Since in the commentaries, you agreed with my proposal, I will say that one operator is enough:void operator +=(pair<string, string> eng_rus) { root = insert(eng_rus.first, eng_rus.second, root); } Which can correctly accept all types of input data you have specified: <char,string>,<string,char>,<char,char>A symbolic type char It's perfectly safe to interpret as a line. std::stringwhich may be initialized by one symbol.The duplication of the host operators is thus eliminated.editexamplestruct S { void operator += (const std::pair<std::string, std::string>& eng_rus) { std::string s; s += eng_rus.first; //root = insert(s, eng_rus.second, root); } }; int main() { S s; s += std::pair<std::string, std::string>(std::string(1, '1'), std::string(1, '2')); return 0; }

I found : this is being implemented by a list of initialization(initializer list) as a reloading of a class designer: my_vector(initializer_list<T> lst) {ваш код. }♪ Class object lst has different methods, such as size, and iters.

It's basic. There are two options:First: simply formed by index to field Array:...Array1.Array[i]... Secondadd the possibility of indexing the class Znumbers:public class Znumbers { ... public int this[int i] { get { return Array[i]; } set { Array[i] = value; } } ... }

If by paragraph:Functions area(), perm(), size() not accepting any parameters, calling them onclick="area(width,height)" It doesn't make sense, because at the time of the challenge, there is no memory.Data, sort of width You're trying to get your own function in the body, causing a blocking function. prompt()that's not exactly the right thing to do. The JavaScript script environment offers many excellent opportunities to obtain user data. For example, with help input type="number"♪ Moreover, the result of the challenge prompt() There's nowhere to be stored. So I wrote:let height = +prompt("height");Where I enter, a local variable for the function, which will accept the return. Drug operator + converts the line (if possible) to number.Telework area() other functions result()but the result of her challenge is not accepted. I'll do it. result() It doesn't come back. Moreover, the function and intended return have the same name. result♪ You could have declared a variable. result in the global field, but in an attempt to give it meaning in the body of a single function (result = prompt("want to see result?")there will be a conflict of names and a violin will end in a mistake. It is therefore better to call a function: getResult, countArea, doSomething, one word. In your version, in part. switch (result) variable result keeps a reference to function as an object. So I'm reintroducing the local variable. resultin which the function is returned getResult() meaning. Although with the same success, one can briefly describe switch(getResult())♪The challenge itself. prompt() to confirm the conclusion of the result is also doubtful. No options proposed 'y' or 'n'that information can only be found from the code. Then you better use it. confirm()♪ Although neither prompt nor confirm nor alert should be fully used under any circumstances.Don't fight hard. I left the author's logic, only led to a minimum working option.function getResult(){ return prompt("want to see result?"); // Возвращается значение, введенное пользователем } function area() { let height = +prompt("height"); // Принимаем значение, преобразовываем в Number и присваиваем переменной let width = +prompt("width?"); let result = getResult(); // Принимаем, возвращаемое функцией, значение switch (result) { case "y": alert(height * width) break; case "n": alert("ok") break; default: break; } }<button onclick="area()">AREA</button> <!--Сократил немного -->

Rewrite one fragment like:for (int k = 0; k < n; k++) { assert(i < this->row && k < this->line); sum +=this->x[i][k]; } And make sure you go outside your matrix.What do we do? Rewrite carefully... I think you're confused that you have a pole, and that the line is a blank, and it's not a big deal: line and row... ("Crate" : line You have a number of lines, row - number of columns.For example, when the matrix is multiplied, a matrix with a number of lines, like in the first, and columns, like in the second, and you have int m = v.row; int p = this->line; c.CrateMatrix(m, p); It's like the other way around. ♪ ♪

#include <iostream> class Test { struct Impl { int a; auto operator[](int b) { return a * b; } }; public: auto operator[](int a) { return Impl{a}; } }; int main() { Test data; std::cout << data[7][6] << std::endl; return 0; } https://ideone.com/cUYsv2

Almost all right, just a few moments:Copying designers (with displacement and without) need no matching of addresses.In the copying constituators (with displacement and without) of the class, the heir shall be required to call class-pre-accounters instead of the assigning operator.Use a list of initializations in designers to initialize members.In the grading operators of the heirloom class, call the proprietor to assign the class probe inside if♪ExampleOne::ExampleOne(const ExampleOne& other) : val1(other.val1), val2(other.val2), str1(other.str1) { } ExampleOne::ExampleOne(ExampleOne&& other) : val1(other.val1), val2(other.val2), str1(std::move(other.str1)) { } ExampleTwo& ExampleTwo::operator=(const ExampleTwo& other) { if (&other != this) { ExampleOne::operator=(other); val3 = other.val3; str2 = other.str2; } return *this; } ExampleTwo& ExampleTwo::operator=(ExampleTwo&& other) { if (&other != this) { ExampleOne::operator=(other); val3 = other.val3; str2 = std::move(other.str2); } return *this; } ExampleTwo::ExampleTwo(const ExampleTwo& other) : ExampleOne(other), val3(other.val3), str2(other.str2) { } ExampleTwo::ExampleTwo(ExampleTwo&& other) : ExampleOne(other), val3(other.val3), str2(std::move(other.str2)) { }

You see, there's no "in general case."What's the difference between words ++i and i++ for simplicity int? Where the meaning of the expression is not used, so the whole effect is only to increase the variable by 1. ♪And if it's more serious, the prefix just increases the meaning, and then the postfix has to keep the old meaning and then return it. Feel the difference? :Like an example, even for a simple one. int Without optimization, VC++ is treated differently:; 15 : cout << ++i; mov ecx, DWORD PTR _i$[ebp] add ecx, 1 mov DWORD PTR _i$[ebp], ecx mov edx, DWORD PTR _i$[ebp] push edx mov ecx, OFFSET ?cout@std@@3V?$basic_ostream@DU?$char_traits@D@std@@@1@A ; std::cout call ??6?$basic_ostream@DU?$char_traits@D@std@@@std@@QAEAAV01@H@Z ; std::basic_ostream&lt;char,std::char_traits&lt;char&gt; &gt;::operator&lt;&lt; ; 16 : cout << i++; mov eax, DWORD PTR _i$[ebp] mov DWORD PTR tv72[ebp], eax mov ecx, DWORD PTR _i$[ebp] add ecx, 1 mov DWORD PTR _i$[ebp], ecx mov edx, DWORD PTR tv72[ebp] push edx mov ecx, OFFSET ?cout@std@@3V?$basic_ostream@DU?$char_traits@D@std@@@1@A ; std::cout call ??6?$basic_ostream@DU?$char_traits@D@std@@@std@@QAEAAV01@H@Z ; std::basic_ostream&lt;char,std::char_traits&lt;char&gt; &gt;::operator&lt;&lt; See, the time variable. tv for storage?What if it's a user class... ♪

Ternar code is an expression. You've got this expression trying to be two different types at the same time, depending on the meaning. a[z]that's impossible.Can I help you? if (a[z]) cout << z; else cout << " "; May a[z] ? cout << z : cout << " "; Or use a common type with an attachment to it, like @Learpcs.

Decimal Decimal::operator+=(const Decimal &d){ this->numeration *= d.denomination; // 3/4 += 1/5 => 3/4 + 1/5 = (3*5 + 1*4)/(4*5) this->numeration += this->denomination * d.numeration; this->denomination *= d.denomination; return *this; } Decimal Decimal::operator-=(const Decimal &d){ this->numeration *= d.denomination; //(3*5 - 1*4)/(4*5) this->numeration -= this->denomination * d.numeration; this->denomination *= d.denomination; return *this; }

The operator's support? started with the phpStorm 2019.3, so you just had to download a new version of IDE. https://intellij-support.jetbrains.com/hc/en-us/community/posts/360006014139-Nullish-Coalescing-operator-support-for-upcoming-TypeScript-release

Maybe we should set up a vault, and keep multipliers in it, depending on the probability?Map<String, Double> data = new HashMap<>(); data.put("50%", 1.5); data.put("40%", 2.506); int number = 4; double result = number * data.get("50%"); System.out.println(result); //6.0

Like... ♪bool T::operator==(const T& a); and T::T(int); There's no way you can do the test.T t; if (5 == t) Whereas for free operator, it's easy, thanks to a type-approval. ♪ ♪Here's a live example:struct T { int i; T(int i):i(i){} bool operator==(const T&amp; t) { return t.i == i; } }; // bool operator==(const T& t, const T& u) { return t.i == u.i; } int main() { T t(5); cout << (4 == t); } Try scoping, it won't work. But get the comment, and it'll work.Is that clear?The operator, as a member of the class, requires that the first argument be the unambiguous object of this class, without any indication of type etc.

Somehow:MyClass & operator += (MyClass &a, int b) { a.value += b; return a; } MyClass operator + (MyClass a, int b) { return a += b; } MyClass operator + (int b, MyClass a) { return a += b; } I mean, you have to decide twice, at least one body?Yeah, we need two operators. But there's a second such thing in the front.MyClass operator + (int b, const MyClass &a) { return a + b; }

It's not quite clear that there's a limitation on the number of options. A friend can be defined both inside and outside the class. The definition outside the class may be both inline and not inline. Why is it all about "friending a function within class"? Why is it inline that is the way out of class? Nevertheless:First of all, whatever. Second, depending on whether you need a friend-access. Need or not?Usually, they prefer an alternate friend even when a friend-access is not necessary, just because it allows all the definition of a overloadable function to be placed in a class. Someone thinks the code looks more comfortable. It's for small functions.Big functions, in my opinion, are better defined outside the class. And here is the question of whether the function will be inline and whether the function of the friend will be determined by extraneous factors specific to each function that you have not mentioned in your question.To classrooms, all this has nothing to do with. With only the comment that in such a case there is one thinness: to declare the non-obsessive overload in such a case can only be as a friend in the definition of a class template. (See. https://ru.stackoverflow.com/questions/849912/%D0%9F%D0%B5%D1%80%D0%B5%D0%B3%D1%80%D1%83%D0%B7%D0%BA%D0%B0-%D1%88%D0%B0%D0%B1%D0%BB%D0%BE%D0%BD%D0%BD%D1%8B%D1%85-%D0%BE%D0%BF%D0%B5%D1%80%D0%B0%D1%82%D0%BE%D1%80%D0%BE%D0%B2-%D1%81-%D1%80%D0%B0%D0%B7%D0%B4%D0%B5%D0%BB%D0%B5%D0%BD%D0%B8%D0%B5%D0%BC-%D0%BD%D0%B0-%D0%BE%D0%BF%D0%B8%D1%81%D0%B0%D0%BD%D0%B8%D0%B5-%D0%B8-%D0%B8%D0%BC%D0%BF%D0%BB%D0%B5%D0%BC%D0%B5%D0%BD%D1%82%D0%B0%D1%86%D0%B8%D1%8E )

I don't know how you're holding a line (size its length, or a memorialized memory, it's unclear how to count the end of the sub-construction - that the second argument points to the last symbol of the substrate or for the construction. I think it's the last symbol of the underground.Then it's like,MyString operator ()(int b,int e) { if (b < 0 || e < 0 || b > e || e >= strlen(str)) throw exception("Неверные данные"); char * s = new char[e-b+2]; for(int i = b; i <= e; i++) s[i-b] = str[i]; s[e-b+1] = 0; MyString ret(s); delete[]s; return ret; }

On the grid the properties of the Denominator, you attribute the same value. It causes endless recurrence and StackOverflowException. You should set up a closed (private) field. denominator and attach value to him

We need to replace the phrase. elif color in rainbow == True: on the right elif color in rainbow:♪That's why the code didn't work out of the question, although at first sight it seemed right:rainbow = "red orange yellow green blue indigo violet" color = 'red' Правильно проверять наличие так: print(color in rainbow) # True is и == в контексте булевой проверки взаимозаменяемы print(color in rainbow is True) # False print((color in rainbow) is True) # True I was wondering why this is how it works, and then I found the answer. http://howtucode.com/python39s-in-contains-operator-returns-a-bool-whose-value-is-neither-true-nor-false-113935.html ♪According to him, when we write the expression color in rainbow is Trueit turns into:(color in rainbow) and (rainbow is True) Okay. (rainbow is True) returns False♪ that's the expression ♪ False♪ But there would be no such problem if the brackets were correct: (color in rainbow) is True♪Well, of course, it's best to write right away:color in rainbow

The operator "," to evaluate the first expression and return the second both must be between parentesis ( ) thus:c = ( a >= b, 50 ); What was happening is that it only took the true or false of the first expression, avoiding the second as it was not taking it into account by not being between parentesis.The same question was also answered in English: https://stackoverflow.com/questions/52550/what-does-the-comma-operator-do I hope it serves you!

You have two lump errors in your code:Function without bracketssumaEnteros.mostrarResultado; mostrarResultado It is a function and therefore needs parenthesis:sumaEnteros.mostrarResultado(); Incorrect IncludeNote that in the main you are including the cpp instead of the header:#include "Operaciones.cpp" The expected include is this:#include "Operaciones.h" Since the header is what the compiler needs to match the size of the object and ćomo are to be invoked to the member functions (if they are static, virtual, ...)