问题
What is the effective way to replace all occurrences of a character with another character in std::string?
回答1:
std::string doesn't contain such function but you could use stand-alone replace function from algorithm header.
#include <algorithm>
#include <string>
void some_func() {
std::string s = "example string";
std::replace( s.begin(), s.end(), 'x', 'y'); // replace all 'x' to 'y'
}
回答2:
I thought I'd toss in the boost solution as well:
#include <boost/algorithm/string/replace.hpp>
// in place
std::string in_place = "blah#blah";
boost::replace_all(in_place, "#", "@");
// copy
const std::string input = "blah#blah";
std::string output = boost::replace_all_copy(input, "#", "@");
回答3:
The question is centered on character replacement, but, as I found this page very useful (especially Konrad's remark), I'd like to share this more generalized implementation, which allows to deal with substrings as well:
std::string ReplaceAll(std::string str, const std::string& from, const std::string& to) {
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(); // Handles case where 'to' is a substring of 'from'
}
return str;
}
Usage:
std::cout << ReplaceAll(string("Number Of Beans"), std::string(" "), std::string("_")) << std::endl;
std::cout << ReplaceAll(string("ghghjghugtghty"), std::string("gh"), std::string("X")) << std::endl;
std::cout << ReplaceAll(string("ghghjghugtghty"), std::string("gh"), std::string("h")) << std::endl;
Outputs:
Number_Of_Beans
XXjXugtXty
hhjhugthty
EDIT:
The above can be implemented in a more suitable way, in case performances are of your concern, by returning nothing (void) and performing the changes directly on the string str given as argument, passed by address instead of by value. This would avoid useless and costly copy of the original string, while returning the result. Your call, then...
Code :
static inline void ReplaceAll2(std::string &str, const std::string& from, const std::string& to)
{
// Same inner code...
// No return statement
}
Hope this will be helpful for some others...
回答4:
Imagine a large binary blob where all 0x00 bytes shall be replaced by "\1\x30" and all 0x01 bytes by "\1\x31" because the transport protocol allows no \0-bytes.
In cases where:
- the replacing and the to-replaced string have different lengths,
- there are many occurences of the to-replaced string within the source string and
- the source string is large,
the provided solutions cannot be applied (because they replace only single characters) or have a performance problem, because they would call string::replace several times which generates copies of the size of the blob over and over. (I do not know the boost solution, maybe it is OK from that perspective)
This one walks along all occurrences in the source string and builds the new string piece by piece once:
void replaceAll(std::string& source, const std::string& from, const std::string& to)
{
std::string newString;
newString.reserve(source.length()); // avoids a few memory allocations
std::string::size_type lastPos = 0;
std::string::size_type findPos;
while(std::string::npos != (findPos = source.find(from, lastPos)))
{
newString.append(source, lastPos, findPos - lastPos);
newString += to;
lastPos = findPos + from.length();
}
// Care for the rest after last occurrence
newString += source.substr(lastPos);
source.swap(newString);
}
回答5:
A simple find and replace for a single character would go something like:
s.replace(s.find("x"), 1, "y")
To do this for the whole string, the easy thing to do would be to loop until your s.find starts returning npos. I suppose you could also catch range_error to exit the loop, but that's kinda ugly.
回答6:
If you're looking to replace more than a single character, and are dealing only with std::string, then this snippet would work, replacing sNeedle in sHaystack with sReplace, and sNeedle and sReplace do not need to be the same size. This routine uses the while loop to replace all occurrences, rather than just the first one found from left to right.
while(sHaystack.find(sNeedle) != std::string::npos) {
sHaystack.replace(sHaystack.find(sNeedle),sNeedle.size(),sReplace);
}
回答7:
As Kirill suggested, either use the replace method or iterate along the string replacing each char independently.
Alternatively you can use the find method or find_first_of depending on what you need to do. None of these solutions will do the job in one go, but with a few extra lines of code you ought to make them work for you. :-)
回答8:
#include <iostream>
#include <string>
using namespace std;
// Replace function..
string replace(string word, string target, string replacement){
int len, loop=0;
string nword="", let;
len=word.length();
len--;
while(loop<=len){
let=word.substr(loop, 1);
if(let==target){
nword=nword+replacement;
}else{
nword=nword+let;
}
loop++;
}
return nword;
}
//Main..
int main() {
string word;
cout<<"Enter Word: ";
cin>>word;
cout<<replace(word, "x", "y")<<endl;
return 0;
}
回答9:
This works! I used something similar to this for a bookstore app, where the inventory was stored in a CSV (like a .dat file). But in the case of a single char, meaning the replacer is only a single char, e.g.'|', it must be in double quotes "|" in order not to throw an invalid conversion const char.
#include <iostream>
#include <string>
using namespace std;
int main()
{
int count = 0; // for the number of occurences.
// final hold variable of corrected word up to the npos=j
string holdWord = "";
// a temp var in order to replace 0 to new npos
string holdTemp = "";
// a csv for a an entry in a book store
string holdLetter = "Big Java 7th Ed,Horstman,978-1118431115,99.85";
// j = npos
for (int j = 0; j < holdLetter.length(); j++) {
if (holdLetter[j] == ',') {
if ( count == 0 )
{
holdWord = holdLetter.replace(j, 1, " | ");
}
else {
string holdTemp1 = holdLetter.replace(j, 1, " | ");
// since replacement is three positions in length,
// must replace new replacement's 0 to npos-3, with
// the 0 to npos - 3 of the old replacement
holdTemp = holdTemp1.replace(0, j-3, holdWord, 0, j-3);
holdWord = "";
holdWord = holdTemp;
}
holdTemp = "";
count++;
}
}
cout << holdWord << endl;
return 0;
}
// result:
Big Java 7th Ed | Horstman | 978-1118431115 | 99.85
Uncustomarily I am using CentOS currently, so my compiler version is below . The C++ version (g++), C++98 default:
g++ (GCC) 4.8.5 20150623 (Red Hat 4.8.5-4)
Copyright (C) 2015 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
回答10:
If you're willing to use std::strings, you can use this sample-app's strsub function as-is, or update it if you want it to take a different type or set of parameters to achieve roughly the same goal. Basically, it uses the properties and functionalities of std::string to quickly erase the matching set of characters, and insert the desired characters directly within the std::string. Every time it does this replacement operation, the offset updates if it can still find matching chars to replace, and if it can't due to nothing more to replace, it returns the string in its state from the last update.
#include <iostream>
#include <string>
std::string strsub(std::string stringToModify,
std::string charsToReplace,
std::string replacementChars);
int main()
{
std::string silly_typos = "annoiiyyyng syyyllii tiipos.";
std::cout << "Look at these " << silly_typos << std::endl;
silly_typos = strsub(silly_typos, "yyy", "i");
std::cout << "After a little elbow-grease, a few less " << silly_typos << std::endl;
silly_typos = strsub(silly_typos, "ii", "y");
std::cout << "There, no more " << silly_typos << std::endl;
return 0;
}
std::string strsub(std::string stringToModify,
std::string charsToReplace,
std::string replacementChars)
{
std::string this_string = stringToModify;
std::size_t this_occurrence = this_string.find(charsToReplace);
while (this_occurrence != std::string::npos)
{
this_string.erase(this_occurrence, charsToReplace.size());
this_string.insert(this_occurrence, replacementChars);
this_occurrence = this_string.find(charsToReplace,
this_occurrence + replacementChars.size());
}
return this_string;
}
If you don't want to rely on using std::strings as your parameters so you can pass in C-style strings instead, you can see the updated sample below:
#include <iostream>
#include <string>
std::string strsub(const char * stringToModify,
const char * charsToReplace,
const char * replacementChars,
uint64_t sizeOfCharsToReplace,
uint64_t sizeOfReplacementChars);
int main()
{
std::string silly_typos = "annoiiyyyng syyyllii tiipos.";
std::cout << "Look at these " << silly_typos << std::endl;
silly_typos = strsub(silly_typos.c_str(), "yyy", "i", 3, 1);
std::cout << "After a little elbow-grease, a few less " << silly_typos << std::endl;
silly_typos = strsub(silly_typos.c_str(), "ii", "y", 2, 1);
std::cout << "There, no more " << silly_typos << std::endl;
return 0;
}
std::string strsub(const char * stringToModify,
const char * charsToReplace,
const char * replacementChars,
uint64_t sizeOfCharsToReplace,
uint64_t sizeOfReplacementChars)
{
std::string this_string = stringToModify;
std::size_t this_occurrence = this_string.find(charsToReplace);
while (this_occurrence != std::string::npos)
{
this_string.erase(this_occurrence, sizeOfCharsToReplace);
this_string.insert(this_occurrence, replacementChars);
this_occurrence = this_string.find(charsToReplace,
this_occurrence + sizeOfReplacementChars);
}
return this_string;
}
回答11:
For simple situations this works pretty well without using any other library then std::string (which is already in use).
Replace all occurences of character a with character b in some_string:
for (size_t i = 0; i < some_string.size(); ++i) {
if (some_string[i] == 'a') {
some_string.replace(i, 1, "b");
}
}
If the string is large or multiple calls to replace is an issue, you can apply the technique mentioned in this answer: https://stackoverflow.com/a/29752943/3622300
回答12:
Old School :-)
std::string str = "H:/recursos/audio/youtube/libre/falta/";
for (int i = 0; i < str.size(); i++) {
if (str[i] == '/') {
str[i] = '\\';
}
}
std::cout << str;
Result:
H:\recursos\audio\youtube\libre\falta\
回答13:
What about Abseil StrReplaceAll? From the header file:
// This file defines `absl::StrReplaceAll()`, a general-purpose string
// replacement function designed for large, arbitrary text substitutions,
// especially on strings which you are receiving from some other system for
// further processing (e.g. processing regular expressions, escaping HTML
// entities, etc.). `StrReplaceAll` is designed to be efficient even when only
// one substitution is being performed, or when substitution is rare.
//
// If the string being modified is known at compile-time, and the substitutions
// vary, `absl::Substitute()` may be a better choice.
//
// Example:
//
// std::string html_escaped = absl::StrReplaceAll(user_input, {
// {"&", "&"},
// {"<", "<"},
// {">", ">"},
// {"\"", """},
// {"'", "'"}});
来源：https://stackoverflow.com/questions/2896600/how-to-replace-all-occurrences-of-a-character-in-string