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I have to create a function in C++ that would remove all the words from a string that start with a certain character inputted by a user. For example, if I were to have a string "She made up her mind to meet up with him in the morning" and a substring "m", I would like my string to be "She up her to up with him in the".
I believe I would need to find the occurrences of "m", erase it and all the characters after it till the space " ". Would that be the right approach and if so what would be the best methods to use in this case?
With your kind help I have altered and added code a little bit. The first function 'GetNextWord' seems to be working alright, however, there is definitely something wrong with my function, which is supposed to strip the words, as I am not getting any output. Here is the code:
string GetNextWord(string& s, size_t pos) {
string word;
char del = ' ';
int i = 0;
for (int i = 0; i < s.length(); i++) {
if (s[i] != del) {
word += s[i];
}
else break;
}
return word;
}
string StripWordsThatBeginWithLetter(string& s, char c) {
string result;
string word;
size_t pos = 0;
while (true)
{
word = GetNextWord(s, pos);
pos += word.size() + 1;
if (word.size() == 0)
{
break;
}
if (word[0] == c) {
size_t inx = 0;
inx = s.find(word[0]);
s.erase(inx, word.length());
}
else result = s;
}
return result;
}
Here's a hint. I'm guessing this is a homework problem. And I'm probably giving too much away.
std::string GetNextWord(const std::string &s, size_t pos)
{
std::string word;
// your code goes here to return a string that includes all the chars starting from s[pos] until the start of the next word (including trailing whitespace)
// return an empty string if at the end of the string
return word;
}
std::string StripWordsThatBeginWithLetter(const std::string& s, char c)
{
std::string result;
std::string word;
size_t pos = 0;
while (true)
{
word = GetNextWord(s, pos);
pos += word.size();
if (word.size() == 0)
{
break;
}
// your code on processing "word" goes here with respect
// to "c" goes here
}
return result;
}
Simple example in french. You are a gentleman and dont want to say "merde" too often, and so decided not to say any word starting with 'm'.
This program will help you :
"je suis beau merde je sais" becomes "je suis beau je sais"
#include <string>
#include <algorithm>
int main(){
std::string str ("je suis beau merde je le sais");
const auto forbiden_start ((const char) 'm');
std::cout << "initial rude string (word starting with \'" << forbiden_start << "\') : " << str << std::endl;
auto i (str.begin ());
auto wait (false);
std::for_each (str.begin (), str.end (), [&i, &forbiden_start, &wait] (const auto& c) {
if (wait) {
if (c == ' ') {
wait = false; return;
}
}
else {
if (c == forbiden_start) {
wait = true;
}
else *i++ = c;
}
});
if (i != str.end ()) str.erase (i, str.end ());
std::cout << "polite string : " << str << std::endl;
return 0;
}
All is not tested (separator is " "), but it is the idea
I have a char array:
char arr[] = "%RED**dsa0x0%xcа%wq233SSS%(dsa........";
The array must be read from the % sign to the next %. That is, the message begins with % and has an unknown length (there can be no % in the middle of the message).
Here is the code:
void SaveMessToVec(const string& str)
{
tail = tail + str;
if (tail.empty()) return;
assert(tail[0] == 0x24);
while (true)
{
size_t f = tail.find(0x24, 1);
if (f == string::npos) return;
string message = tail.substr(0, f);
//cout << message << '\n';
vecForAP.push_back(message);
tail.erase(0, f);
}
}
However, my controller can't work with string and vector. How to rewrite this code to use only char?
If you can use vector and string you can use this function.
std::vector<std::string> split(const std::string& s, char delimiter) {
std::vector<std::string> tokens;
std::string token;
std::istringstream tokenStream(s);
while (std::getline(tokenStream, token, delimiter)) {
if(token.size()!=0)
tokens.push_back(token);
}
return tokens;
}
int main() {
std::string str = "%RED**dsa0x0%xcа%wq233SSS%(dsa";
std::vector<std::string> vec = split(str,'%');
for(auto &v : vec)
std::cout << v << std::endl;
return 0;
}
If you cannot use string and vector you can use strtok(). strtok can tokenize a char array
int main() {
char arr[] = "%RED**dsa0x0%xcа%wq233SSS%(dsa";
char* token = strtok(arr,"%");
while(token != NULL) {
std::cout << token << std::endl;
token = strtok(NULL,"%");
}
return 0;
}
How you tried to do this is not the appropiate way to tokenize a string. But if you want to do this way I show you a solution for you.
I do not understand your whole code procisely but maybe this would be good enough for you.
vector<string> vecForAP; // global variable to store data for you
void SaveMessToVec(string& str) {
str.erase(str.begin()); //because the first letter is %
while (true){
size_t f = str.find('%');
if (f == string::npos) {
vecForAP.push_back(str);
return;
}
string message = str.substr(0,f);
vecForAP.push_back(message);
str.erase(0, f+1);
}
}
How do I replace part of a string with another string using the standard C++ libraries?
QString s("hello $name"); // Example using Qt.
s.replace("$name", "Somename");
There's a function to find a substring within a string (find), and a function to replace a particular range in a string with another string (replace), so you can combine those to get the effect you want:
bool replace(std::string& str, const std::string& from, const std::string& to) {
size_t start_pos = str.find(from);
if(start_pos == std::string::npos)
return false;
str.replace(start_pos, from.length(), to);
return true;
}
std::string string("hello $name");
replace(string, "$name", "Somename");
In response to a comment, I think replaceAll would probably look something like this:
void replaceAll(std::string& str, const std::string& from, const std::string& to) {
if(from.empty())
return;
size_t start_pos = 0;
while((start_pos = str.find(from, start_pos)) != std::string::npos) {
str.replace(start_pos, from.length(), to);
start_pos += to.length(); // In case 'to' contains 'from', like replacing 'x' with 'yx'
}
}
With C++11 you can use std::regex like so:
#include <regex>
...
std::string string("hello $name");
string = std::regex_replace(string, std::regex("\\$name"), "Somename");
The double backslash is required for escaping an escape character.
Using std::string::replace:
s.replace(s.find("$name"), sizeof("$name") - 1, "Somename");
To have the new string returned use this:
std::string ReplaceString(std::string subject, const std::string& search,
const std::string& replace) {
size_t pos = 0;
while ((pos = subject.find(search, pos)) != std::string::npos) {
subject.replace(pos, search.length(), replace);
pos += replace.length();
}
return subject;
}
If you need performance, here is an optimized function that modifies the input string, it does not create a copy of the string:
void ReplaceStringInPlace(std::string& subject, const std::string& search,
const std::string& replace) {
size_t pos = 0;
while ((pos = subject.find(search, pos)) != std::string::npos) {
subject.replace(pos, search.length(), replace);
pos += replace.length();
}
}
Tests:
std::string input = "abc abc def";
std::cout << "Input string: " << input << std::endl;
std::cout << "ReplaceString() return value: "
<< ReplaceString(input, "bc", "!!") << std::endl;
std::cout << "ReplaceString() input string not modified: "
<< input << std::endl;
ReplaceStringInPlace(input, "bc", "??");
std::cout << "ReplaceStringInPlace() input string modified: "
<< input << std::endl;
Output:
Input string: abc abc def
ReplaceString() return value: a!! a!! def
ReplaceString() input string not modified: abc abc def
ReplaceStringInPlace() input string modified: a?? a?? def
string.replace(string.find("%s"), string("%s").size(), "Something");
You could wrap this in a function but this one-line solution sounds acceptable.
The problem is that this will change the first occurence only, you might want to loop over it, but it also allows you to insert several variables into this string with the same token (%s).
Yes, you can do it, but you have to find the position of the first string with string's find() member, and then replace with it's replace() member.
string s("hello $name");
size_type pos = s.find( "$name" );
if ( pos != string::npos ) {
s.replace( pos, 5, "somename" ); // 5 = length( $name )
}
If you are planning on using the Standard Library, you should really get hold of a copy of the book The C++ Standard Library which covers all this stuff very well.
I use generally this:
std::string& replace(std::string& s, const std::string& from, const std::string& to)
{
if(!from.empty())
for(size_t pos = 0; (pos = s.find(from, pos)) != std::string::npos; pos += to.size())
s.replace(pos, from.size(), to);
return s;
}
It repeatedly calls std::string::find() to locate other occurrences of the searched for string until std::string::find() doesn't find anything. Because std::string::find() returns the position of the match we don't have the problem of invalidating iterators.
If all strings are std::string, you'll find strange problems with the cutoff of characters if using sizeof() because it's meant for C strings, not C++ strings. The fix is to use the .size() class method of std::string.
sHaystack.replace(sHaystack.find(sNeedle), sNeedle.size(), sReplace);
That replaces sHaystack inline -- no need to do an = assignment back on that.
Example usage:
std::string sHaystack = "This is %XXX% test.";
std::string sNeedle = "%XXX%";
std::string sReplace = "my special";
sHaystack.replace(sHaystack.find(sNeedle),sNeedle.size(),sReplace);
std::cout << sHaystack << std::endl;
This could be even better to use
void replace(string& input, const string& from, const string& to)
{
auto pos = 0;
while(true)
{
size_t startPosition = input.find(from, pos);
if(startPosition == string::npos)
return;
input.replace(startPosition, from.length(), to);
pos += to.length();
}
}
wstring myString = L"Hello $$ this is an example. By $$.";
wstring search = L"$$";
wstring replace = L"Tom";
for (int i = myString.find(search); i >= 0; i = myString.find(search))
myString.replace(i, search.size(), replace);
If you want to do it quickly you can use a two scan approach.
Pseudo code:
first parse. find how many matching chars.
expand the length of the string.
second parse. Start from the end of the string when we get a match we replace, else we just copy the chars from the first string.
I am not sure if this can be optimized to an in-place algo.
And a C++11 code example but I only search for one char.
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
void ReplaceString(string& subject, char search, const string& replace)
{
size_t initSize = subject.size();
int count = 0;
for (auto c : subject) {
if (c == search) ++count;
}
size_t idx = subject.size()-1 + count * replace.size()-1;
subject.resize(idx + 1, '\0');
string reverseReplace{ replace };
reverse(reverseReplace.begin(), reverseReplace.end());
char *end_ptr = &subject[initSize - 1];
while (end_ptr >= &subject[0])
{
if (*end_ptr == search) {
for (auto c : reverseReplace) {
subject[idx - 1] = c;
--idx;
}
}
else {
subject[idx - 1] = *end_ptr;
--idx;
}
--end_ptr;
}
}
int main()
{
string s{ "Mr John Smith" };
ReplaceString(s, ' ', "%20");
cout << s << "\n";
}
What about the boost solution:
boost::replace_all(value, "token1", "token2");
std::string replace(std::string base, const std::string from, const std::string to) {
std::string SecureCopy = base;
for (size_t start_pos = SecureCopy.find(from); start_pos != std::string::npos; start_pos = SecureCopy.find(from,start_pos))
{
SecureCopy.replace(start_pos, from.length(), to);
}
return SecureCopy;
}
My own implementation, taking into account that string needs to be resized only once, then replace can happen.
template <typename T>
std::basic_string<T> replaceAll(const std::basic_string<T>& s, const T* from, const T* to)
{
auto length = std::char_traits<T>::length;
size_t toLen = length(to), fromLen = length(from), delta = toLen - fromLen;
bool pass = false;
std::string ns = s;
size_t newLen = ns.length();
for (bool estimate : { true, false })
{
size_t pos = 0;
for (; (pos = ns.find(from, pos)) != std::string::npos; pos++)
{
if (estimate)
{
newLen += delta;
pos += fromLen;
}
else
{
ns.replace(pos, fromLen, to);
pos += delta;
}
}
if (estimate)
ns.resize(newLen);
}
return ns;
}
Usage could be for example like this:
std::string dirSuite = replaceAll(replaceAll(relPath.parent_path().u8string(), "\\", "/"), ":", "");
I'm just now learning C++, but editing some of the code previously posted, I'd probably use something like this. This gives you the flexibility to replace 1 or multiple instances, and also lets you specify the start point.
using namespace std;
// returns number of replacements made in string
long strReplace(string& str, const string& from, const string& to, size_t start = 0, long count = -1) {
if (from.empty()) return 0;
size_t startpos = str.find(from, start);
long replaceCount = 0;
while (startpos != string::npos){
str.replace(startpos, from.length(), to);
startpos += to.length();
replaceCount++;
if (count > 0 && replaceCount >= count) break;
startpos = str.find(from, startpos);
}
return replaceCount;
}
Here is a one liner that uses c++'s standard library.
The replacement better not have the old string in it (ex: replacing , with ,,), otherwise you have an INFINITE LOOP. Moreso, it is slow for large strings compared to other techniques because the find operations start at the begining of the string call every time. Look for better solutions if you're not too lazy. I put this in for completeness and inspiration for others. You've been warned.
while(s.find(old_s) != string::npos) s.replace(s.find(old_s), old_s.size(), new_s);
And a lambda option
auto replaceAll = [](string& s, string o, string n){ while(s.find(o) != string::npos) s.replace(s.find(o), o.size(), n); };
// EXAMPLES:
// Used like
string text = "hello hello world";
replaceAll(text, "hello", "bye"); // Changes text to "bye bye world"
// Do NOT use like
string text = "hello hello world";
replaceAll(text, "hello", "hello hello"); // Loops forever
You can use this code for remove subtring and also replace , and also remove extra white space .
code :
#include<bits/stdc++.h>
using namespace std;
void removeSpaces(string &str)
{
int n = str.length();
int i = 0, j = -1;
bool spaceFound = false;
while (++j <= n && str[j] == ' ');
while (j <= n)
{
if (str[j] != ' ')
{
if ((str[j] == '.' || str[j] == ',' ||
str[j] == '?') && i - 1 >= 0 &&
str[i - 1] == ' ')
str[i - 1] = str[j++];
else str[i++] = str[j++];
spaceFound = false;
}
else if (str[j++] == ' ')
{
if (!spaceFound)
{
str[i++] = ' ';
spaceFound = true;
}
}
}
if (i <= 1)
str.erase(str.begin() + i, str.end());
else str.erase(str.begin() + i - 1, str.end());
}
int main()
{
string s;
cin >> s;
for(int i = s.find("WUB"); i >= 0; i = s.find("WUB"))
s.replace(i,3," ");
removeSpaces(s);
cout << s << endl;
return 0;
}
I am working on a algorithm where I am trying the following output:
Given values/Inputs:
char *Var = "1-5,10,12,15-16,25-35,67,69,99-105";
int size = 29;
Here "1-5" depicts a range value, i.e. it will be understood as "1,2,3,4,5" while the values with just "," are individual values.
I was writing an algorithm where end output should be such that it will give complete range of output as:
int list[]=1,2,3,4,5,10,12,15,16,25,26,27,28,29,30,31,32,33,34,35,67,69,99,100,101,102,103,104,105;
If anyone is familiar with this issue then the help would be really appreciated.
Thanks in advance!
My initial code approach was as:
if(NULL != strchr((char *)grp_range, '-'))
{
int_u8 delims[] = "-";
result = (int_u8 *)strtok((char *)grp_range, (char *)delims);
if(NULL != result)
{
start_index = strtol((char*)result, (char **)&end_ptr, 10);
result = (int_u8 *)strtok(NULL, (char *)delims);
}
while(NULL != result)
{
end_index = strtol((char*)result, (char**)&end_ptr, 10);
result = (int_u8 *)strtok(NULL, (char *)delims);
}
while(start_index <= end_index)
{
grp_list[i++] = start_index;
start_index++;
}
}
else if(NULL != strchr((char *)grp_range, ','))
{
int_u8 delims[] = ",";
result = (unison_u8 *)strtok((char *)grp_range, (char *)delims);
while(result != NULL)
{
grp_list[i++] = strtol((char*)result, (char**)&end_ptr, 10);
result = (int_u8 *)strtok(NULL, (char *)delims);
}
}
But it only works if I have either "0-5" or "0,10,15". I am looking forward to make it more versatile.
Here is a C++ solution for you to study.
#include <vector>
#include <string>
#include <sstream>
#include <iostream>
using namespace std;
int ConvertString2Int(const string& str)
{
stringstream ss(str);
int x;
if (! (ss >> x))
{
cerr << "Error converting " << str << " to integer" << endl;
abort();
}
return x;
}
vector<string> SplitStringToArray(const string& str, char splitter)
{
vector<string> tokens;
stringstream ss(str);
string temp;
while (getline(ss, temp, splitter)) // split into new "lines" based on character
{
tokens.push_back(temp);
}
return tokens;
}
vector<int> ParseData(const string& data)
{
vector<string> tokens = SplitStringToArray(data, ',');
vector<int> result;
for (vector<string>::const_iterator it = tokens.begin(), end_it = tokens.end(); it != end_it; ++it)
{
const string& token = *it;
vector<string> range = SplitStringToArray(token, '-');
if (range.size() == 1)
{
result.push_back(ConvertString2Int(range[0]));
}
else if (range.size() == 2)
{
int start = ConvertString2Int(range[0]);
int stop = ConvertString2Int(range[1]);
for (int i = start; i <= stop; i++)
{
result.push_back(i);
}
}
else
{
cerr << "Error parsing token " << token << endl;
abort();
}
}
return result;
}
int main()
{
vector<int> result = ParseData("1-5,10,12,15-16,25-35,67,69,99-105");
for (vector<int>::const_iterator it = result.begin(), end_it = result.end(); it != end_it; ++it)
{
cout << *it << " ";
}
cout << endl;
}
Live example
http://ideone.com/2W99Tt
This is my boost approach :
This won't give you array of ints, instead a vector of ints
Algorithm used: (nothing new)
Split string using ,
Split the individual string using -
Make a range low and high
Push it into vector with help of this range
Code:-
#include<iostream>
#include<vector>
#include <boost/algorithm/string.hpp>
#include <boost/lexical_cast.hpp>
int main(){
std::string line("1-5,10,12,15-16,25-35,67,69,99-105");
std::vector<std::string> strs,r;
std::vector<int> v;
int low,high,i;
boost::split(strs,line,boost::is_any_of(","));
for (auto it:strs)
{
boost::split(r,it,boost::is_any_of("-"));
auto x = r.begin();
low = high =boost::lexical_cast<int>(r[0]);
x++;
if(x!=r.end())
high = boost::lexical_cast<int>(r[1]);
for(i=low;i<=high;++i)
v.push_back(i);
}
for(auto x:v)
std::cout<<x<<" ";
return 0;
}
You're issue seems to be misunderstanding how strtok works. Have a look at this.
#include <string.h>
#include <stdio.h>
int main()
{
int i, j;
char delims[] = " ,";
char str[] = "1-5,6,7";
char *tok;
char tmp[256];
int rstart, rend;
tok = strtok(str, delims);
while(tok != NULL) {
for(i = 0; i < strlen(tok); ++i) {
//// range
if(i != 0 && tok[i] == '-') {
strncpy(tmp, tok, i);
rstart = atoi(tmp);
strcpy(tmp, tok + i + 1);
rend = atoi(tmp);
for(j = rstart; j <= rend; ++j)
printf("%d\n", j);
i = strlen(tok) + 1;
}
else if(strchr(tok, '-') == NULL)
printf("%s\n", tok);
}
tok = strtok(NULL, delims);
}
return 0;
}
Don't search. Just go through the text one character at a time. As long as you're seeing digits, accumulate them into a value. If the digits are followed by a - then you're looking at a range, and need to parse the next set of digits to get the upper bound of the range and put all the values into your list. If the value is not followed by a - then you've got a single value; put it into your list.
Stop and think about it: what you actually have is a comma
separated list of ranges, where a range can be either a single
number, or a pair of numbers separated by a '-'. So you
probably want to loop over the ranges, using recursive descent
for the parsing. (This sort of thing is best handled by an
istream, so that's what I'll use.)
std::vector<int> results;
std::istringstream parser( std::string( var ) );
processRange( results, parser );
while ( isSeparator( parser, ',' ) ) {
processRange( results, parser );
}
with:
bool
isSeparator( std::istream& source, char separ )
{
char next;
source >> next;
if ( source && next != separ ) {
source.putback( next );
}
return source && next == separ;
}
and
void
processRange( std::vector<int>& results, std::istream& source )
{
int first = 0;
source >> first;
int last = first;
if ( isSeparator( source, '-' ) ) {
source >> last;
}
if ( last < first ) {
source.setstate( std::ios_base::failbit );
}
if ( source ) {
while ( first != last ) {
results.push_back( first );
++ first;
}
results.push_back( first );
}
}
The isSeparator function will, in fact, probably be useful in
other projects in the future, and should be kept in your
toolbox.
First divide whole string into numbers and ranges (using strtok() with "," delimiter), save strings in array, then, search through array looking for "-", if it present than use sscanf() with "%d-%d" format, else use sscanf with single "%d" format.
Function usage is easily googling.
One approach:
You need a parser that identifies 3 kinds of tokens: ',', '-', and numbers. That raises the level of abstraction so that you are operating at a level above characters.
Then you can parse your token stream to create a list of ranges and constants.
Then you can parse that list to convert the ranges into constants.
Some code that does part of the job:
#include <stdio.h>
// Prints a comma after the last digit. You will need to fix that up.
void print(int a, int b) {
for (int i = a; i <= b; ++i) {
printf("%d, ", i);
}
}
int main() {
enum { DASH, COMMA, NUMBER };
struct token {
int type;
int value;
};
// Sample input stream. Notice the sentinel comma at the end.
// 1-5,10,
struct token tokStream[] = {
{ NUMBER, 1 },
{ DASH, 0 },
{ NUMBER, 5 },
{ COMMA, 0 },
{ NUMBER, 10 },
{ COMMA, 0 } };
// This parser assumes well formed input. You have to add all the error
// checking yourself.
size_t i = 0;
while (i < sizeof(tokStream)/sizeof(struct token)) {
if (tokStream[i+1].type == COMMA) {
print(tokStream[i].value, tokStream[i].value);
i += 2; // skip to next number
}
else { // DASH
print(tokStream[i].value, tokStream[i+2].value);
i += 4; // skip to next number
}
}
return 0;
}
If s is a std::string, then is there a function like the following?
s.replace("text to replace", "new text");
Replace first match
Use a combination of std::string::find and std::string::replace.
Find the first match:
std::string s;
std::string toReplace("text to replace");
size_t pos = s.find(toReplace);
Replace the first match:
s.replace(pos, toReplace.length(), "new text");
A simple function for your convenience:
void replace_first(
std::string& s,
std::string const& toReplace,
std::string const& replaceWith
) {
std::size_t pos = s.find(toReplace);
if (pos == std::string::npos) return;
s.replace(pos, toReplace.length(), replaceWith);
}
Usage:
replace_first(s, "text to replace", "new text");
Demo.
Replace all matches
Define this O(n) method using std::string as a buffer:
void replace_all(
std::string& s,
std::string const& toReplace,
std::string const& replaceWith
) {
std::string buf;
std::size_t pos = 0;
std::size_t prevPos;
// Reserves rough estimate of final size of string.
buf.reserve(s.size());
while (true) {
prevPos = pos;
pos = s.find(toReplace, pos);
if (pos == std::string::npos)
break;
buf.append(s, prevPos, pos - prevPos);
buf += replaceWith;
pos += toReplace.size();
}
buf.append(s, prevPos, s.size() - prevPos);
s.swap(buf);
}
Usage:
replace_all(s, "text to replace", "new text");
Demo.
Boost
Alternatively, use boost::algorithm::replace_all:
#include <boost/algorithm/string.hpp>
using boost::replace_all;
Usage:
replace_all(s, "text to replace", "new text");
Do we really need a Boost library for seemingly such a simple task?
To replace all occurences of a substring use this function:
std::string ReplaceString(std::string subject, const std::string& search,
const std::string& replace) {
size_t pos = 0;
while ((pos = subject.find(search, pos)) != std::string::npos) {
subject.replace(pos, search.length(), replace);
pos += replace.length();
}
return subject;
}
If you need performance, here is an optimized function that modifies the input string, it does not create a copy of the string:
void ReplaceStringInPlace(std::string& subject, const std::string& search,
const std::string& replace) {
size_t pos = 0;
while ((pos = subject.find(search, pos)) != std::string::npos) {
subject.replace(pos, search.length(), replace);
pos += replace.length();
}
}
Tests:
std::string input = "abc abc def";
std::cout << "Input string: " << input << std::endl;
std::cout << "ReplaceString() return value: "
<< ReplaceString(input, "bc", "!!") << std::endl;
std::cout << "ReplaceString() input string not modified: "
<< input << std::endl;
ReplaceStringInPlace(input, "bc", "??");
std::cout << "ReplaceStringInPlace() input string modified: "
<< input << std::endl;
Output:
Input string: abc abc def
ReplaceString() return value: a!! a!! def
ReplaceString() input string not modified: abc abc def
ReplaceStringInPlace() input string modified: a?? a?? def
Yes: replace_all is one of the boost string algorithms:
Although it's not a standard library, it has a few things on the standard library:
More natural notation based on ranges rather than iterator pairs. This is nice because you can nest string manipulations (e.g., replace_all nested inside a trim). That's a bit more involved for the standard library functions.
Completeness. This isn't hard to be 'better' at; the standard library is fairly spartan. For example, the boost string algorithms give you explicit control over how string manipulations are performed (i.e., in place or through a copy).
#include <iostream>
#include <string>
using namespace std;
int main ()
{
string str("one three two four");
string str2("three");
str.replace(str.find(str2),str2.length(),"five");
cout << str << endl;
return 0;
}
Output
one five two four
like some say boost::replace_all
here a dummy example:
#include <boost/algorithm/string/replace.hpp>
std::string path("file.gz");
boost::replace_all(path, ".gz", ".zip");
Not exactly that, but std::string has many replace overloaded functions.
Go through this link to see explanation of each, with examples as to how they're used.
Also, there are several versions of string::find functions (listed below) which you can use in conjunction with string::replace.
find
rfind
find_first_of
find_last_of
find_first_not_of
find_last_not_of
Also, note that there are several versions of replace functions available from <algorithm> which you can also use (instead of string::replace):
replace
replace_if
replace_copy
replace_copy_if
// replaced text will be in buffer.
void Replace(char* buffer, const char* source, const char* oldStr, const char* newStr)
{
if(buffer==NULL || source == NULL || oldStr == NULL || newStr == NULL) return;
int slen = strlen(source);
int olen = strlen(oldStr);
int nlen = strlen(newStr);
if(olen>slen) return;
int ix=0;
for(int i=0;i<slen;i++)
{
if(oldStr[0] == source[i])
{
bool found = true;
for(int j=1;j<olen;j++)
{
if(source[i+j]!=oldStr[j])
{
found = false;
break;
}
}
if(found)
{
for(int j=0;j<nlen;j++)
buffer[ix++] = newStr[j];
i+=(olen-1);
}
else
{
buffer[ix++] = source[i];
}
}
else
{
buffer[ix++] = source[i];
}
}
}
Here's the version I ended up writing that replaces all instances of the target string in a given string. Works on any string type.
template <typename T, typename U>
T &replace (
T &str,
const U &from,
const U &to)
{
size_t pos;
size_t offset = 0;
const size_t increment = to.size();
while ((pos = str.find(from, offset)) != T::npos)
{
str.replace(pos, from.size(), to);
offset = pos + increment;
}
return str;
}
Example:
auto foo = "this is a test"s;
replace(foo, "is"s, "wis"s);
cout << foo;
Output:
thwis wis a test
Note that even if the search string appears in the replacement string, this works correctly.
void replace(char *str, char *strFnd, char *strRep)
{
for (int i = 0; i < strlen(str); i++)
{
int npos = -1, j, k;
if (str[i] == strFnd[0])
{
for (j = 1, k = i+1; j < strlen(strFnd); j++)
if (str[k++] != strFnd[j])
break;
npos = i;
}
if (npos != -1)
for (j = 0, k = npos; j < strlen(strRep); j++)
str[k++] = strRep[j];
}
}
int main()
{
char pst1[] = "There is a wrong message";
char pfnd[] = "wrong";
char prep[] = "right";
cout << "\nintial:" << pst1;
replace(pst1, pfnd, prep);
cout << "\nfinal : " << pst1;
return 0;
}
void replaceAll(std::string & data, const std::string &toSearch, const std::string &replaceStr)
{
// Get the first occurrence
size_t pos = data.find(toSearch);
// Repeat till end is reached
while( pos != std::string::npos)
{
// Replace this occurrence of Sub String
data.replace(pos, toSearch.size(), replaceStr);
// Get the next occurrence from the current position
pos =data.find(toSearch, pos + replaceStr.size());
}
}
More CPP utilities: https://github.com/Heyshubham/CPP-Utitlities/blob/master/src/MEString.cpp#L60
is there a function like the following?
One other(in addition to using boost and other methods given in different answers) possible way of doing this is using std::regex_replace as shown below:
std::string s{"my name is my name and not my name mysometext myto"}; //this is the original line
std::string replaceThis = "my";
std::string replaceWith = "your";
std::regex pattern("\\b" + replaceThis + "\\b");
std::string replacedLine = std::regex_replace(s, pattern, replaceWith);
std::cout<<replacedLine<<std::endl;