I have a single-space delimited string and I want to replace field x.
I can repeatedly use find to locate the x - 1 and x spaces, then use substr to grab the two strings on either side, then concatenate the two sub strings and my replacement text.
But man that seems like an awful lot of work for something that should be simple. Is there a better solution-- one that doesn't require Boost?
Answer
I've cleaned up #Domenic Lokies answer below:
sting fieldReplace( const string input, const string outputField, int index )
{
vector< char > stringIndex( numeric_limits< int >::digits10 + 2 );
_itoa_s( index, stringIndex.begin()._Ptr, stringIndex.size(), 10 );
const string stringRegex( "^((?:\\w+ ){" ); //^((?:\w+ ){$index})\w+
return regex_replace( input, regex( stringRegex + stringIndex.begin()._Ptr + "})\\w+" ), "$1" + outputField );
}
(_itoa_s and _Ptr are MSVS only I believe, so you'll need to clean those up if you want code portability. )
You can do it using one of the string::replace methods:
Locate the position of the x-1-st space. You can do it by calling string::find repeatedly
Locate the position of the x-th space by calling string::find one more time
Calculate the length of the word being replaced by subtracting the first index from the second one
Call string::replace passing the first index, the length, and the replacement string.
Here is how you can implement this:
#include <iostream>
#include <string>
using namespace std;
int main() {
string s = "quick brown frog jumps over the lazy dog";
size_t start = -1;
int cnt = 3; // Word number three
do {
start = s.find(' ', start+1);
} while (start != string::npos && --cnt > 1);
size_t end = s.find(' ', start+1);
s.replace(start+1, end-start-1, "fox");
cout << s << endl;
return 0;
}
Demo on ideone.
Since C++11 you should use a Regular Expression for your purposes. If you are not using a compiler which supports C++11, you can take a look at Boost.Regex.
Never combine std::string::find with std::string::replace, that is just not a good style in a language like C++.
I have written a short example for you to show you how to use Regular Expressions in C++.
#include <string>
#include <regex>
#include <iostream>
int main()
{
std::string subject = "quick brown frog jumps over the lazy dog";
std::regex pattern("frog");
std::cout << std::regex_replace(subject, pattern, "fox");
}
Related
I am writing what amounts to a tiny DSL in which each script is read from a single string, like this:
"func1;func2;func1;4*func3;func1"
I need to expand the loops, so that the expanded script is:
"func1;func2;func1;func3;func3;func3;func3;func1"
I have used the C++ standard regex library with the following regex to find those loops:
regex REGEX_SIMPLE_LOOP(":?[0-9]+)\\*([_a-zA-Z][_a-zA-Z0-9]*;");
smatch match;
bool found = std::regex_search(*this, match, std::regex(REGEX_SIMPLE_LOOP));
Now, it's not too difficult to read out the loop multiplier and print the function N times, but how do I then replace the original match with this string? I want to do this:
if (found) match[0].replace(new_string);
But I don't see that the library can do this.
My backup place is to regex_search, then construct the new string, and then use regex_replace, but it seems clunky and inefficient and not nice to essentially do two full searches like that. Is there a cleaner way?
You can also NOT use regex, the parsing isn't too difficult.
So regex might be overkill. Demo here : https://onlinegdb.com/RXLqLtrUQ-
(and yes my output gives an extra ; at the end)
#include <string>
#include <sstream>
#include <iostream>
int main()
{
std::istringstream is{ "func1;func2;func1;4*func3;func1" };
std::string split;
// use getline to split
while (std::getline(is, split, ';'))
{
// assume 1 repeat
std::size_t count = 1;
// if split part starts with a digit
if (std::isdigit(split.front()))
{
// look for a *
auto pos = split.find('*');
// the first part of the string contains the repeat count
auto count_str = split.substr(0, pos);
// convert that to a value
count = std::stoi(count_str);
// and keep the rest ("funcn")
split = split.substr(pos + 1, split.size() - pos - 1);
}
// now use the repeat count to build the output string
for (std::size_t n = 0; n < count; ++n)
{
std::cout << split << ";";
}
}
// TODO invalid input string handling.
return 0;
}
Using the C++ Standard Template Library function regex_replace(), how do I remove non-numeric characters from a std::string and return a std::string?
This question is not a duplicate of question
747735
because that question requests how to use TR1/regex, and I'm
requesting how to use standard STL regex, and because the answer given
is merely some very complex documentation links. The C++ regex
documentation is extremely hard to understand and poorly documented,
in my opinion, so even if a question pointed out the standard C++
regex_replace
documentation,
it still wouldn't be very useful to new coders.
// assume #include <regex> and <string>
std::string sInput = R"(AA #-0233 338982-FFB /ADR1 2)";
std::string sOutput = std::regex_replace(sInput, std::regex(R"([\D])"), "");
// sOutput now contains only numbers
Note that the R"..." part means raw string literal and does not evaluate escape codes like a C or C++ string would. This is very important when doing regular expressions and makes your life easier.
Here's a handy list of single-character regular expression raw literal strings for your std::regex() to use for replacement scenarios:
R"([^A-Za-z0-9])" or R"([^A-Za-z\d])" = select non-alphabetic and non-numeric
R"([A-Za-z0-9])" or R"([A-Za-z\d])" = select alphanumeric
R"([0-9])" or R"([\d])" = select numeric
R"([^0-9])" or R"([^\d])" or R"([\D])" = select non-numeric
Regular expressions are overkill here.
#include <algorithm>
#include <iostream>
#include <iterator>
#include <string>
inline bool not_digit(char ch) {
return '0' <= ch && ch <= '9';
}
std::string remove_non_digits(const std::string& input) {
std::string result;
std::copy_if(input.begin(), input.end(),
std::back_inserter(result),
not_digit);
return result;
}
int main() {
std::string input = "1a2b3c";
std::string result = remove_non_digits(input);
std::cout << "Original: " << input << '\n';
std::cout << "Filtered: " << result << '\n';
return 0;
}
The accepted answer if fine for the specifics of the given sample.
But it will fail for a number such as "-12.34" (it would result in "1234").
(note how the sample could be negative numbers)
Then the regex should be:
(-|\+)?(\d)+(.(\d)+)*
explanation: (optional ( "-" or "+" )) with (a number, repeated 1 to n times) with (optionally end's with: ( a "." followed by (a number, repeated 1 to n times) )
A bit over-reaching, but I was looking for this and the page showed up first in my search, so I'm adding my answer for future searches.
I am making a roman numeral converter. I have everything figured out except there is one problem at the end.
The string looks like IVV
I need to make it IX
I have split the string at each new letter, then appended them back on, then using an if statement to see if it contains 2 "V"s. I want to know if there is a simpler way to do this.
Using std::string should help you tremendously as you can leverage its search and replace functionality. You'll want to start with the find function which allows you to search for a character or a string and returns an index where what you are searching for exists or npos if the search fails.
You can then call replace passing it the index returned by find, the number of characters you want to replace and what replace the range with.
The code below should help you get started.
#include <string>
#include <iostream>
int main()
{
std::string roman("IVV");
// Search for the string you want to replace
std::string::size_type loc = roman.find("VV");
// If the substring is found replace it.
if (loc != std::string::npos)
{
// replace 2 characters staring at position loc with the string "X"
roman.replace(loc, 2, "X");
}
std::cout << roman << std::endl;
return 0;
}
You could use std string find and rfind operations, these find the position of the first and the last occurrence of the entered parameter, check if these are not equal and you will know
Answer updated
#include <string>
int main()
{
std::string x1 = "IVV";
if (x1.find('V') !=x1.rfind('V'))
{
x1.replace(x1.find('V'), 2, 'X');
}
return 0;
}
Is there any inbuilt function available two get string between two delimiter string in C/C++?
My input look like
_STARTDELIMITER_0_192.168.1.18_STOPDELIMITER_
And my output should be
_0_192.168.1.18_
Thanks in advance...
You can do as:
string str = "STARTDELIMITER_0_192.168.1.18_STOPDELIMITER";
unsigned first = str.find(STARTDELIMITER);
unsigned last = str.find(STOPDELIMITER);
string strNew = str.substr (first,last-first);
Considering your STOPDELIMITER delimiter will occur only once at the end.
EDIT:
As delimiter can occur multiple times, change your statement for finding STOPDELIMITER to:
unsigned last = str.find_last_of(STOPDELIMITER);
This will get you text between the first STARTDELIMITER and LAST STOPDELIMITER despite of them being repeated multiple times.
I have no idea how the top answer received so many votes that it did when the question clearly asks how to get a string between two delimiter strings, and not a pair of characters.
If you would like to do so you need to account for the length of the string delimiter, since it will not be just a single character.
Case 1: Both delimiters are unique:
Given a string _STARTDELIMITER_0_192.168.1.18_STOPDELIMITER_ that you want to extract _0_192.168.1.18_ from, you could modify the top answer like so to get the desired effect. This is the simplest solution without introducing extra dependencies (e.g Boost):
#include <iostream>
#include <string>
std::string get_str_between_two_str(const std::string &s,
const std::string &start_delim,
const std::string &stop_delim)
{
unsigned first_delim_pos = s.find(start_delim);
unsigned end_pos_of_first_delim = first_delim_pos + start_delim.length();
unsigned last_delim_pos = s.find(stop_delim);
return s.substr(end_pos_of_first_delim,
last_delim_pos - end_pos_of_first_delim);
}
int main() {
// Want to extract _0_192.168.1.18_
std::string s = "_STARTDELIMITER_0_192.168.1.18_STOPDELIMITER_";
std::string s2 = "ABC123_STARTDELIMITER_0_192.168.1.18_STOPDELIMITER_XYZ345";
std::string start_delim = "_STARTDELIMITER";
std::string stop_delim = "STOPDELIMITER_";
std::cout << get_str_between_two_str(s, start_delim, stop_delim) << std::endl;
std::cout << get_str_between_two_str(s2, start_delim, stop_delim) << std::endl;
return 0;
}
Will print _0_192.168.1.18_ twice.
It is necessary to add the position of the first delimiter in the second argument to std::string::substr as last - (first + start_delim.length()) to ensure that the it would still extract the desired inner string correctly in the event that the start delimiter is not located at the very beginning of the string, as demonstrated in the second case above.
See the demo.
Case 2: Unique first delimiter, non-unique second delimiter:
Say you want to get a string between a unique delimiter and the first non unique delimiter encountered after the first delimiter. You could modify the above function get_str_between_two_str to use find_first_of instead to get the desired effect:
std::string get_str_between_two_str(const std::string &s,
const std::string &start_delim,
const std::string &stop_delim)
{
unsigned first_delim_pos = s.find(start_delim);
unsigned end_pos_of_first_delim = first_delim_pos + start_delim.length();
unsigned last_delim_pos = s.find_first_of(stop_delim, end_pos_of_first_delim);
return s.substr(end_pos_of_first_delim,
last_delim_pos - end_pos_of_first_delim);
}
If instead you want to capture any characters in between the first unique delimiter and the last encountered second delimiter, like what the asker commented above, use find_last_of instead.
Case 3: Non-unique first delimiter, unique second delimiter:
Very similar to case 2, just reverse the logic between the first delimiter and second delimiter.
Case 4: Both delimiters are not unique:
Again, very similar to case 2, make a container to capture all strings between any of the two delimiters. Loop through the string and update the first delimiter's position to be equal to the second delimiter's position when it is encountered and add the string in between to the container. Repeat until std::string:npos is reached.
To get a string between 2 delimiter strings without white spaces.
string str = "STARTDELIMITER_0_192.168.1.18_STOPDELIMITER";
string startDEL = "STARTDELIMITER";
// this is really only needed for the first delimiter
string stopDEL = "STOPDELIMITER";
unsigned firstLim = str.find(startDEL);
unsigned lastLim = str.find(stopDEL);
string strNew = str.substr (firstLim,lastLim);
//This won't exclude the first delimiter because there is no whitespace
strNew = strNew.substr(firstLim + startDEL.size())
// this will start your substring after the delimiter
I tried combining the two substring functions but it started printing the STOPDELIMITER
Hope that helps
Hope you won't mind I'm answering by another question :)
I would use boost::split or boost::split_iter.
http://www.boost.org/doc/libs/1_54_0/doc/html/string_algo/usage.html#idp166856528
For example code see this SO question:
How to avoid empty tokens when splitting with boost::iter_split?
Let's say you need to get 5th argument (brand) from output below:
zoneid:zonename:state:zonepath:uuid:brand:ip-type:r/w:file-mac-profile
You cannot use any "str.find" function, because it is in the middle, but you can use 'strtok'. e.g.
char *brand;
brand = strtok( line, ":" );
for (int i=0;i<4;i++) {
brand = strtok( NULL, ":" );
}
This is a late answer, but this might work too:
string strgOrg= "STARTDELIMITER_0_192.168.1.18_STOPDELIMITER";
string strg= strgOrg;
strg.replace(strg.find("STARTDELIMITER"), 14, "");
strg.replace(strg.find("STOPDELIMITER"), 13, "");
Hope it works for others.
void getBtwString(std::string oStr, std::string sStr1, std::string sStr2, std::string &rStr)
{
int start = oStr.find(sStr1);
if (start >= 0)
{
string tstr = oStr.substr(start + sStr1.length());
int stop = tstr.find(sStr2);
if (stop >1)
rStr = oStr.substr(start + sStr1.length(), stop);
else
rStr ="error";
}
else
rStr = "error"; }
or if you are using Windows and have access to c++14, the following,
void getBtwString(std::string oStr, std::string sStr1, std::string sStr2, std::string &rStr)
{
using namespace std::literals::string_literals;
auto start = sStr1;
auto end = sStr2;
std::regex base_regex(start + "(.*)" + end);
auto example = oStr;
std::smatch base_match;
std::string matched;
if (std::regex_search(example, base_match, base_regex)) {
if (base_match.size() == 2) {
matched = base_match[1].str();
}
rStr = matched;
}
}
Example:
string strout;
getBtwString("it's_12345bb2","it's","bb2",strout);
getBtwString("it's_12345bb2"s,"it's"s,"bb2"s,strout); // second solution
Headers:
#include <regex> // second solution
#include <string.h>
Ok, so I need some info parsed and I would like to know what would be the best way to do it.
Ok so here is the string that I need to parse. The delimeter is the "^"
John Doe^Male^20
I need to parse the string into name, gender, and age variables. What would be the best way to do it in C++? I was thinking about looping and set the condition to while(!string.empty()
and then assign all characters up until the '^' to a string, and then erase what I have already assigned. Is there a better way of doing this?
You can use getline in C++ stream.
istream& getline(istream& is,string& str,char delimiter=’\n’)
change delimiter to '^'
You have a few options. One good option you have, if you can use boost, is the split algorithm they provide in their string library. You can check out this so question to see the boost answer in action: How to split a string in c
If you cannot use boost, you can use string::find to get the index of a character:
string str = "John Doe^Male^20";
int last = 0;
int cPos = -1;
while ((cPos = str.find('^', cPos + 1)) != string::npos)
{
string sub = str.substr(last, cPos - last);
// Do something with the string
last = cPos + 1;
}
#include <stdio.h>
#include <string.h>
int main ()
{
char str[] = "This is a sample string";
char * pch;
printf ("Looking for the 's' character in \"%s\"...\n",str);
pch=strchr(str,'s');
while (pch!=NULL)
{
printf ("found at %d\n",pch-str+1);
pch=strchr(pch+1,'s');
}
return 0;
}
Do something like this in an array.
You have a number of choices but I would use strtok(), myself. It would make short work of this.