avoid empty token in cpp

Question

I have a string:

s = "server ('m1.labs.terada')ta.com') username ('user5') password('user)5') dbname ('default')";

I am extracting the argument names: such as server, username..., dbname.

To do this I am using the following regex:

regex re("\\(\'[!-~]+\'\\)");
sregex_token_iterator i(s.begin(), s.end(), re, -1);
sregex_token_iterator j;
unsigned count = 0;
while(i != j)
{
    string str1 = *i;
    cout <<"token = "<<str1<< endl;
    i++;
    count++;
}
cout << "There were " << count << " tokens found." << endl

For this the output I am getting is :

token = server 
token =  username 
token =  password
token =  dbname 
token =  
There were 5 tokens found.

How shall I avoid the 5th token formed. Am I missing out on anything?

Are you sure you provided the same input in the question as you are using for test? [I get 4 tokens as a result.](https://ideone.com/NPf1BD) There must be some trailing whitespace in the string, [like here](https://ideone.com/vRMHVc). — Wiktor Stribiżew, Jun 20 '17 at 07:50
Yes. I am sure. Is there any way I can avoid the 5th token formed.? — hydra123, Jun 20 '17 at 08:03
I tested your code, @WiktorStribiżew, still I am getting the 5th token as a space — hydra123, Jun 20 '17 at 08:05
I suspect the tokens you need to extract are composed of ASCII letters, digits or `_`. Try using `regex re("\\W*\\(\'[!-~]+\'\\)\\W*");`. If not, replace `\W*` with `\s*`: `re("\\s*\\(\'[!-~]+\'\\)\\s*");` to check if there are just whitespaces at the end. — Wiktor Stribiżew, Jun 20 '17 at 08:09

score 1 · Accepted Answer · answered Jun 20 '17 at 11:37

It turns out there are some non-word chars at the end of the string. You may match them with \W* (zero or more non-word chars). Since your tokens are only composed of word chars, you may safely wrap your pattern with \W* pattern:

regex re("\\W*\\(\'[!-~]+\'\\)\\W*");

See the C++ online demo

Result:

token = server
token = username
token = password
token = dbname
There were 4 tokens found.

avoid empty token in cpp

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