Deleting two specific Characters - c++

Question

So I have a little problem, I want to achieve this in C++, but I don't know how to do it:
Given is a string containing random numbers, symbols, and letters:
std::string = "1653gbdtsr362g2v3f3t52bv^hdtvsbjj;hdfuue,9^1dkkns";
Now I'm trying to find all ^ characters, and if those are followed by a number between 0 and 9, delete the ^ and the number, so:
"^1ghhu^7dndn^g"
becomes:
"ghhudndn^g"
I know how to find and replace/erase chars from a string, but I don't know how to check if it's followed by a number in a not hard coded way. Any help is appreciated.

Answer 1

std::string s = "^1ghhu^7dndn^g";
for (int i = 0; i < s.length() - 1; ++i)
{
if (s[i] == '^' && std::isdigit(s[i + 1]))
{
s.erase(i, 2);
--i;
}
}
This needs these includes:
#include <string>
#include <cctype>

Answer 2

I would do it this way:
#include <iostream>
#include <string>
#include <utility>
#include <iterator>
template<class Iter, class OutIter>
OutIter remove_escaped_numbers(Iter first, Iter last, OutIter out) {
for ( ; first != last ; )
{
auto c = *first++;
if (c == '^' && first != last)
{
c = *first++;
if (std::isdigit(c))
continue;
else {
*out++ = '^';
*out++ = c;
}
}
else {
*out++ = c;
}
}
return out;
}
int main()
{
using namespace std::literals;
auto input = "^1ghhu^7dndn^g"s;
auto output = std::string{};
remove_escaped_numbers(input.begin(), input.end(), std::back_inserter(output));
std::cout << output << std::endl;
}
or this way:
#include <iostream>
#include <regex>
int main()
{
using namespace std::literals;
auto input = "^1ghhu^7dndn^g"s;
static const auto repl = std::regex { R"___(\^\d)___" };
auto output = std::regex_replace(input, repl, "");
std::cout << output << std::endl;
}

Answer 3

A solution using std::stringstream, and returning the input string cleared of caret-digit's.
#include <iostream>
#include <sstream>
#include <cctype>
int t404()
{
std::stringstream ss;
std::string inStr("1653gbdtsr362g2v3f3t52bv^hdtvsbjj;hdfuue,9^1dkkns");
for (size_t i = 0; i<inStr.size(); ++i)
{
if(('^' == inStr[i]) && isdigit(inStr[i+1]))
{
i += 1; // skip over caret followed by single digit
}
else
{
ss << inStr[i];
}
}
std::cout << inStr << std::endl; // compare input
std::cout << ss.str() << std::endl; // to results
return 0;
}
Output:
1653gbdtsr362g2v3f3t52bv^hdtvsbjj;hdfuue,9^1dkkns
1653gbdtsr362g2v3f3t52bv^hdtvsbjj;hdfuue,9dkkns

Answer 4

you can simply loop over the string and copy it while skipping the undesired chars. Here is a possible function to do it:
std::string filterString (std::string& s) {
std::string result = "";
std::string::iterator it = s.begin();
char c;
while (it != s.end()) {
if (*it == '^' && it != s.end() && (it + 1) != s.end()) {
c = *(it + 1);
if(c >= '0' && c <= '9') {
it += 2;
continue;
}
}
result.push_back(*it);
++ it;
}
return result;
}

Answer 5

A robust solution would be to use the regex library that C++11 brings in.
std::string input ("1653gbdtsr362g2v3f3t52bv^hdtvsbjj;hdfuue,9^1dkkns");
std::regex rx ("[\\^][\\d]{1}"); // "[\^][\d]{1}"
std::cout << std::regex_replace(input,rx,"woot");
>> 1653gbdtsr362g2v3f3t52bv^hdtvsbjj;hdfuue,9wootdkkns
This locates a "^" character ([\^]) followed by 1 ({1}) digit ([\d]) and replaces all occurances with "woot".

Answer 6

Hope this code can solve your problem:
#include <iostream>
#include <string>
int main()
{
std::string str = "^1ghhu^7dndn^g";
std::string::iterator first, last;
for ( std::string::iterator it=str.begin(); it!=str.end(); ++it)
{
if(*it == '^')
{
first = it;
it++;
while(isdigit(*it))
{
it++;
}
last = it - 1;
if(first != last)
{
if((last + 1) != str.end())
{
str.erase(first, last + 1);
}
else
{
str.erase(first, str.end());
break;
}
}
}
}
std::cout<< str << std::endl;
return 0;
}
The output:
$ ./erase
ghhudndn^g