C++ split string to vector<string>

Question

The function I am using is

vector<string> tokenise(string s){

}

Firstly, I intend to split the string into substrings, in which case the string is always arithmetic expression (e.g. "100+5") and there could be some whitespaces.

"100+5" is needed to convert to "100", "+", "5"

After conversion, the substrings will be stored in a vector and return it. I am struggling with the fist step and using the subscript to loop over a string. The type of the value returned is char, so there is no way to put it in the vector.

You can convert a char to a string using std::string str(1, ch); — Ajay Brahmakshatriya
– Ajay Brahmakshatriya, Commented May 3, 2017 at 4:49
If you only use strings with one character, you may as well use char — user7881131
– user7881131, Commented May 3, 2017 at 4:51
@InternetAussie As I asked, I have to use strings, which leaves me no options. — Joey LU
– Joey LU, Commented May 3, 2017 at 4:57
@AjayBrahmakshatriya Can you explain further in actual code? — Joey LU
– Joey LU, Commented May 3, 2017 at 4:59

user1944429 · Accepted Answer · 2017-05-03 06:10:07Z

2

You could just call the string's substring method, after figuring out the range of characters that are either digits, arithmetic characters, or unwanted.

answered May 3, 2017 at 6:10

user1944429

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1 Comment

Joey LU Over a year ago

It doesn't work for me, coz the range of string to distinguish digits and operators is uncertain.

Ajay Brahmakshatriya · Accepted Answer · 2017-05-03 06:29:08Z

1

You mentioned - The type of the value returned is char, so there is no way to put it in the vector.

You have some function that returns a character. You want to then insert the equivalent string into the vector.

Assuming your vector is defined as

std::vector<std::string> broken_strings;

So you can do it as follows.

char ch = ...; // Here comes the character that you get from the function.
std::string str(1, ch);
broken_strings.push_back(str);

Then you can return broken_strings.

Edit:

OP mentions that he wants to tokenize algebraic expressions. So it will have to be done it a different way.

Following is a simple approach.

 std::vector<std::string> broken;
 std::string temp;
 for ( int i = 0; i<s.length() ;i++){
     char ch = s[i];
     if (ch == ' ')
         continue;
     else if (ch >= '0' && ch <='9')
         temp += ch;
     else{
         if (temp.length() != 0)
             broken.push_back(temp);
         temp = "";
         temp += ch;
         broken.push_back(temp);
         temp = "";
     }
 } 
 if (temp.length() != 0)
     broken.push_back(token);
 return broken;

You can see the demo of the same here

Ideone

edited May 3, 2017 at 6:29

answered May 3, 2017 at 5:11

Ajay Brahmakshatriya

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7 Comments

n.caillou Over a year ago

That, or with string::substr

Ajay Brahmakshatriya Over a year ago

You have completely changed your question. Now you need some logic to identify where a token ends. You might want to look at lexing tools like lex or flex.

Joey LU Over a year ago

sorry for that, is there any easy method to solve this.

Ajay Brahmakshatriya Over a year ago

Assuming you have set of characters that can be the operators, it can be done. I will edit my answer to do the same.

Joey LU Over a year ago

Where do you declare the "s"

|

sameerkn · Accepted Answer · 2017-05-03 06:57:19Z

0

vector<string> tokenise(string s)
{
    vector<string> v;

    string number;

    for(int i = 0; i < s.length(); ++i)
    {
        if((s[i] >= '0') && (s[i] <= '9'))
        {
            number += string(1, s[i]);
        }
        else if(s[i] == '.')
        {
            number += string(1, s[i]);
        }
        else if((s[i] == '+') || (s[i] == '-') || (s[i] == '*') || (s[i] == '/'))
        {
            if(number.size())
            {
                v.push_back(number);
                number.clear();
            }
            v.push_back(string(1, c));
        }
    }

    if(number.size())
    {
        v.push_back(number);
        number.clear();
    }

    return v;
}

answered May 3, 2017 at 6:57

sameerkn

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1 Comment

Joey LU Over a year ago

I like your solution, almost the same as my idea

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