C++ Remove punctuation from String

Question

I got a string and I want to remove all the punctuations from it. How do I do that? I did some research and found that people use the ispunct() function (I tried that), but I cant seem to get it to work in my code. Anyone got any ideas?

#include <string>

int main() {

string text = "this. is my string. it's here."

if (ispunct(text))
text.erase();

return 0;
}

Answer 1

i got it.

size_t found = text.find('.');
text.erase(found, 1);

Answer 2

Using algorithm remove_copy_if :-

string text,result;
std::remove_copy_if(text.begin(), text.end(),            
                        std::back_inserter(result), //Store output           
                        std::ptr_fun<int, int>(&std::ispunct)  
                       );

Answer 3

POW already has a good answer if you need the result as a new string. This answer is how to handle it if you want an in-place update.

The first part of the recipe is std::remove_if, which can remove the punctuation efficiently, packing all the non-punctuation as it goes.

std::remove_if (text.begin (), text.end (), ispunct)

Unfortunately, std::remove_if doesn't shrink the string to the new size. It can't because it has no access to the container itself. Therefore, there's junk characters left in the string after the packed result.

To handle this, std::remove_if returns an iterator that indicates the part of the string that's still needed. This can be used with strings erase method, leading to the following idiom...

text.erase (std::remove_if (text.begin (), text.end (), ispunct), text.end ());

I call this an idiom because it's a common technique that works in many situations. Other types than string provide suitable erase methods, and std::remove (and probably some other algorithm library functions I've forgotten for the moment) take this approach of closing the gaps for items they remove, but leaving the container-resizing to the caller.

Answer 4

#include <string>
#include <iostream>
#include <cctype>

int main() {

    std::string text = "this. is my string. it's here.";

    for (int i = 0, len = text.size(); i < len; i++)
    {
        if (ispunct(text[i]))
        {
            text.erase(i--, 1);
            len = text.size();
        }
    }

    std::cout << text;
    return 0;
}

Output

this is my string its here

When you delete a character, the size of the string changes. It has to be updated whenever deletion occurs. And, you deleted the current character, so the next character becomes the current character. If you don't decrement the loop counter, the character next to the punctuation character will not be checked.

Answer 5

ispunct takes a char value not a string.

you can do like

for (auto c : string)
     if (ispunct(c)) text.erase(text.find_first_of(c));

This will work but it is a slow algorithm.

Answer 6

Pretty good answer by Steve314. I would like to add a small change :

text.erase (std::remove_if (text.begin (), text.end (), ::ispunct), text.end ());

Adding the :: before the function ispunct takes care of overloading .

Answer 7

The problem here is that ispunct() takes one argument being a character, while you are trying to send a string. You should loop over the elements of the string and erase each character if it is a punctuation like here:

for(size_t i = 0; i<text.length(); ++i)
  if(ispunct(text[i]))
    text.erase(i--, 1);

Answer 8

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main() {
    string str = "this. is my string. it's here.";

    transform(str.begin(), str.end(), str.begin(), [](char ch)
    {
        if( ispunct(ch) )
            return '\0';
        return ch;
    });
}

Answer 9

Another way you could do this would be as follows:

#include <ctype.h> //needed for ispunct()
string onlyLetters(string str){
    string retStr = "";

    for(int i = 0; i < str.length(); i++){
        if(!ispunct(str[i])){
            retStr += str[i];
        }
    }
    return retStr;

This ends up creating a new string instead of actually erasing the characters from the old string, but it is a little easier to wrap your head around than using some of the more complex built in functions.

Answer 10

 #include <iostream>
 #include <string>

 using namespace std;

 int main()
 {
   string s;//string is defined here.

  cout << "Please enter a string with punctuation's: " << endl;//Asking for users input

  getline(cin, s);//reads in a single string one line at a time

/* ERROR Check: The loop didn't run at first because a semi-colon was placed at the end 
                of the statement.  Remember not to add it for loops. */
        for(auto &c : s)  //loop checks every character 
        {       
             if (ispunct(c)) //to see if its a punctuation
              {
               c=' ';       //if so it replaces it with a blank space.(delete) 
              }

        }

        cout <<  s << endl; 


   system("pause");
   return 0;
   }

Answer 11

I tried to apply @Steve314's answer but couldn't get it to work until I came across this note here on cppreference.com:

Notes

Like all other functions from <cctype>, the behavior of std::ispunct is undefined if the argument's value is neither representable as unsigned char nor equal to EOF. To use these functions safely with plain chars (or signed chars), the argument should first be converted to unsigned char.

By studying the example it provides, I am able to make it work like this:

#include <string>
#include <iostream>
#include <cctype>
#include <algorithm>

int main()
{
    std::string text = "this. is my string. it's here.";
    std::string result;
    text.erase(std::remove_if(text.begin(),
                              text.end(),
                              [](unsigned char c) { return std::ispunct(c); }),
               text.end());
    std::cout << text << std::endl;
}

Answer 12

Try to use this one, it will remove all the punctuation on the string in the text file oky.

str.erase(remove_if(str.begin(), str.end(), ::ispunct), str.end());

please reply if helpful