mobile application and games

Qt framework is quit good because it’s cross platform

hmm how can i drop my code?

hmm how can i drop my code?

I have a template class Type Sort(Type pt, int n); How can I check that the pt variable accepts a certain data type, for example, char?

I have a template class Type Sort(Type pt, int n); How can I check that the pt variable accepts a certain data type, for example, char?

Do you want to have a separate specialization for specific type? Or you want to prohibit of using other types?

string proverkaper = ""; proverkaper = typeid(pt).name(); if (proverkaper == "int *" || proverkaper == "double *") I was able to determine the data type this way.

I can't figure out how to return an array from a template

Error C2440 return: unable to convert "Type*" to "Type" There is such an error If I do return *pt; Only the first element of the array is returned

string proverkaper = ""; proverkaper = typeid(pt).name(); if (proverkaper == "int *" || proverkaper == "double *") I was able to determine the data type this way.

Can you show your template declaration?

template <class Type> Type Sort(Type *pt, int n) {} int main() { double* arr_double = new double[count]; int* ar_int_1 = new int[count]; char* ar_char_2 = new char[count]; if (fin) { for (int i = 0; i != count; i++) { fin >> arr_double[i]; ar_int_1[i] = (int)arr_double[i]; } } Sort(ar_char_2, count); }

What should Sort return? It doesn't seem that you use the returned value

Type Sort(Type *pt, int n) { string proverkaper = ""; proverkaper = typeid(pt).name(); if (proverkaper == "int *" || proverkaper == "double *") { for (int startIndex = 0; startIndex < n - 1; ++startIndex) { int smallestIndex = startIndex; for (int currentIndex = startIndex + 1; currentIndex < n; ++currentIndex) { if (pt[currentIndex] < pt[smallestIndex]) smallestIndex = currentIndex; } std::swap(pt[startIndex], pt[smallestIndex]); } } else { cout << "Simvoli nelazya razmestit po vosrastaniu esli tolko ne preobrazovat ih"; } return *pt; }

string proverkaper = ""; proverkaper = typeid(pt).name(); if (proverkaper == "int *" || proverkaper == "double *") I was able to determine the data type this way.

What should happen if some other type has been passed? Do you want this code to still compile but return runtime error?

sorry return *pt;

If the program receives a char array. There should be a message that it cannot be sorted

Why don't you want to make it a compile error? Since the type is always known at compile time you can save time to the person who tries to use your function in a wrong way

You can make Type* as return type, but I would rather make it void and not return anything, since you returning the array passed as the argument anyway

Yes, I would do that, but the task is)

Ok, there's std::is_same_v<Type, int*>, that I think will return true only if the type is the same

template <class Type> Type Sort(Type *pt, int n) {} int main() { double* arr_double = new double[count]; int* ar_int_1 = new int[count]; char* ar_char_2 = new char[count]; if (fin) { for (int i = 0; i != count; i++) { fin >> arr_double[i]; ar_int_1[i] = (int)arr_double[i]; } } Sort(ar_char_2, count); }

Is this c programming?

Yes, I would do that, but the task is)

So I need to return the array after sorting. I receive it, sort it, return it

с++)

Oh, can i ask what was the purpose of putting '*' in the double type?

Then change the return type from Type to Type*

Oh, can i ask what was the purpose of putting '*' in the double type?

double* arr_double = new double[count]; int* ar_int_1 = new int[count]; char* ar_char_2 = new char[count]; if (fin) { for (int i = 0; i != count; i++) { fin >> arr_double[i]; ar_int_1[i] = (int)arr_double[i]; } } fin.close();

Yea what was the purpose of '*'?

Yea what was the purpose of '*'?

Im new in c++ so i dont know what is it

Im new in c++ so i dont know what is it

im change, but it return only the first element of the array is returned

If you return pointer and the pointer is to the first element it is the same as returning the array

And how do I get it back completely? I need to write it to a file

Since cpp2x modules are in experimental state and compilers don’t support it completely, is it safe to write my small project on it? “Because, it’s been a long while I’m waiting for it and what if I prepare my codebase for it?”

Writing your own modules is well supported by all major compilers. It is only when it comes to using the std module that major compilers are found lacking. If you write and use your own modules and continue using the standard headers then you are safe for now. But most of the people and organisations still use C++14 only with very few having moved to C++17. So if you intend your library to be used by a majority of folks, then using C++20 features is not a good idea. It would be another 2 or 3 years before major organizations even think of adopting C++20

Hello everyone Is there anything similar to getch() that used for string instead of char!?

std::getline() maybe?

Since cpp2x modules are in experimental state and compilers don’t support it completely, is it safe to write my small project on it? “Because, it’s been a long while I’m waiting for it and what if I prepare my codebase for it?”

I want something similar to getch() Because getch doesn't show the character And I want to do the same thing on string or for a line

Hi guys, so I was studying my Professor's notes, and, at some point, I found the type "DWORD", which should stand for "double word". Ok, but after I found another type: "DWORD WINAPI", and I don't know what to do with it. I don't even know if it's a real thing or simply a mistake by the Professor. Could you please explain me what it is? Thank you!

anyone can give me a document or video about C++11 lambda?

#include <bits/stdc++.h> using namespace std; int max(int a, int b) { return (a > b)? a : b; } int knapSack(int W, int wt[], int val[], int n) { if (n == 0 || W == 0) return 0; if (wt[n-1] > W) return knapSack(W, wt, val, n-1); else return max( val[n-1] + knapSack(W-wt[n-1], wt, val, n-1), knapSack(W, wt, val, n-1) ); } int main() { int val[] = {60, 100, 120}; int wt[] = {10, 20, 30}; int W = 50; int n = sizeof(val)/sizeof(val[0]); cout<<knapSack(W, wt, val, n); return 0; }

I need this source code but I do not understand two things The first is what int n does and the next question is what line else return max (val [n-1] + knapSack (W-wt [n-1], wt, val, n-1), knapSack (W, wt, val, n-1)); What does it do? This program is related to backpack problem solving 0.1

Is your English good enough to understand what sizeof means?

I need this source code but I do not understand two things The first is what int n does and the next question is what line else return max (val [n-1] + knapSack (W-wt [n-1], wt, val, n-1), knapSack (W, wt, val, n-1)); What does it do? This program is related to backpack problem solving 0.1

can anyone tell me why this outputs 0? double x,a,b; a=50,b=60; x= 11/100*a; cout << x;

can anyone tell me why this outputs 0? double x,a,b; a=50,b=60; x= 11/100*a; cout << x;

11 / 100 = 0 0 * 50 = 0

double x,a,b; a=50,b=60; double n1,n2; n1 = 101; n2 = 100; x= n1/n2*a; cout << x ;

Why is 11/100 not 0.11

Because 11 and 100 have type “int” by default. (int) / (int) = (int) If you want to get result type as double you have to cast one or both of operands to “double”: (double) / (int) = (double) (int) / (double) = (double) (double) / (double) = (double) So double x,a,b; a=50,b=60; x= (double)11/100*a; cout << x;

can anyone tell me why this outputs 0? double x,a,b; a=50,b=60; x= 11/100*a; cout << x;

double x,a,b; a=50,b=60; x = 11.0 / 100.0 * a; cout << x; You either have to cast it with (double) or add .0 to it to tell the compiler it is a double or float rather than int.

Hello, My code cannot read input because of getline() but number of "pair" words is unknown, any suggestions on how to fix it? // test #include <bits/stdc++.h> using namespace std; int main () { int n; cin >> n; // no. of "int" keywords cout << n << "\n"; string s; getline(cin, s); cout << s << "\n"; return 0; } /* in: 3 pair pair int int int */