How to remove certain characters from a string in C++?

Question

For example I have a user input a phone number.

cout << "Enter phone number: ";
INPUT: (555) 555-5555
cin >> phone;

I want to remove the "(", ")", and "-" characters from the string. I've looked at the string remove, find and replace functions however I only see that they operate based on position.

Is there a string function that I can use to pass a character, "(" for example, and have it remove all instances within a string?

Answer 1

   string str("(555) 555-5555");

   char chars[] = "()-";

   for (unsigned int i = 0; i < strlen(chars); ++i)
   {
      // you need include <algorithm> to use general algorithms like std::remove()
      str.erase (std::remove(str.begin(), str.end(), chars[i]), str.end());
   }

   // output: 555 5555555
   cout << str << endl;

To use as function:

void removeCharsFromString( string &str, char* charsToRemove ) {
   for ( unsigned int i = 0; i < strlen(charsToRemove); ++i ) {
      str.erase( remove(str.begin(), str.end(), charsToRemove[i]), str.end() );
   }
}
//example of usage:
removeCharsFromString( str, "()-" );

Answer 2

I want to remove the "(", ")", and "-" characters from the string.

You can use the std::remove_if() algorithm to remove only the characters you specify:

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

bool IsParenthesesOrDash(char c)
{
    switch(c)
    {
    case '(':
    case ')':
    case '-':
        return true;
    default:
        return false;
    }
}

int main()
{
    std::string str("(555) 555-5555");
    str.erase(std::remove_if(str.begin(), str.end(), &IsParenthesesOrDash), str.end());
    std::cout << str << std::endl; // Expected output: 555 5555555
}

The std::remove_if() algorithm requires something called a predicate, which can be a function pointer like the snippet above.

You can also pass a function object (an object that overloads the function call () operator). This allows us to create an even more general solution:

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

class IsChars
{
public:
    IsChars(const char* charsToRemove) : chars(charsToRemove) {};

    bool operator()(char c)
    {
        for(const char* testChar = chars; *testChar != 0; ++testChar)
        {
            if(*testChar == c) { return true; }
        }
        return false;
    }

private:
    const char* chars;
};

int main()
{
    std::string str("(555) 555-5555");
    str.erase(std::remove_if(str.begin(), str.end(), IsChars("()- ")), str.end());
    std::cout << str << std::endl; // Expected output: 5555555555
}

You can specify what characters to remove with the "()- " string. In the example above I added a space so that spaces are removed as well as parentheses and dashes.

Answer 3

remove_if() has already been mentioned. But, with C++0x, you can specify the predicate for it with a lambda instead.

Below is an example of that with 3 different ways of doing the filtering. "copy" versions of the functions are included too for cases when you're working with a const or don't want to modify the original.

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

string& remove_chars(string& s, const string& chars) {
    s.erase(remove_if(s.begin(), s.end(), [&chars](const char& c) {
        return chars.find(c) != string::npos;
    }), s.end());
    return s;
}
string remove_chars_copy(string s, const string& chars) {
    return remove_chars(s, chars);
}

string& remove_nondigit(string& s) {
    s.erase(remove_if(s.begin(), s.end(), [](const char& c) {
        return !isdigit(c);
    }), s.end());
    return s;
}
string remove_nondigit_copy(string s) {
    return remove_nondigit(s);
}

string& remove_chars_if_not(string& s, const string& allowed) {
    s.erase(remove_if(s.begin(), s.end(), [&allowed](const char& c) {
        return allowed.find(c) == string::npos;
    }), s.end());
    return s;
}
string remove_chars_if_not_copy(string s, const string& allowed) {
    return remove_chars_if_not(s, allowed);
}

int main() {
    const string test1("(555) 555-5555");
    string test2(test1);
    string test3(test1);
    string test4(test1);
    cout << remove_chars_copy(test1, "()- ") << endl;
    cout << remove_chars(test2, "()- ") << endl;
    cout << remove_nondigit_copy(test1) << endl;
    cout << remove_nondigit(test3) << endl;
    cout << remove_chars_if_not_copy(test1, "0123456789") << endl;
    cout << remove_chars_if_not(test4, "0123456789") << endl;
}

Answer 4

Here is a different solution for anyone interested. It uses the new For range in c++11

string str("(555) 555-5555");
string str2="";

for (const auto c: str){

    if(!ispunct(c)){

        str2.push_back(c);
    }
}

str = str2;
//output: 555 5555555
cout<<str<<endl;

Answer 5

I'm afraid there is no such a member for std::string, but you can easily program that kind of functions. It may not be the fastest solution but this would suffice:

std::string RemoveChars(const std::string& source, const std::string& chars) {
   std::string result="";
   for (unsigned int i=0; i<source.length(); i++) {
      bool foundany=false;
      for (unsigned int j=0; j<chars.length() && !foundany; j++) {
         foundany=(source[i]==chars[j]);
      }
      if (!foundany) {
         result+=source[i];
      }
   }
   return result;
}

EDIT: Reading the answer below, I understood it to be more general, not only to detect digit. The above solution will omit every character passed in the second argument string. For example:

std::string result=RemoveChars("(999)99-8765-43.87", "()-");

Will result in

99999876543.87

Answer 6

boost::is_any_of

Strip for all characters from one string that appear in another given string:

#include <cassert>

#include <boost/range/algorithm/remove_if.hpp>
#include <boost/algorithm/string/classification.hpp>

int main() {
    std::string str = "a_bc0_d";
    str.erase(boost::remove_if(str, boost::is_any_of("_0")), str.end());
    assert((str == "abcd"));
}

Tested in Ubuntu 16.04, Boost 1.58.

Answer 7

using namespace std;


// c++03
string s = "(555) 555-5555";
s.erase(remove_if(s.begin(), s.end(), not1(ptr_fun(::isdigit))), s.end());

// c++11
s.erase(remove_if(s.begin(), s.end(), ptr_fun(::ispunct)), s.end());

Note: It's posible you need write ptr_fun<int, int> rather than simple ptr_fun

Answer 8

Yes, you can use the isdigit() function to check for a digits :)

Here you go:

#include <iostream>
#include <cctype>
#include <string.h>

using namespace std;

int main(){

  char *str = "(555) 555-5555";
  int len = strlen(str);

  for (int i=0; i<len; i++){
      if (isdigit(*(str+i))){
        cout << *(str+i);
      }
  }

  cout << endl;


return 0;   
}

Hope it helps :)

Answer 9

If you have access to a compiler that supports variadic templates, you can use this:

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

template<char ... CharacterList>
inline bool check_characters(char c) {
    char match_characters[sizeof...(CharacterList)] = { CharacterList... };
    for(int i = 0; i < sizeof...(CharacterList); ++i) {
        if(c == match_characters[i]) {
            return true;
        }
    }
    return false;
}

template<char ... CharacterList>
inline void strip_characters(std::string & str) {
    str.erase(std::remove_if(str.begin(), str.end(), &check_characters<CharacterList...>), str.end());
}

int main()
{
    std::string str("(555) 555-5555");
    strip_characters< '(',')','-' >(str);
    std::cout << str << std::endl;
}

Answer 10

I'm new, but some of the answers above are insanely complicated, so here's an alternative.

NOTE: As long as 0-9 are contiguous (which they should be according to the standard), this should filter out all other characters but numbers and ' '. Knowing 0-9 should be contiguous and a char is really an int, we can do the below.

EDIT: I didn't notice the poster wanted spaces too, so I altered it...

#include <cstdio>
#include <cstring>

void numfilter(char * buff, const char * string)
{
  do
  { // According to standard, 0-9 should be contiguous in system int value.
    if ( (*string >= '0' && *string <= '9') || *string == ' ')
      *buff++ = *string;
  } while ( *++string );
  *buff++ = '\0'; // Null terminate
}

int main()
{
  const char *string = "(555) 555-5555";
  char buff[ strlen(string) + 1 ];

  numfilter(buff, string);
  printf("%s\n", buff);

return 0;
}

Below is to filter supplied characters.

#include <cstdio>
#include <cstring>

void cfilter(char * buff, const char * string, const char * toks)
{
  const char * tmp;  // So we can keep toks pointer addr.
  do
  {
    tmp = toks;
    *buff++ = *string; // Assume it's correct and place it.
    do                 // I can't think of a faster way.
    {
      if (*string == *tmp)
      {
        buff--;  // Not correct, pull back and move on.
        break;
      }
    }while (*++tmp);
  }while (*++string);

  *buff++ = '\0';  // Null terminate
}

int main()
{
  char * string = "(555) 555-5555";
  char * toks = "()-";
  char buff[ strlen(string) + 1 ];

  cfilter(buff, string, toks);
  printf("%s\n", buff);

  return 0;
}

Answer 11

For those of you that prefer a more concise, easier to read lambda coding style...

This example removes all non-alphanumeric and white space characters from a wide string. You can mix it up with any of the other ctype.h helper functions to remove complex-looking character-based tests.

(I'm not sure how these functions would handle CJK languages, so walk softly there.)

    // Boring C loops: 'for(int i=0;i<str.size();i++)' 
    // Boring C++ eqivalent: 'for(iterator iter=c.begin; iter != c.end; ++iter)'

See if you don't find this easier to understand than noisy C/C++ for/iterator loops:

TSTRING label = _T("1.   Replen & Move  RPMV");
TSTRING newLabel = label;
set<TCHAR> badChars; // Use ispunct, isalpha, isdigit, et.al. (lambda version, with capture list parameter(s) example; handiest thing since sliced bread)
for_each(label.begin(), label.end(), [&badChars](TCHAR n){
    if (!isalpha(n) && !isdigit(n))
        badChars.insert(n);
});

for_each(badChars.begin(), badChars.end(), [&newLabel](TCHAR n){
    newLabel.erase(std::remove(newLabel.begin(), newLabel.end(), n), newLabel.end());
});

newLabel results after running this code: "1ReplenMoveRPMV"

This is just academic, since it would clearly be more precise, concise and efficient to combine the 'if' logic from lambda0 (first for_each) into the single lambda1 (second for_each), if you have already established which characters are the "badChars".

Answer 12

Using std::wstring and wchar_t (requires the Unicode header):

//#include <tchar.h>
std::wstring phone(L"(555) 555-5555");

...fancy static range initializer next; not necessary to setup badChars2 this exact same way. It's overkill; more academic than anything else:

const wchar_t *tmp = L"()-"; 
const std::set<wchar_t> badChars2(tmp,tmp + sizeof(tmp)-1);

Simple, concise lambda:

1. Uses phone in the lambda capture list.
2. Uses Erase-remove idiom

3. Removes all bad characters from phone

   for_each(badChars2.begin(), badChars2.end(), [&phone](wchar_t n){
        phone.erase(std::remove(phone.begin(), phone.end(), n), phone.end());
   });
   wcout << phone << endl;
   

Output: "555 5555555"

Answer 13

Here's yet another alternative:

template<typename T>
void Remove( std::basic_string<T> & Str, const T * CharsToRemove )
{
    std::basic_string<T>::size_type pos = 0;
    while (( pos = Str.find_first_of( CharsToRemove, pos )) != std::basic_string<T>::npos )
    {
        Str.erase( pos, 1 ); 
    }
}

std::string a ("(555) 555-5555");
Remove( a, "()-");

Works with std::string and std::wstring

Answer 14

Lot of good answers, here is another way to clean up a string of numbers, is not deleting chars but by moving the numbers out.

string str("(555) 555-5555"), clean;
for (char c : str)
    if (c >= 48 and c <= 57)
        clean.push_back(c);

Answer 15

From C++20, you can use erase/erase_if for std::basic_string, which is basically a convenience wrapper for the erase-remove idiom

std::erase(phone, '('); 

and

std::erase_if(phone, [](char x) { 
                        return x == '(' or x == ')' or x == '-'; 
                     });

Note that these functions return the count of how many characters were erased as well.