Truncating numbers at end of string

I need to check whether a string of letters contains any numbers at
the end of the string, and if so, remove all numbers. There won't ever
be numbers at the beginning or middle of the string. Any ideas on how
to go about this? It's a strange requirement for a program but i can't
get around it. Thanks.

re: Truncating numbers at end of string

Alex Leach wrote:Simple (and slow):
While String has characters and the last characer is a number
Erase the last character from the string

A bit more complicated (but faster):
Find the last character in the string that is not a number
If found,
Erase all characters after the found one

That's pseudo-code that you can simply convert to C++ if you know
simple things like 'while', 'isdigit', 'std::string' and its members..

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

re: Truncating numbers at end of string

On Feb 7, 3:42 pm, "Alex Leach" <ajle...@gmail.comwrote:
=cat trunc.cxx
#include <iostream>
#include <string>
#include <algorithm>
#include <cctype>

using namespace std;

bool ifNumber ( char c )
{
if (isdigit(c))
return true;
return false;
}

int main()
{
string s = "pankaj123";
cout << "Before erase: " << s << endl;
s.erase(find_if(s.begin(), s.end(), ifNumber), s.end());
cout << "After erase: " << s << endl;
return 0;
}

=g++ trunc.cxx
=./a.out
Before erase: pankaj123
After erase: pankaj

/P

re: Truncating numbers at end of string

Alex Leach wrote:
You could find the last non-digit in the string and cut it off after that
position. There is a function There is a member function
find_last_not_of() that might help.


Best

Kai-Uwe Bux

re: Truncating numbers at end of string

Alex Leach wrote:how about using std::string::find_last_not_of()?

re: Truncating numbers at end of string

Alex Leach a écrit :Use the string::find_last_not_of() method and erase character if return
value is not npos.
size_t pos=str.find_last_not_of("0123456789");
if(pos!=std::string::npos)
{
str.erase(pos+1);
}

Or use the find algorithm with the erase function of string in the same
line (a little more obfuscating)
str.erase( find_if( str.begin(),
str.end(),
ptr_fun(isdigit) ) );

Michael