remove characters from a string

[robert pond]

I have a string variable.

string mystring = "str?in.g!";

How can i remove the punctuation (? . !) from the
string and keep the rest of the characters?

[Mike Wahler]

robert pond <rp...@vt.edu> wrote in message
news:8lq38n$cl0$1...@solaris.cc.vt.edu...

Define a 'filter' as a predicate for std::remove_if() :

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

bool pfilter(std::string::value_type v)
{
std::string punct = "?!.";
std::string::iterator it = punct.begin();

while(it != punct.end())
if(v == *it++)
return true;

return false;
}

int main()
{
std::string s;

while(true)
{
std::cout << "Enter a string <Enter to quit>: ";
std::getline(std::cin, s);

if(s.empty())
break;

s.erase(std::remove_if(s.begin(), s.end(), pfilter),
s.end());

std::cout << "with punctuation removed: "
<< s << std::endl << std::endl;
}

return 0;
}

Test run:

Enter a string <Enter to quit>: str?in.g!
with punctuation removed: string

Enter a string <Enter to quit>: ?s.tri!ng
with punctuation removed: string

Enter a string <Enter to quit>: !s.?tri?ng!
with punctuation removed: string

Enter a string <Enter to quit>: ??str.ing
with punctuation removed: string

Enter a string <Enter to quit>: !?..??
with punctuation removed:

Enter a string <Enter to quit>: string
with punctuation removed: string

Enter a string <Enter to quit>:

[back to OS]

Reference: Josuttis, "The C++ Standard Library
-- A Tutorial and Reference"
(Addison Wesley Longman, 1999)

A very, very, very good book, imo. :-)

HTH,
-Mike

[John Harrison]

"robert pond" <rp...@vt.edu> wrote in message
news:8lq38n$cl0$1...@solaris.cc.vt.edu...
> I have a string variable.
>
> string mystring = "str?in.g!";
>
> How can i remove the punctuation (? . !) from the
> string and keep the rest of the characters?
>

A little for loop calling ispunct wouldn't be too difficult but if you
prefer here's the one line obscure STL solution

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

int main()
{
string str = "str?in.g!";
str.erase(remove_if(str.begin(), str.end(), ispunct), str.end());
cout << str << endl;
return 0;
}

john

[robert pond]

How can I filter out " " quotation marks though, by putting a \ in front of
them?

> > Define a 'filter' as a predicat
e for std::remove_if() :
>

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

[Homer Meyer]

"Mike Wahler" <mkwa...@mkwahler.net> wrote in message
news:8lq7ql$q85$1...@slb0.atl.mindspring.net...

> robert pond <rp...@vt.edu> wrote in message
> news:8lq38n$cl0$1...@solaris.cc.vt.edu...
> > I have a string variable.
> >
> > string mystring = "str?in.g!";
> >
> > How can i remove the punctuation (? . !) from the
> > string and keep the rest of the characters?
>

> Define a 'filter' as a predicate for std::remove_if() :

You don't have to work that hard.

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

std::string s = "str?in.g!";
s.erase(std::remove_if(s.begin(),s.end(),std::ispunct),s.end());

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[Mike Wahler]

robert pond <rp...@vt.edu> wrote in message

news:8lqage$fta$1...@solaris.cc.vt.edu...

> How can I filter out " " quotation marks though, by putting a \ in front
of
> them?

Have you tried it?

But yes, that's how you express an embedded quote
in a string literal. This should be in your C++ text.

std::string punct = "?.!\"";

-Mike

[robert pond]

"Homer Meyer" <ho...@cqg.com> wrote in message
news:so1cqt...@news.supernews.com...

>
>
> You don't have to work that hard.
>
> #include <cctype>
> #include <string>
> #include <algorithm>
>
> std::string s = "str?in.g!";
> s.erase(std::remove_if(s.begin(),s.end(),std::ispunct),s.end());
>
> --

Actually I did, I couldnt use ispunct because I had to keep hyphens and
apostrophes in the string.

Thanks for all the help guys.