用另一个子字符串 C++ 替换子字符串
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【中文标题】用另一个子字符串 C++ 替换子字符串【英文标题】:Replace substring with another substring C++ 【发布时间】:2011-06-06 07:45:19 【问题描述】: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
【问题讨论】:
几乎是 ***.com/questions/3418231/… 的副本,在接受的答案中有更强大的解决方案。 【参考方案1】:C++ 中没有一个内置函数可以做到这一点。如果您想用另一个子字符串替换一个子字符串的所有实例,您可以通过混合调用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,我将index
增加了插入字符串的长度。在这个特定的示例中 - 用 "def"
替换 "abc"
- 这实际上不是必需的。但是,在更一般的设置中,跳过刚刚被替换的字符串很重要。例如,如果您想用"abcabc"
替换"abc"
,而不跳过新替换的字符串段,此代码将不断替换部分新替换的字符串,直到内存耗尽。独立地,无论如何跳过这些新字符可能会稍微快一些,因为这样做可以通过 string::find
函数节省一些时间和精力。
希望这会有所帮助!
【讨论】:
我认为您不需要增加索引,因为您已经替换了数据,因此无论如何它都不会拾取它。 @Aidiakapi 如果将其转换为通用函数,它不会陷入无限循环,因为它会将搜索位置 (index
) 推进到被替换的字符串部分.
@TimR。你说得对,我在回复 rossb83,他说索引的增加是不必要的。只是想防止错误信息。所以对于其他所有人:将索引增加替换字符串的长度(在本例中为 3)是必要的。不要从代码示例中删除它。
@rossb83 cmets 需要清理或说明。有 5 个赞成的评论说您不必增加索引,然后一个人说您需要以粗体表示。这对来这里学习的人没有帮助。
@JulianCienfuegos 我刚刚更新了解决这个问题的答案 - 感谢您指出这一点! (另外,Aidiakapi 是其他人……不确定那是谁。)【参考方案2】:
如果您确定字符串中存在所需的子字符串,那么这会将第一次出现的"abc"
替换为"hij"
test.replace( test.find("abc"), 3, "hij");
如果你在测试中没有“abc”,它会崩溃,所以小心使用它。
【讨论】:
【参考方案3】: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,因此它不允许像您的示例中那样漂亮的语法,但以下将做等效:
test = string_replace( string_replace( test, "abc", "hij"), "def", "klm")
【讨论】:
【参考方案4】: 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;
我认为这符合您的要求,只需很少的代码!
【讨论】:
您没有考虑多次出现/替换【参考方案5】: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"
);
【讨论】:
杰伊。我需要 boost 来替换所有子字符串。 Boost 大多是矫枉过正。【参考方案6】:替换子字符串应该没有那么难。
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,这里有一个优化的函数,它修改输入字符串,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();
测试:
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
【讨论】:
需要添加检查if (search.empty()) return;
以避免在传递空“搜索”时出现无限循环。
尝试了 ReplaceString 功能 - 无效。但回答如下: str.replace(str.find(str2),str2.length(),str3);很简单,效果很好。【参考方案7】:
如果替换字符串的长度与要替换的字符串的长度不同,我认为所有解决方案都会失败。 (搜索“abc”并替换为“xxxxxx”) 一般的方法可能是:
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 );
【讨论】:
【参考方案8】:概括 rotmax 的答案,这里是搜索和替换字符串中所有实例的完整解决方案。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 );
【讨论】:
谢谢。使用几年没有问题;然而,最终需要将oldString
和newString
都变成const
参数。【参考方案9】:
str.replace(str.find(str2),str2.length(),str3);
在哪里
str
是基本字符串
str2
是要查找的子字符串
str3
是替换子串
【讨论】:
这只会替换第一次出现,不是吗? 我建议确保 str.find(str2) 的结果不等于 std::string::npos auto found = str.find(str2); if(found != std::string::npos) str.replace(found, str2.length(), str3); 我不打算用这个来编写整个应用程序,但是在没有对输入进行任何检查的情况下,有些情况是未定义的......【参考方案10】:@Czarek Tomczak 改进的版本。
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();
【讨论】:
【参考方案11】:在c++11,你可以使用std::regex_replace
:
#include <string>
#include <regex>
std::string test = "abc def abc def";
test = std::regex_replace(test, std::regex("def"), "klm"); // replace 'def' -> 'klm'
// test = "abc klm abc klm"
【讨论】:
如果我们有 c++11 那就太好了!! 小心这不能很好地概括,你最终可能会传递一些std:regex
解释不同的东西,比如 std::regex_replace(test, std::regex("."), "klm")
...【参考方案12】:
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;
;
【讨论】:
【参考方案13】:这是一个使用递归的解决方案,它用另一个子字符串替换所有出现的子字符串。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);
使用示例:
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;
【讨论】:
【参考方案14】: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);
【讨论】:
【参考方案15】:std::string replace(std::string str, std::string substr1, std::string substr2)
for (size_t index = str.find(substr1, 0); index != std::string::npos && substr1.length(); index = str.find(substr1, index + substr2.length() ) )
str.replace(index, substr1.length(), substr2);
return str;
不需要任何额外库的简短解决方案。
【讨论】:
这个问题还有 14 个其他答案。为什么不解释一下为什么你的更好? 看起来是迄今为止最优雅的答案,没有任何过度设计【参考方案16】: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;
【讨论】:
【参考方案17】:#include <string>
第一:
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;
【讨论】:
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