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Finding the value of the function

authera

введите сюда описание изображения

Simple assignment - write a C+++. Baseline: x=5, a = 3.3

#include "stdafx.h"
#include <iostream>
#include <conio.h>
#include <math.h>
using namespace::std;
int main()

{

setlocale(LC_ALL, "Russian");

double x, y;

const double a = 3.3;

cout << "Введите x: " << endl;

cin >> x;

y = 12.7 * pow(asin(3 * pow(x, 2)), 3) / pow(log(sqrt(x + exp(a))), 2);

cout << "y = " << y;

_getch();

return 0;

}

The result is right (2.814), but my source is nan(ind). I tried to take it underneath the root. What did I miss?

emmalee last edited by

So it's most likely that the number of 0.5 was recorded in the mission without the lead 0-- ♪ 5
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First from the latter.Function and Function objectsAll functions in Function are represented by Function objects. Function Object new Function() You can also use "function statement" and "function expression" which are dedicated syntax in addition to the generation. http://www.ecma-international.org/ecma-262/5.1/#sec-4.3.24 http://www.ecma-international.org/ecma-262/5.1/#sec-13 Even if it is a native function, it is a Function object.document.write(alert instanceof Function);ConstructorBecause the constructor is a function (a function object), it can be said to be a type of function.The use of "constructors" in isnew Calling a command line function is expressed as a constructor. If you think from here, a function is a constructor, it is a constructor.newI think it is possible to call)For example, for Date objects, the following description can be found in the standard. When Date is new It is a new object. --- http://www.ecma-international.org/ecma-262/5.1/#sec-15.9.3 More QuotesIn this way, new I think that a function is called "constructor" under the premise that it is called (as a constructor).More Array() Like new If you do not use the same object,When Array is called as a function rather than as a constructor, it creates a new Array object. --- http://www.ecma-international.org/ecma-262/5.1/#sec-15.4.1 More QuotesI think that it is a function that produces an object like a constructor, although it is not a constructor strictly speaking. Class-based languages new Hoge() static methods to internally call Hoge.create() the same as not calling it a constructor even if it exists.
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Like this:persons.stream() .filter(p -> p.getFirstName().length() + p.getLastName().length() < 15) .max(Comparator.comparing(Person::getAge)) .ifPresent(p -> System.out.println(p.getFirstName() + " " + p.getLastName()));
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For example:Option even with multiple numbers:string revs(const vector<int>&n) { return accumulate(n.rbegin(),n.rend(),r, [](string r, int x) { return r += to_string(x); } ); } If these numerical numbers also need to be deployed, well, it's harder:string revs(const vector<int>&n) { string r; return accumulate(n.rbegin(),n.rend(),r, [](string r, int x) { string t = to_string(x); reverse(t.begin(),t.end()); return r += t; } ); } See work programme https://ideone.com/adFtqp ♪Well, if you need a number (although you're writing in the title)in line"Well, you can just use the line stoll♪ It's easier, but it's not clear how to deal with multiple numbers. If you specify what a vector should be, say, {1, 23, 45670, 5} - let's talk.For only single-digit numbers, the method @ARHovsepyan is ideal in both the sense of simplicity and productivity, which I would also rewrite in one line:int revs(const vector<int>&n) { return accumulate(n.rbegin(),n.rend(),0, [](int res, int x) { return res = res*10+x; } ); } mainly because, in such a case, it is applicable to virtually any container (except for forward_list and all unorderedand the latter simply because they have no idea of the order.The options for the previous one are mine and @Qwertiy for numerical numbers, but both suffer from an unnecessary transformation - in my line, in @Qwertiy - in the line, in the flow and again. ♪ ♪
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In your first example:function get(s){ var f = function(r){ // code // code return response; } someobject.somefunc(arg1, {arg2:arg3}, f); return f(r); } In your same option, in the first case, a function get Never return anything. In the second case, only reference to function someobject.somefunc♪ It's hard to know that this is your function. someobject.somefunc, carries out by itself the function that you give to the third argument and does what it needs. If someobject.somefunc Don't bring anything back, make her give you the right meaning.In the second case, it was like,return someobject.somefunc(arg1, {arg2:arg3}, function(r){ // code // code return response; });
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A common idea. Let us face the challenge. U Over time Tu Then we need to maximise the amount. solve(0,Tu) + Cu + solve(Tu+Su,inf)♪ It's already clear that it's a naive solution to take over the first task we're gonna face. I'd say we take over the last task that we're gonna face. 0..NThen we'll figure out the results for her. But it won't be a long time, so the tree will go, but remember the memory.Pseudocod (tasks already quashed at start):for (last_task = 0; last_task < N; last_task ++){ Tree.update(T[last_task] + S[last_task], Tree.max(0,T[last_task]) + C[last_task] ); } cout << Tree.max(0, MAX_INT) << endl; Restoring the answer is no longer a problem. To this end, a tree needs to be changed to recognize not only the value but also the number + the position of the maximum element. Let the tree return {value,time, position}♪ Then:vector<int> ans; int MaxTime = MAX_INT; while (true){ auto tmp = Tree.getFullMax(0,MaxTime); if (tmp.value == 0) break; ans.push_back(tmp.position); MaxTime = tmp.time; } cout << ans.size()<<endl; for (int i=ans.size()-1;i>=0;i--) cout << ans[i]<<" "; It's like without a tree:struct Task{ int t,s,c,n; }; Task task[100000]; long long bestSum[100010]; int goSum[100010]; bool cmp(Task a, Task b){ return a.t + a.s < b.t + b.s; } int main() { freopen("olympiad.in","r",stdin); freopen("olympiad.out","w",stdout); int N; cin >> N; for (int i=0;i<N;i++){ cin >> task[i].t >> task[i].s >> task[i].c; task[i].n = i+1; } sort(task,task + N, cmp); bestSum[0] = 0; Task empt; empt.s = 0; empt.c = 0; empt.n = 0; for (int i=0;i<N;i++){ bestSum[i + 1] = bestSum[i]; goSum[i + 1] = -1; empt.t = task[i].t; int prev = upper_bound(task,task+N,empt,cmp) - task; long long nn = bestSum[prev] + task[i].c; if (nn > bestSum[i + 1]){ goSum[i + 1] = prev; bestSum[i + 1] = nn; } } cout << bestSum[N]<<endl; vector<int> ans; while (N) if (goSum[N] == -1) N--; else { ans.push_back(task[N-1].n); N = goSum[N]; } cout << ans.size() << endl; for (int i=ans.size() -1; i>=0; i--) cout << ans[i]<< " "; cout << endl; } Thus, the author ' s idea is generally being pursued. ♪ ♪
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How to change the value of a variable that is within a function through another function?
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Try this:class Stack: def __init__(self, name): self.stack = [] self.name = name self.index = 0 def push_(self, string_): self.index += 1 self.stack.insert(self.index, string_) def pop_(self): if self.index <= 0: return print('шпиль пуст, нечего удалять!') else: self.index -= 1 print('удаляем верхнее кольцо', self.stack[self.index]) return self.stack.remove(self.stack[self.index]) stack_remove = self.stack[self.index] # + self.stack.remove(stack_remove) # + return stack_remove # + def move(source, destination): element = source.pop_() print(element) destination.push_(element) Называем шпили и присваиваем им класс Stack first = Stack('Первый шпиль') second = Stack('Второй шпиль') third = Stack('Третий шпиль') Заполняем шпиль first big_ring = 'Большое кольцо' middle_ring = 'Среднее кольцо' small_ring = 'Маленькое кольцо' first.push_(big_ring) first.push_(middle_ring) first.push_(small_ring) print('СТАРТОВОЕ СОСТОЯНИЕ') print(f'Башня: {first.name}, Содержание {first.stack}') print(f'Башня: {second.name}, Содержание {second.stack}') print(f'Башня: {third.name}, Содержание {third.stack}') print('-----------------------------------') Решаем задачу move(first, third) print(f'Башня: {first.name}, Содержание {first.stack}') print(f'Башня: {second.name}, Содержание {second.stack}') print(f'Башня: {third.name}, Содержание {third.stack}') move(first, second) print(f'Башня: {first.name}, Содержание {first.stack}') print(f'Башня: {second.name}, Содержание {second.stack}') print(f'Башня: {third.name}, Содержание {third.stack}') move(third, second) print(f'Башня: {first.name}, Содержание {first.stack}') print(f'Башня: {second.name}, Содержание {second.stack}') print(f'Башня: {third.name}, Содержание {third.stack}') move(first, third) print(f'Башня: {first.name}, Содержание {first.stack}') print(f'Башня: {second.name}, Содержание {second.stack}') print(f'Башня: {third.name}, Содержание {third.stack}') move(second, first) print(f'Башня: {first.name}, Содержание {first.stack}') print(f'Башня: {second.name}, Содержание {second.stack}') print(f'Башня: {third.name}, Содержание {third.stack}') move(second, third) print(f'Башня: {first.name}, Содержание {first.stack}') print(f'Башня: {second.name}, Содержание {second.stack}') print(f'Башня: {third.name}, Содержание {third.stack}') move(first, third) print(f'Башня: {first.name}, Содержание {first.stack}') print(f'Башня: {second.name}, Содержание {second.stack}') print(f'Башня: {third.name}, Содержание {third.stack}')
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We don't have to go through the processors, but through a change in dom element.On the Elements of his right button, Break on... ♪ attribute modifications
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How do you write a spinBox variable and then put it in another shape?
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If you need the value at the time of the shape, you can use the design parameters:class ksp_ecology : public QWidget{ double _Mkrt; public: explicit ksp_ecology(double Mkrt, QWidget *parent = 0): QWidget(parent), _Mkrt(Mkrt) {} //... }; And in the show, do it like that:void MainWindow::on_pushButton_clicked(){ double Mkrt = ui->spinBox->value(); ksp_ecology *ksp_e= new ksp_ecology(Mkrt, this); ksp_e->setAttribute(Qt::WA_DeleteOnClose); ksp_e->show(); } Pay attention to this: ksp_e->setAttribute(Qt::WA_DeleteOnClose); Without that, you have a memory leak. For each application of the button, the shape shall be created and never removed. So she'll be self-destructive when she's locked up.
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You're not the first person to be interested in this problem; this issue has even been on the official list of frequently asked questions on C+++.Known https://isocpp.org/wiki/faq/operator-overloading#matrix-subscript-op advised to use operator ():Use it. operator()Not operator[]♪If you have a multi-dimensional mass, the most pure solution is to use operator()Not operator[]♪ The reason is that operator[] Always accepts one parameter while operator() Accepts any number of them (in the case of a rectangular matrix, you need two). https://isocpp.org/wiki/faq/operator-overloading#matrix-array-of-array and https://isocpp.org/wiki/faq/operator-overloading#matrix-c-style-subscript Explanation why the brackets are better than two pairs of square.Summary:The square brackets require the establishment of an intermediate object representing the table line, which limits the physical location of the data in the main class.round brackets are clearly not worse under any circumstances, and sometimes betteruse of antique for C+++ syntaxis is betterchecking the input parameters is easier to improve the quality of the code
Starting the process in the endless cycle
Software Programming • c++ c language linux • • Analeea

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how to make the process work in an endless cycleTwo things should be done:(1) Replace the lineexecl("/bin/ls", "ls", (char *) 0); Endless cycle. Well, something like:while (1) sleep(10);(2) Remove the expectation of completion of the subsidiary process in the main:waitpid(pid, &status, 0) Otherwise, the basic program will never end.Re-launch the programmeInstead of a lineprintf("Child id: %i\n", pid); Fill the pid in some text file. And your main program, at the beginning of your work, should verify the existence of this file (see pid file) and, if it exists, should read the pid value, send a signal to the kill process:kill(pid, SIGTERM); and remove the pid file.Also, I'm interested in one thing that does that line.execl("/bin/ls", "ls", (char *) 0);This challenge in no way creates a new process. He, inside the existing target, replaces the current process with others. That's how it says in man:The exec() family of functions replaces the current process image with a new process image.In order to start the process of programme B from process A, it is necessary first to establish the FIPS of process A (fork) and then to replace process A with process B with one of the exec family functions.That is, this line, in the target area, a copy of your main process launches the programme ls♪ Parameter * ) 0 Functions execl that the programme ls The BEZ is running a command line.The family ' s responsibilities are the same, but the mode of transfer to the new process of variable command line and ENV-replaced.
Function does not return value
Software Programming • php mysql • • Demir

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It takes into account that return will only be called after the for have run.That is, the cycle for will run in the integral, the value of $c will be re-written until the cycle is over. Then, when the cycle ends, the last value that $c has it is 9 and it is that value that will be passed return.When you join one echo within the cycle for you are registering (make echo) of each value $i take and go $cSo the difference. The final result is the same, the returned by return It's the same. Note that in the version with echo within the cycle for you are sure 2 times the number 9, the second return of the function.
In what cases do you use the reference class index and in which heir?
Software Programming • c++ polymorphysics • • Laycee

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Usually, when polymorphysics are spoken in programming, they attempt to define the word in the language of programming.However, in my view, a better definition of polymorphism is given in biology. Just the word организм should be replaced by объект in programming terms:)Polymorphism in biology (dr.-greek) The ability of some organisms to exist in conditions with different species internal or externalYour demonstration program defines a type-indicator. Base *i.e. the static type of objects addressed by this indicator is the type Base♪Base *pBase; When the indicator is assigned the address of the production class object, the object to which the indicator is applied shall be considered to be the object of the type Base♪ In fact, all information that the object is in fact a production class rather than a base is lost, as the static type of indicator is lost. Base *♪However, the existence of a tool such as virtual functions allows for a variety of behaviour of the object addressed by the indicator, depending on the type in which the object is actually addressed. Thus, through virtual functions, it is possible to distinguish the real type of objects to be addressed and to treat them uniquely. And that's what your programme is showing in the example of this proposal.Base *pBase = &derived;//Объявили указатель на базовый класс, присвоив ему ссылку на объект класса наследника pBase->Virtual_method(); // Вызов виртуального метода pBase->Nonvirtual_method(); // Вызов не виртуального метода As seen by the type-indicator Base * It's actually a class object. But if it's not a virtual methodpBase->Nonvirtual_method(); // Вызов не виртуального метода all information about the production class is not available because the compiler causes functions according to the static type of object addressed by the indicator, i.e., by type Base♪ Compiler in search of the name of the function to be called the definition of the base class. He doesn't know anything about derivative classes.If you cause a virtual functionpBase->Virtual_method(); // Вызов виртуального метода the compiler also looks at the definition of the reference class when looking for the name of the function. But there's some trick going on. The production class changed the definition of a virtual function in the base class to define the function. Technically, this is done as follows. If the class declares a virtual function, it shall establish a table of indexes for virtual functions as stated in its definition. The same production classes, which redesign virtual functions, place their definitions of functions in this table. Therefore, the function at which the address is located in the virtual function is dynamically raised during the programme. As for your question.In which cases the heirloom index should be used, What class?When you announce the heir to the class, you lose polymorphys, because you can only work with the subjects of the heir class (I assume that the heir does not inherit other classes in turn). And when you want to achieve a polymorphysicist behavior,(1) The base class shall contain virtual functions t (2) An index or reference bearing the static type of indicator or reference to the objects of the reference class should be declared and the objects of the derivative classes should be initiated. And then you get it. The ability of some objects (organisms) to exist in conditions different internal structures or different external formsNote♪ If, when you call a virtual function, you give her a skilled name, "virtuality" disappears. The class function defined in the same class shall be performed.Example.#include <iostream> struct Base { virtual ~Base() { } virtual void virtual_function() const { std::cout << "Base::virtual_function() is called" << std::endl; } }; struct Derived : Base { void virtual_function() const override { std::cout << "Derived::virtual_function() is called" << std::endl; } }; int main() { Derived d; Base &rBase = d;; rBase.virtual_function(); rBase.Base::virtual_function(); return 0; } Its conclusion to the console will be as follows:Derived::virtual_function() is called Base::virtual_function() is called In the first case, polymorphysics plays. Reassigned virtual function of the production class Derivedalthough the reference type is a class Base♪In the second case, when a skilled name is used, the virtuality of the poppy is gone, and the function of the basic class is called Base♪
After the decommissioning of the function, it gives an finde
Software Programming • javascript jquery op • • emmalee

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Validate accepts only one parameter function Validate(elem1) { But they are trying to transfer two: new Validate(elem1, elem2)♪function inputLength returns result of implementation console.logresults of this function undefined♪ Since nothing else comes back, it's a legal result: undefined It's already in the calling code.It's just enough to get back to the decision. error♪

Finding the value of the function

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