#include <iostream>
#include <regex>
int main() {
std::wstring str = LR"(
bst.enable_adb_access="1"
)";
std::wregex re(L"(?<=bst\\.enable_adb_access.*?)\\d ");
str = std::regex_replace(str, re, L"0");
std::wcout << str << std::endl;
}
error:
terminate called after throwing an instance of 'std::regex_error'
what(): Invalid special open parenthesis.
https://regex101.com/r/a33eFL/1
Whats wrong with the parenthesis?
CodePudding user response：
Well, this is one illustration why the plural of "regex" is "regrets"...
C accepts several flavours of regexes, but none of them seems to understand lookbehinds. Default modified ECMAScript flavour only accepts lookaheads. I'm not 100% sure about POSIX, awk and grep flavours, but none of them seems to have any lookarounds whatsoever.
Fortunately, you can get the same effect without lookarounds, using capturing group. I had to change format string rules to sed, because default ECMAScript rules allow for two-digit backreferences.
#include <iostream>
#include <regex>
int main() {
std::wstring str = LR"(
bst.enable_adb_access="1"
)";
std::wregex re(L"(bst\\.enable_adb_access.*?)\\d ");
str = std::regex_replace(str, re, L"\\10", std::regex_constants::format_sed);
std::wcout << str << std::endl;
}
CodePudding user response：
You don't need to use a lookbehind for this situation. Simply use a normal capturing group and include it in the replacement string:
#include <iostream>
#include <regex>
using namespace std;
int main() {
std::wstring str = LR"(
bst.enable_adb_access="1"
)";
std::wregex re(L"(bst\\.enable_adb_access.*?)\\d ");
str = std::regex_replace(str, re, L"$010");
std::wcout << str << std::endl;
}
Output:
bst.enable_adb_access="0"
Note that because the substitution for the capturing group is followed by a digit, we need to use the $nn format for the group number (hence $010), otherwise $10 could - dependent on the compiler - be interpreted as replacing with capture group 10.