如何用C ++中的另一个子字符串替换字符串中的子字符串，我可以使用哪些函数？
eg: string test = "abc def abc def";
test.replace("abc", "hij").replace("def", "klm"); //replace occurrence of abc and def with other substring
答案 0 :(得分：63)
C ++中没有一个内置函数可以做到这一点。如果您想将一个子字符串的所有实例替换为另一个子字符串，可以通过混合调用string::find和string::replace来实现。例如：
size_t index = 0;
while (true) {
/* Locate the substring to replace. */
index = str.find("abc", index);
if (index == std::string::npos) break;
/* Make the replacement. */
str.replace(index, 3, "def");
/* Advance index forward so the next iteration doesn't pick it up as well. */
index += 3;
}
在这段代码的最后一行中，我将index增加了插入字符串的字符串的长度。在此特定示例中 - 将"abc"替换为"def" - 这实际上不是必需的。但是，在更一般的设置中，跳过刚刚替换的字符串非常重要。例如，如果要将"abc"替换为"abcabc"，而不跳过新替换的字符串段，则此代码将持续替换部分新替换的字符串，直到内存耗尽为止。无论如何，跳过这些新角色可能会稍微快一些，因为这样做可以节省string::find函数的时间和精力。
希望这有帮助！
答案 1 :(得分：57)
Boost String Algorithms Library方式：
#include <boost/algorithm/string/replace.hpp>
{ // 1.
string test = "abc def abc def";
boost::replace_all(test, "abc", "hij");
boost::replace_all(test, "def", "klm");
}
{ // 2.
string test = boost::replace_all_copy
( boost::replace_all_copy<string>("abc def abc def", "abc", "hij")
, "def"
, "klm"
);
}
答案 2 :(得分：35)
我认为如果替换字符串的长度与要替换的字符串的长度不同，则所有解决方案都将失败。 （搜索“abc”并替换为“xxxxxx”） 一般方法可能是：
void replaceAll( string &s, const string &search, const string &replace ) {
for( size_t pos = 0; ; pos += replace.length() ) {
// Locate the substring to replace
pos = s.find( search, pos );
if( pos == string::npos ) break;
// Replace by erasing and inserting
s.erase( pos, search.length() );
s.insert( pos, replace );
}
}
答案 3 :(得分：29)
在C ++ 11中，您可以使用UICollectionViewDelegateFlowLayout：
regex_replace答案 4 :(得分：25)
str.replace(str.find(str2),str2.length(),str3);
其中
str是基本字符串str2是要查找的子字符串str3是替换子字符串答案 5 :(得分：15)
替换子串不应该那么难。
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;
}
如果你需要性能，这里是一个修改输入字符串的优化函数，它不会创建字符串的副本：
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();
}
}
试验：
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 changed: "
<< input << std::endl;
ReplaceStringInPlace(input, "bc", "??");
std::cout << "ReplaceStringInPlace() input string modified: "
<< input << std::endl;
输出：
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
答案 6 :(得分：5)
using std::string;
string string_replace( string src, string const& target, string const& repl)
{
// handle error situations/trivial cases
if (target.length() == 0) {
// searching for a match to the empty string will result in
// an infinite loop
// it might make sense to throw an exception for this case
return src;
}
if (src.length() == 0) {
return src; // nothing to match against
}
size_t idx = 0;
for (;;) {
idx = src.find( target, idx);
if (idx == string::npos) break;
src.replace( idx, target.length(), repl);
idx += repl.length();
}
return src;
}
由于它不是string类的成员，因此它不允许使用与示例中相同的语法，但以下内容将执行相同的操作：
test = string_replace( string_replace( test, "abc", "hij"), "def", "klm")
答案 7 :(得分：2)
如果您确定字符串中存在必需的子字符串，那么这将替换"abc"到"hij"
test.replace( test.find("abc"), 3, "hij");
如果你在测试中没有“abc”，它会崩溃，所以要小心使用它。
答案 8 :(得分：2)
概括了rotmax的答案，这是一个完整的搜索和解决方案。替换字符串中的所有实例。如果两个子字符串的大小不同，则使用string :: erase和string :: insert。替换子字符串，否则使用更快的string :: replace。
void FindReplace(string& line, string& oldString, string& newString) {
const size_t oldSize = oldString.length();
// do nothing if line is shorter than the string to find
if( oldSize > line.length() ) return;
const size_t newSize = newString.length();
for( size_t pos = 0; ; pos += newSize ) {
// Locate the substring to replace
pos = line.find( oldString, pos );
if( pos == string::npos ) return;
if( oldSize == newSize ) {
// if they're same size, use std::string::replace
line.replace( pos, oldSize, newString );
} else {
// if not same size, replace by erasing and inserting
line.erase( pos, oldSize );
line.insert( pos, newString );
}
}
}
答案 9 :(得分：1)
这是我使用构建器策略编写的解决方案：
#include <string>
#include <sstream>
using std::string;
using std::stringstream;
string stringReplace (const string& source,
const string& toReplace,
const string& replaceWith)
{
size_t pos = 0;
size_t cursor = 0;
int repLen = toReplace.length();
stringstream builder;
do
{
pos = source.find(toReplace, cursor);
if (string::npos != pos)
{
//copy up to the match, then append the replacement
builder << source.substr(cursor, pos - cursor);
builder << replaceWith;
// skip past the match
cursor = pos + repLen;
}
}
while (string::npos != pos);
//copy the remainder
builder << source.substr(cursor);
return (builder.str());
}
测试：
void addTestResult (const string&& testId, bool pass)
{
...
}
void testStringReplace()
{
string source = "123456789012345678901234567890";
string toReplace = "567";
string replaceWith = "abcd";
string result = stringReplace (source, toReplace, replaceWith);
string expected = "1234abcd8901234abcd8901234abcd890";
bool pass = (0 == result.compare(expected));
addTestResult("567", pass);
source = "123456789012345678901234567890";
toReplace = "123";
replaceWith = "-";
result = stringReplace(source, toReplace, replaceWith);
expected = "-4567890-4567890-4567890";
pass = (0 == result.compare(expected));
addTestResult("start", pass);
source = "123456789012345678901234567890";
toReplace = "0";
replaceWith = "";
result = stringReplace(source, toReplace, replaceWith);
expected = "123456789123456789123456789";
pass = (0 == result.compare(expected));
addTestResult("end", pass);
source = "123123456789012345678901234567890";
toReplace = "123";
replaceWith = "-";
result = stringReplace(source, toReplace, replaceWith);
expected = "--4567890-4567890-4567890";
pass = (0 == result.compare(expected));
addTestResult("concat", pass);
source = "1232323323123456789012345678901234567890";
toReplace = "323";
replaceWith = "-";
result = stringReplace(source, toReplace, replaceWith);
expected = "12-23-123456789012345678901234567890";
pass = (0 == result.compare(expected));
addTestResult("interleaved", pass);
source = "1232323323123456789012345678901234567890";
toReplace = "===";
replaceWith = "-";
result = utils_stringReplace(source, toReplace, replaceWith);
expected = source;
pass = (0 == result.compare(expected));
addTestResult("no match", pass);
}
答案 10 :(得分：0)
@Czarek Tomczak提供的版本。
同时允许std::string和std::wstring。
template <typename charType>
void ReplaceSubstring(std::basic_string<charType>& subject,
const std::basic_string<charType>& search,
const std::basic_string<charType>& replace)
{
if (search.empty()) { return; }
typename std::basic_string<charType>::size_type pos = 0;
while((pos = subject.find(search, pos)) != std::basic_string<charType>::npos) {
subject.replace(pos, search.length(), replace);
pos += replace.length();
}
}
答案 11 :(得分：0)
string & replace(string & subj, string old, string neu)
{
size_t uiui = subj.find(old);
if (uiui != string::npos)
{
subj.erase(uiui, old.size());
subj.insert(uiui, neu);
}
return subj;
}
我认为这很少符合您的要求！
答案 12 :(得分：0)
std::string replace(const std::string & in
, const std::string & from
, const std::string & to){
if(from.size() == 0 ) return in;
std::string out = "";
std::string tmp = "";
for(int i = 0, ii = -1; i < in.size(); ++i) {
// change ii
if ( ii < 0 && from[0] == in[i] ) {
ii = 0;
tmp = from[0];
} else if( ii >= 0 && ii < from.size()-1 ) {
ii ++ ;
tmp = tmp + in[i];
if(from[ii] == in[i]) {
} else {
out = out + tmp;
tmp = "";
ii = -1;
}
} else {
out = out + in[i];
}
if( tmp == from ) {
out = out + to;
tmp = "";
ii = -1;
}
}
return out;
};
答案 13 :(得分：0)
这是一个使用递归的解决方案，它将所有出现的子字符串替换为另一个子字符串。无论字符串的大小如何，这都有效。
std::string ReplaceString(const std::string source_string, const std::string old_substring, const std::string new_substring)
{
// Can't replace nothing.
if (old_substring.empty())
return source_string;
// Find the first occurrence of the substring we want to replace.
size_t substring_position = source_string.find(old_substring);
// If not found, there is nothing to replace.
if (substring_position == std::string::npos)
return source_string;
// Return the part of the source string until the first occurance of the old substring + the new replacement substring + the result of the same function on the remainder.
return source_string.substr(0,substring_position) + new_substring + ReplaceString(source_string.substr(substring_position + old_substring.length(),source_string.length() - (substring_position + old_substring.length())), old_substring, new_substring);
}
用法示例：
std::string my_cpp_string = "This string is unmodified. You heard me right, it's unmodified.";
std::cout << "The original C++ string is:\n" << my_cpp_string << std::endl;
my_cpp_string = ReplaceString(my_cpp_string, "unmodified", "modified");
std::cout << "The final C++ string is:\n" << my_cpp_string << std::endl;
答案 14 :(得分：0)
size_t index = 0;
std::string str = "T X T", substr1=" " /*To replace*/, substr2="-" /*replace with this*/;
for (index = str.find(substr1, index); index != std::string::npos; index = str.find(substr1, index + strlen(substr1.c_str())) )
str.replace(index, strlen(substr1.c_str()), substr2);
答案 15 :(得分：0)
std::string replace(std::string str, const std::string& sub1, const std::string& sub2)
{
if (sub1.empty())
return str;
std::size_t pos;
while ((pos = str.find(sub1)) != std::string::npos)
str.replace(pos, sub1.size(), sub2);
return str;
}
答案 16 :(得分：0)
#include <string>
首先：
void replace_first(std::string& text, const std::string& from,
const std::string& to)
{
const auto at = text.find(from, 0);
if (at != std::string::npos)
text.replace(at, from.length(), to);
}
全部：
void replace_all(std::string& text, const std::string& from,
const std::string& to)
{
for (auto at = text.find(from, 0); at != std::string::npos;
at = text.find(from, at + to.length()))
{
text.replace(at, from.length(), to);
}
}
计数：
size_t replace_count(std::string& text,
const std::string& from, const std::string& to)
{
size_t count = 0;
for (auto at = text.find(from, 0); at != std::string::npos;
at = text.find(from, at + to.length()))
{
++count;
text.replace(at, from.length(), to);
}
return count;
}
复制：
std::string replace_all_copy(const std::string& text,
const std::string& from, const std::string& to)
{
auto copy = text;
replace_all(copy, from, to);
return copy;
}