Lexical Analyzer Project - Vector not outputting correctly - c++

Question

I have the following code which is part of a larger project. What this code is supposed to do is go through the line character by character looking for "tokens." The token I am looking for in this code is an ID. Which is defined as a letter followed by zero or more numbers or letters.
When a letter is detected it goes into the inner loop and loops through the next few characters, adding each character or letter to the idstring, until it finds the end of ID character(which is defined in the code) and then adds that idstring to a vector. At the end of the line it should output each element of the vector. Im not getting the output I need. I hope this is enough information to understand what is going on in the code. If someone could help me fix this problem I would be very great full. Thank you!
The output I need: ab : ab
What I get: a : a
#include <iostream>
#include <regex>
#include <string>
#include <vector>
int main()
{
std::vector<std::string> id;
std::regex idstart("[a-zA-Z]");
std::regex endID("[^a-z]|[^A-Z]|[^0-9]");
std::string line = "ab ab";
//Loops character by character through the line
//Adding each recognized token to the appropriate vector
for ( int i = 0; i<line.length(); i++ )
{
std::string tempstring(1,line[i]);
//Character is letter
if ( std::regex_match(tempstring,idstart) )
{
std::string tempIDString = tempstring;
int lineInc = 0;
for ( int j = i + 1; j<line.length(); j++)
{
std::string tempstring2(1,line[j]);
//Checks next character for end of potential ID
if ( std::regex_match(tempstring2,endID) )
{
i+=lineInc+1;
break;
}
else
{
tempIDString+=tempstring2;
lineInc++;
}
}
id.push_back(tempIDString);
}
}
std::cout << id.at(0) << " : " << id[1] << std::endl;
return 0;
}

Answer 1

The question is 2.5 year old and now you will maybe laugh seeing it. You break; the inner for when finding the second charcter that matches and so you will never assign tempstring2 to tempstring1.
But let's forget about that code. There is no good design here.
You had a good idea to use std::regex, but you did not know, how it worked.
So lets have a look at the correct implementation:
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <regex>
// Our test data (raw string). So, containing also \n and so on
std::string testData(
R"#( :-) IDcorrect1 _wrongID I2DCorrect
3FALSE lowercasecorrect Underscore_not_allowed
i3DCorrect,i4 :-)
}
)#");
std::regex re("(\\b[a-zA-Z][a-zA-Z0-9]*\\b)");
int main(void)
{
// Define the variable id as vector of string and use the range constructor to read the test data and tokenize it
std::vector<std::string> id{ std::sregex_token_iterator(testData.begin(), testData.end(), re, 1), std::sregex_token_iterator() };
// For debug output. Print complete vector to std::cout
std::copy(id.begin(), id.end(), std::ostream_iterator<std::string>(std::cout, "\n"));
return 0;
}
This does all the work in the definition of the variable and by calling the range constructor. So, a typical one-liner.
Hope somebody can learn from this code . . .