c ++ problem with std :: vector

You do not enter characters, as you think, but numbers. So if you type something different from an integer, then, respectively ...

 vector<char> str; char a; while(cin >> a) str.push_back(a); for(vector<char>::const_iterator i = str.begin(); i != str.end(); ++i) cout << *i; 

This is how it will work, only the input you need to complete by pressing Ctrl-Z on Windows (on Linux Ctrl-D, it seems ... I don’t remember exactly) - since you check the state of the cin stream.

But why don't you work with string strings?

To get the expected result, you have to enter by a separate digit, such as 1 2 3 4 5 6 7 8 9 0, and convert these integer values ​​into character codes of the numbers, as shown below in the demo program

 #include <iostream> #include <vector> int main() { std::vector<char> str; int a; while ( std::cin >> a ) str.push_back( a + '0' ); for ( char c : str ) std::cout << c; std::cout << std::endl; return 0; } 

If you enter not a single digit, e some number, the result will no longer be predictable.

There is a standard function std::to_string , which translates a number into an object of type std::string .

For example, you could write

 #include <iostream> #include <vector> #include <string> int main() { std::vector<char> str; int x; while (std::cin >> x) { std::string tmp = std::to_string(x); if (!str.empty()) str.push_back(' '); str.insert(str.end(), tmp.begin(), tmp.end()); } for (char c : str) std::cout << c; std::cout << std::endl; } 

 int a; 

 char a; 

 ... str.begin(); i != str.end(); ++i 

 ... str.cbegin(); i != str.cend(); ++i // ^__________________^--------------- константный