std::string
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LCI wrote: So i am using std::string but i have never used this before and am so confused This is one good reason to prefer the standard library, you'll know it when you need it :-) What you really need is regex, I wonder if you're able to use the regex implimentation that boost offers ? Otherwise :
string s = "xxxcdfdftbvbfggbnghghg Name: Todd Shields ccfgfgnfgnrri DOB:091580 hgertrtreggeegegggggggggggg"; size_t n = s.find_first_of("Name : ") + 7; // ( 7 is the length of the string we searched for size_t nLast = s.find_first_of(" ", n); nLast = s.find_first_of(" ", nLast + 1); // Need to search twice because of the space in the name string name = s.substr(n, nLast - n);This is REALLY fragile. Ideally, you'd know the string that comes after ( ccfgfgnfgnrri in this case ), so you can search for the item AFTER the name, instead of having to count on there being a certain number of whitespace chars IN the name. The date is obviously easier, but either way, CString wouldn't help much, regex is the real way to deal with this. Christian Graus - Microsoft MVP - C++Great.. thanks for all your input. I also wrote a small method that will strip all the spaces from the end of an basic_string up until is reaches text. So a string like "This is a test " .. would be "This is a test" once the method is complete. Something is not working correctly though. typedef std::string::size_type size_type; std::string AnyString; char MySpace = ' '; size_type CHAR_INDEX; CHAR_INDEX = AnyString.size();//AnyString.rfind(MySpace); for (size_type i = CHAR_INDEX;AnyString[i]== MySpace; i--) AnyString.erase(i,1); The above source only works when stripping one space, like from "2003 " to "2003". If i have "2003 ", it does not work. Can anyone tell me what i am doing wrong here?
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I have a .dll that i am writing and cannot use standard MFC classes such as CString to parse a blob of data. So i am using std::string but i have never used this before and am so confused. I need to search a string for a key word. Once i find the keyword, i have to parse the data afterwards to pick out the necessary data that i need. I am using the find method but just stuck and confused to accomplish my goal. Can anyone offer some examples or help? E.g. If i have a string that looks like xxxcdfdftbvbfggbnghghg Name: Todd Shields ccfgfgnfgnrri DOB:091580 hgertrtreggeegegggggggggggg But i want to get the name and the DOB. There is no standard in terms of how much space is allocated for name or DOB. I just have to find it and get the name and DOB from the string. Any input is appreciated.
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Great.. thanks for all your input. I also wrote a small method that will strip all the spaces from the end of an basic_string up until is reaches text. So a string like "This is a test " .. would be "This is a test" once the method is complete. Something is not working correctly though. typedef std::string::size_type size_type; std::string AnyString; char MySpace = ' '; size_type CHAR_INDEX; CHAR_INDEX = AnyString.size();//AnyString.rfind(MySpace); for (size_type i = CHAR_INDEX;AnyString[i]== MySpace; i--) AnyString.erase(i,1); The above source only works when stripping one space, like from "2003 " to "2003". If i have "2003 ", it does not work. Can anyone tell me what i am doing wrong here?
LCI wrote: for (size_type i = CHAR_INDEX;AnyString[i]== MySpace; i--) AnyString.erase(i,1); The above source only works when stripping one space, like from "2003 " to "2003". If i have "2003 ", it does not work. Can anyone tell me what i am doing wrong here? You probably want a 'trim_back()' function. It must also work for 'unusual' cases like empty strings and strings that contain only white space. Also note that string::size_type is unsigned (cannot be less than 0).
#include <string>
#include <iostream>std::string& trim_back (std::string& str)
{
using namespace std;
while (str.size()) {
if (isspace (str[str.size()-1])) {
str.erase(str.size()-1);
} else {
break;
}
}
return str;
}int main()
{
using namespace std;string test ("2003 "),
whitespaceOnly (" \n\t"),
empty;cout << '|' << trim_back(test) << '|' << endl;
cout << '|' << trim_back(whitespaceOnly) << '|' << endl;
cout << '|' << trim_back(empty) << '|' << endl;return 0;
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Great.. thanks for all your input. I also wrote a small method that will strip all the spaces from the end of an basic_string up until is reaches text. So a string like "This is a test " .. would be "This is a test" once the method is complete. Something is not working correctly though. typedef std::string::size_type size_type; std::string AnyString; char MySpace = ' '; size_type CHAR_INDEX; CHAR_INDEX = AnyString.size();//AnyString.rfind(MySpace); for (size_type i = CHAR_INDEX;AnyString[i]== MySpace; i--) AnyString.erase(i,1); The above source only works when stripping one space, like from "2003 " to "2003". If i have "2003 ", it does not work. Can anyone tell me what i am doing wrong here?
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Great.. thanks for all your input. I also wrote a small method that will strip all the spaces from the end of an basic_string up until is reaches text. So a string like "This is a test " .. would be "This is a test" once the method is complete. Something is not working correctly though. typedef std::string::size_type size_type; std::string AnyString; char MySpace = ' '; size_type CHAR_INDEX; CHAR_INDEX = AnyString.size();//AnyString.rfind(MySpace); for (size_type i = CHAR_INDEX;AnyString[i]== MySpace; i--) AnyString.erase(i,1); The above source only works when stripping one space, like from "2003 " to "2003". If i have "2003 ", it does not work. Can anyone tell me what i am doing wrong here?
Try
inline string trim_right(const string& source, const string& t = " ")
{
string str = source;
return str.erase(source.find_last_not_of(t) + 1);
}inline string trim_left(const string& source, const string& t = " ")
{
string str = source;
return str.erase(0, source.find_first_not_of(t));
}inline string trim(const string& source, const string& t = " ")
{
return trim_left(trim_right(source, t), t);
}Kevin
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Try
inline string trim_right(const string& source, const string& t = " ")
{
string str = source;
return str.erase(source.find_last_not_of(t) + 1);
}inline string trim_left(const string& source, const string& t = " ")
{
string str = source;
return str.erase(0, source.find_first_not_of(t));
}inline string trim(const string& source, const string& t = " ")
{
return trim_left(trim_right(source, t), t);
}Kevin
Kevin McFarlane wrote: Try Don't! inline string trim_right(const string& source, const string& t = " "){ string str = source; return str.erase(source.find_last_not_of(t) + 1);} 1. You search only for " ", not for white space (e.g. '\t') 2. find_last_not_of() returns npos when no match is found (which is not an appropriate input for erase) 3. You return by value thus duplicating the string unnecessarily. In sum, your 'solution' is incorrect and inefficient, but fine otherwise. :suss:
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Kevin McFarlane wrote: Try Don't! inline string trim_right(const string& source, const string& t = " "){ string str = source; return str.erase(source.find_last_not_of(t) + 1);} 1. You search only for " ", not for white space (e.g. '\t') 2. find_last_not_of() returns npos when no match is found (which is not an appropriate input for erase) 3. You return by value thus duplicating the string unnecessarily. In sum, your 'solution' is incorrect and inefficient, but fine otherwise. :suss:
I copied the code from somewhere. It seemed to work for me so I didn't check its details. Still, I think a trim function should have been part of basic_string (if only to prevent incorrect implementations like this!:)). It's a common enough operation. Kevin
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I copied the code from somewhere. It seemed to work for me so I didn't check its details. Still, I think a trim function should have been part of basic_string (if only to prevent incorrect implementations like this!:)). It's a common enough operation. Kevin
Kevin McFarlane wrote: Still, I think a trim function should have been part of basic_string Nah. std::basic_string is already too big and fat[^]
My programming blahblahblah blog. If you ever find anything useful here, please let me know to remove it.
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Kevin McFarlane wrote: Try Don't! inline string trim_right(const string& source, const string& t = " "){ string str = source; return str.erase(source.find_last_not_of(t) + 1);} 1. You search only for " ", not for white space (e.g. '\t') 2. find_last_not_of() returns npos when no match is found (which is not an appropriate input for erase) 3. You return by value thus duplicating the string unnecessarily. In sum, your 'solution' is incorrect and inefficient, but fine otherwise. :suss:
Sloppy Joseph wrote: You search only for " ", not for white space (e.g. '\t') Actually it does handle '\t', as I've only specified a default argument. Sloppy Joseph wrote: find_last_not_of() returns npos when no match is found (which is not an appropriate input for erase) Agreed, but still seems to work anyway. Sloppy Joseph wrote: You return by value thus duplicating the string unnecessarily. OK, how about this?
inline string& trim(string& s, const string& t)
{
string::size_type st1, st2;
st1 = s.find_first_not_of(t);if (st1 != string::npos) { st2 = s.find\_last\_not\_of(t) + 1; s = s.substr(st1, st2 - st1); } else { s.clear(); } return s;}
Kevin
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Sloppy Joseph wrote: You search only for " ", not for white space (e.g. '\t') Actually it does handle '\t', as I've only specified a default argument. Sloppy Joseph wrote: find_last_not_of() returns npos when no match is found (which is not an appropriate input for erase) Agreed, but still seems to work anyway. Sloppy Joseph wrote: You return by value thus duplicating the string unnecessarily. OK, how about this?
inline string& trim(string& s, const string& t)
{
string::size_type st1, st2;
st1 = s.find_first_not_of(t);if (st1 != string::npos) { st2 = s.find\_last\_not\_of(t) + 1; s = s.substr(st1, st2 - st1); } else { s.clear(); } return s;}
Kevin
Kevin McFarlane wrote: Actually it does handle '\t', as I've only specified a default argument. Well, you require the user to define the white spaces for you? This is, at least, not user friendly (answer quickly, which chars are white space? Is '\n' a white space, is '\a'?). Hint: isspace. Agreed, but still seems to work anyway. An questionable answer for a professional software developer. OK, how about this? if (st1 != string::npos) { st2 = s.find_last_not_of(t) + 1; s = s.substr(st1, st2 - st1); } substr() produces a (partial) copy which is assigned (= copied again) to s. You can do it all in place.