How to quickly deploy zsh on an ssh server without root rights?

Question

C++ / C#

Maybe there are already ready-made scripts, since wget pulled it from gist.github.com and you're done?
I really like zsh.

Answer 1

This is very old code, way before 1998 when C++ became the standard.
Now (but not in the know - in C ++ 98 or C ++ 03), a huge number of C ++ headers have been combined into the standard template library (STL). So the code needs to be slightly modified.
1. Instead <iostream.h>of using <iostream>. <stdio.h>, in principle, works, but it is recommended to take <cstdio>.
2. All functions are in the namespace std. That is: std::cout, std::endl, etc. Or, as suggested by D' Normalization , using namespace std;.

Answer 2

1. What is wrong
2. Includes (#include) are empty
3. After the includes should be using namespace std; Or just sign cout and cin with std:: (std::cout, std::cin)

Answer 3

Toaster, what's wrong with you? Stop breaking. Formatted and *working* code (the toaster won't suggest editing):

// Программа для преобразования
// градусов Цельсия в градусы Фаренгейта:
// Fahrenheit = NCelsius * (212 - 32)/100 + 32
//
#include <stdio.h>
#include <iostream>

using namespace std;

int main(int nNumberofArgs, char* pszArgs[])
{
  // Введите температуру в градусах Цельсия
  int nNCelsius;
  cout << "Введите температуру по Цельсию: ";
  cin >> nNCelsius;
  // для приведенной формулы преобразования
  //вычислим преобразующий множитель
  int nNFactor;
  nNFactor = 212 - 32;
  // используем вычисленный множитель для
  // преобразования градусов Цельсия в
  // градусы Фаренгейта
  int nFahrenheit;
  nFahrenheit = nNFactor * nNCelsius/100 + 32;
  // вывод результатов
  cout << "Температура по Фаренгейту: ";
  cout << nFahrenheit;
  return 0;
}

$ g++ 1.cpp
$ ./a.out
Enter Celsius temperature: 22
Fahrenheit temperature: 71

Answer 4

only if you can compile it yourself in a local directory and in .bashrc register the launch of zsh, but this is somehow too crutch .. it’s better to get the rights and put it normally.