I have a problem, the code is easy kompliiruetsya in exe, and the first two inputs work perfectly, but when it comes time for the third variable program starts to not work with the code all odinkaovo
https://www.onlinegdb.com/online_c++_compiler – you can then quickly skompliirovat.

# include & lt; iostream & gt;
INT MAIN ()
{
  STD :: COUT & LT; & LT; "Enter a:"; // ask the user to enter any number
  int a = 0;
  std :: cin & gt; & gt; a; // get the number of the user and store it in a variable
  STD :: COUT & LT; & LT; "Enter b:"; // ask the user to enter any number
  char b = 0;
  std :: cin & gt; & gt; b; // get the number of the user and stores it in the variable b
  STD :: COUT & LT; & LT; "Enter c:"; // ask the user to enter any number
  int c = 0;
  std :: cin & gt; & gt; c; // get the number of the user and store it in a variable
  STD :: COUT & LT; & LT; "Finish" & lt; & lt; STD :: ENDL;
  Return 0;
}

https://prnt.sc/rlb818

The first two variables are correct this is what happens on the

third

Please tell me how to do so that I can introduce a third variable I have this c. If you have any questions to ask.

[23.03.2020 19:22]
Corrected: Changed the names of variables, for greater ease and understanding.

Answer 1, Authority 100%

Well, what do you want?

char dll = 0;
std :: cin & gt; & gt; dll;

Once a char read – Reads one character (you are not surprised concluded D in the picture?). But everything else – LL.dll – comes from the buffer when reading

std :: cin & gt; & gt; lvls;

It is understood that this is not a number – that’s actually in the lvls nothing to read, to cin is set error flag fail (you do not check …).

It’s simple.

Read the line – in line:)