasure, which requires a virtual function call internally, just like std::function. struct OnScopeExit {
OnScopeExit(std::function<void()> f) : f_(f) {};
~OnScopeExit() { f_();};
std::function<void()> f_;
};
Which you'd then use as {
OnScopeExit x([]() { std::cout << "Exited scope!" << std:endl; };
// ...
}"Meta-Failures
2016 This darn CV of Failures has received way more attention than my entire body of academic work"
And I don't know many westerners that would say that $951M is not a lot of money :)
I believe that Bremermann's limit should also apply to quantum computers, but you are right in that example given for breaking cryptography is way off if you allow quantum computers.