#include <iostream> #include <vector> bool findMaxSumOFSubarray(const std::vector<int>& source, int* res) { if (source.empty()) { return false; } else /* left blank intentionally */; int tmp = source[0]; *res = tmp; for (size_t i = 1; i != source.size(); ++i) { tmp = std::max(tmp + source[i], source[i]); *res = std::max(*res, tmp); } return true; } int main() { const std::vector<int> source = {1, -1, 2, 4, -10, 8, 3, -2, -7, 0}; int res; if (findMaxSumOFSubarray(source, &res)) { std::cout << "Result: " << res << "." << std::endl; } else /* left blank intentionally */; return 0; }

Just aware of that ... your alignment here is pure OCD style 😂 your employees must feel painful

Just aware of that ... your alignment here is pure OCD style 😂 your employees must feel painful

BTW, if the source array is a circle, i.e. subarraies could go through the end and the begin of the array but not overlap. How to solve the maxSumOFSubarray problem in O(n) time and O(1) space?

Using cellphone... I guess remembering the beginning point of the max sum subarray could be enough?

Using cellphone... I guess remembering the beginning point of the max sum subarray could be enough?

Oh I didn’t mean that ... I meant based on the previous greedy solution, adding a beginning point could help check if overlapped, which may be equivalent to using a maxSumSubarrayLength and a flag to check if rewinded...

Hello. i have qouestion about kernel linux. What is kernel book-keeping?

Thanks for the help man

destructors usually need a single if clause

+ branching can lead to perf losses

btw how can we assign a function pointer to a functor(function object)?

No. Your compiler will optimize empty branches out.

like struct Obj{ void operator()(){ std::cout «"heeey it's me lunatic\n"; } }; then we have : void(*f)() how can we assign Obj.operator() to the above pointer

obj.operator() calls the pointers "operator"()

like struct Obj{ void operator()(){ std::cout «"heeey it's me lunatic\n"; } }; then we have : void(*f)() how can we assign Obj.operator() to the above pointer

No you cannot assign Obj.operator() to void(*f)(), since Obj.operator()'s type is void(Obj::* f)().

if I can't why I can assign lamdas to function pointers

They are not the same thing.

#include <iostream> struct Foo { void bar(void) { std::cout << "hello world" << std::endl; } static void baz(void) { std::cout << "bye world" << std::endl; } }; int main() { Foo foo; void(Foo::* pbar)() = &Foo::bar; (foo.*pbar)(); void(* pbaz)() = &Foo::baz; pbaz(); return 0; } and for static member functions, they act like normal functions.

the same applies for operator() right?

I once read that lamdas are a functors 🤔

xD i'll take a decade to understand that

No you cannot assign Obj.operator() to void(*f)(), since Obj.operator()'s type is void(Obj::* f)().

I once read that lamdas are a functors 🤔

They are lambdas, the compiler implements them as it wants

On assembler level, they are probably compiled down to a pointer and a stack for states

On assembler level, they are probably compiled down to a pointer and a stack for states

.

Yes, of course. But we are in the world of C++...

This is because: closure type for a lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator..

BTW, this reflects the philosophy of C++: zero-overhead.

we should call it "optimal overhead"

like the best case?

You didn't understand the words. The term overhead here means extra overhead on C++'s abstraction,.

You didn't understand the words. The term overhead here means extra overhead on C++'s abstraction,.

yes. Like using vars on the stack unlike C# which puts every single fucking one of them onto the heap

holy shit is that true automatic variable on the heap?

There are so many concepts in C++ mihgt mislead newbies.

C#does that and is 25% slower

wait a minute user-defined types or primitive?