Trim whitespace from a String

Question

I know there are several ways to do this in Java and C that are nice, but in C++ I can't seem to find a way to easily implement a string trimming function.

This is what I currently have:

string trim(string& str)
{
    size_t first = str.find_first_not_of(' ');
    size_t last = str.find_last_not_of(' ');
    return str.substr(first, (last-first+1));
}

but whenever I try and call

trim(myString);

I get the compiler error

/tmp/ccZZKSEq.o: In function `song::Read(std::basic_ifstream<char, 
std::char_traits<char> >&, std::basic_ifstream<char, std::char_traits<char> >&, char const*, char const*)':
song.cpp:(.text+0x31c): undefined reference to `song::trim(std::string&)'
collect2: error: ld returned 1 exit status

I am trying to find a simple and standard way of trimming leading and trailing whitespace from a string without it taking up 100 lines of code, and I tried using regex, but could not get that to work as well.

I also cannot use Boost.

Answer 1

Your code is fine. What you are seeing is a linker issue.

If you put your code in a single file like this:

#include <iostream>
#include <string>

using namespace std;

string trim(const string& str)
{
    size_t first = str.find_first_not_of(' ');
    if (string::npos == first)
    {
        return str;
    }
    size_t last = str.find_last_not_of(' ');
    return str.substr(first, (last - first + 1));
}

int main() {
    string s = "abc ";
    cout << trim(s);

}

then do g++ test.cc and run a.out, you will see it works.

You should check if the file that contains the trim function is included in the link stage of your compilation process.

Answer 2

Here is how you can do it:

std::string & trim(std::string & str)
{
   return ltrim(rtrim(str));
}

And the supportive functions are implemeted as:

std::string & ltrim(std::string & str)
{
  auto it2 =  std::find_if( str.begin() , str.end() , [](char ch){ return !std::isspace<char>(ch , std::locale::classic() ) ; } );
  str.erase( str.begin() , it2);
  return str;   
}

std::string & rtrim(std::string & str)
{
  auto it1 =  std::find_if( str.rbegin() , str.rend() , [](char ch){ return !std::isspace<char>(ch , std::locale::classic() ) ; } );
  str.erase( it1.base() , str.end() );
  return str;   
}

And once you've all these in place, you can write this as well:

std::string trim_copy(std::string const & str)
{
   auto s = str;
   return ltrim(rtrim(s));
}

Try this

Answer 3

I think that substr() throws an exception if str only contains the whitespace.

I would modify it to the following code:

string trim(string& str)
{
    size_t first = str.find_first_not_of(' ');
    if (first == std::string::npos)
        return "";
    size_t last = str.find_last_not_of(' ');
    return str.substr(first, (last-first+1));
}

Answer 4

Using a regex

#include <regex>
#include <string>

string trim(string s) {
    regex e("^\\s+|\\s+$");   // remove leading and trailing spaces
    return regex_replace(s, e, "");
}

// if you prefer the namespaced version
std::string trim(std::string s) {
    std::regex e("^\\s+|\\s+$"); // remove leading and trailing spaces
    return std::regex_replace(s, e, "");
}

Credit to: https://www.regular-expressions.info/examples.html for the regex

As @o_oTurtle mentioned - regexes are very slow.

An alternate approach, which includes additional whitespace characters:

std::string trim(const std::string& str, const std::string REMOVE = " \n\r\t")
{
    size_t first = str.find_first_not_of(REMOVE);
    if (std::string::npos == first)
    {
        return str;
    }
    size_t last = str.find_last_not_of(REMOVE);
    return str.substr(first, (last - first + 1));
}

Answer 5

#include <vector>
#include <numeric>
#include <sstream>
#include <iterator>

void Trim(std::string& inputString)
{
    std::istringstream stringStream(inputString);
    std::vector<std::string> tokens((std::istream_iterator<std::string>(stringStream)), std::istream_iterator<std::string>());

    inputString = std::accumulate(std::next(tokens.begin()), tokens.end(),
                                 tokens[0], // start with first element
                                 [](std::string a, std::string b) { return a + " " + b; });
}

Answer 6

In addition to answer of @gjha:

inline std::string ltrim_copy(const std::string& str)
{
    auto it = std::find_if(str.cbegin(), str.cend(),
        [](char ch) { return !std::isspace<char>(ch, std::locale::classic()); });
    return std::string(it, str.cend());
}

inline std::string rtrim_copy(const std::string& str)
{
    auto it = std::find_if(str.crbegin(), str.crend(),
        [](char ch) { return !std::isspace<char>(ch, std::locale::classic()); });
    return it == str.crend() ? std::string() : std::string(str.cbegin(), ++it.base());
}

inline std::string trim_copy(const std::string& str)
{
    auto it1 = std::find_if(str.cbegin(), str.cend(),
        [](char ch) { return !std::isspace<char>(ch, std::locale::classic()); });
    if (it1 == str.cend()) {
        return std::string();
    }
    auto it2 = std::find_if(str.crbegin(), str.crend(),
        [](char ch) { return !std::isspace<char>(ch, std::locale::classic()); });
    return it2 == str.crend() ? std::string(it1, str.cend()) : std::string(it1, ++it2.base());
}

Answer 7

A solution with tests

It is somewhat surprising that none of the answers here provide a test function to demonstrate how trim behaves in corner cases. The empty string and strings composed entirely of spaces can both be troublesome.

Here is such a function:

#include <iomanip>
#include <iostream>
#include <string>

void test(std::string const& s)
{
    auto const quote{ '\"' };
    auto const q{ quote + s + quote };
    auto const t{ quote + trim(s) + quote };
    std::streamsize const w{ 6 };
    std::cout << std::left 
        << "s = " << std::setw(w) << q 
        << " : trim(s) = " << std::setw(w) << t 
        << '\n';
}

int main()
{
    for (std::string s : {"", " ", "   ", "a", " a", "a ", " a "})
        test(s);
}

Testing the accepted solution

When I ran this on 2023-Aug-05 against the accepted answer, I was disappointed with how it treated strings composed entirely of spaces. I expected them to be trimmed to the empty string. Instead, they were returned unchanged. The if-statement is the reason why.

// Accepted solution by @Anthony Kong. Copied on 2023-Aug-05.

using namespace std;

string trim(const string& str)
{
    size_t first = str.find_first_not_of(' ');
    if (string::npos == first)
    {
        return str;
    }
    size_t last = str.find_last_not_of(' ');
    return str.substr(first, (last - first + 1));
}

Here is the output from my test:

s = ""     : trim(s) = ""
s = " "    : trim(s) = " "
s = "   "  : trim(s) = "   "
s = "a"    : trim(s) = "a"
s = " a"   : trim(s) = "a"
s = "a "   : trim(s) = "a"
s = " a "  : trim(s) = "a"

Testing the "new and improved" version of trim

To make it work the way I wanted, I changed the if-statement.

While I was at it, I got rid of using namespace std;, and changed the parameter name to s. You know, because I could.

// "New and improved" version of trim. Lol.
std::string trim(std::string const& s)
{
    auto const first{ s.find_first_not_of(' ') };
    if (first == std::string::npos)
        return {};
    auto const last{ s.find_last_not_of(' ') };
    return s.substr(first, (last - first + 1));
}

Now the test routine produces the output I like:

s = ""     : trim(s) = ""
s = " "    : trim(s) = ""
s = "   "  : trim(s) = ""
s = "a"    : trim(s) = "a"
s = " a"   : trim(s) = "a"
s = "a "   : trim(s) = "a"
s = " a "  : trim(s) = "a"