I am trying to perform a regex search, but instead of typing the pattern (e.g std::regex reg = "\.") I want to give regex a string input. Here is my code where I convert string to const char* and pass it to regex_search:
void trigger_regex(string name, string body)
{
map<string,string>::iterator it;
it=test.mydata.find(name);
if( it == test.mydata.end())
{
std::cout<<"REGEX not found"<<endl;
}
else
cout << "Found REGEX with name: " << it->first << " and rule: " << it->second << endl ;
string subject = body;
string rule = it -> second;
const char * pattern = rule.c_str();
regex reg = (pattern);
smatch match;
while (regex_search (subject,match,reg)) {
for (auto x:match) std::cout << x << " ";
std::cout << std::endl;
body = match.suffix().str();
}
}
I get error:
conversion from 'const char*' to non-scalar type 'std::regex {aka
std::basic_regex}' requested
regex reg1 = (rule1);
any help pls?
If you look at e.g. this std::regex constructor reference you will see that the constructor taking just a string (the second constructor in the list) is marked explicit. That means you must explicitly use that constructor, you can't just hope to convert a string to a std::regex object.
In other words, the solution is to define and initialize the object like this
std::regex reg1{pattern};
Related
I have a function with the following if statements:
if (name.length() < 10 || name.length() > 64)
{
return false;
}
if (name.front() == L'.' || name.front() == L' ')
{
return false;
}
I was curious to see if can do this using the following regular expression:
^(?!\ |\.)([A-Za-z]{10,46}$)
to dissect the above expression the first part ^(?!\ |.) preforms a negative look ahead to assert that it is impossible for the string to start with space or dot(.) and the second part should take care of the string length condition. I wrote the following to test the expression out:
std::string randomStrings [] = {" hello",
" hellllloooo",
"\.hello",
".zoidbergvddd",
"asdasdsadadasdasdasdadasdsad"};
std::regex txt_regex("^(?!\ |\.)([A-Za-z]{10,46}$)");
for (const auto& str : randomStrings)
{
std::cout << str << ": " << std::boolalpha << std::regex_match(str, txt_regex) << '\n';
}
I expected the last one to to match since it does not start with space or dot(.) and it meets the length criteria. However, this is what I got:
hello: false
hellllloooo: false
.hello: false
.zoidbergvddd: false
asdasdsadadasdasdasdadasdsad: false
Did I miss something trivial here or this is not possible using regex? It seems like it should be.
Feel free to suggest a better title, I tried to be as descriptive as possible.
Change your regular expression to: "^(?![\\s.])([A-Za-z]{10,46}$)" and it will work.
\s refers to any whitespace and you need to escape the \ inside the string and that's why it becomes \\s.
You can also check this link
You need to turn on compiler warnings. It would have told you that you have an unknown escape sequence in your regex. I recommend using a raw literal.
#include <iostream>
#include <regex>
int main() {
std::string randomStrings[] = { " hello", " hellllloooo", ".hello",
".zoidbergvddd", "asdasdsadadasdasdasdadasdsad" };
std::regex txt_regex(R"foo(^(?!\ |\.)([A-Za-z]{10,46}$))foo");
for (const auto& str : randomStrings) {
std::cout << str << ": " << std::boolalpha
<< std::regex_match(str, txt_regex) << '\n';
}
}
clang++-3.8 gives
hello: false
hellllloooo: false
.hello: false
.zoidbergvddd: false
asdasdsadadasdasdasdadasdsad: true
I'm trying to write a recursive descent parser and am trying to search a match a regex within a string inputted by the user. I am trying to do the following to try to understand the <regex> library offered by C++11, but I'm getting unexpected results.
std::string expression = "2+2+2";
std::regex re("[-+*/()]");
std::smatch m;
std::cout << "My expression is " << expression << std::endl;
if(std::regex_search(expression, re)) {
std::cout << "Found a match!" << std::endl;
}
std::regex_match(expression, m, re);
std::cout << "matches:" << std::endl;
for (auto it = m.begin(); it!=m.end(); ++it) {
std::cout << *it << std::endl;
}
So based on my regular expression, I expect it to output
Found a match!
matches:
+
+
However, the output I get is:
My expression is 2+2+2
Found a match!
matches:
I feel like I'm making a stupid mistake, but I can't seem to figure out why there's a discrepancy between the outputs.
Thanks,
erip
You've got a few issues. First, let's look at some working code:
#include <regex>
#include <iostream>
int main() {
std::string expr = "2+2+2";
std::regex re("[+\\-*/()]");
const auto operators_begin = std::sregex_iterator(expr.begin(), expr.end(), re);
const auto operators_end = std::sregex_iterator();
std::cout << "Count: " << std::distance(operators_begin, operators_end) << "\n";
for (auto i = operators_begin; i != operators_end; ++i) {
std::smatch match = *i;
std::cout << match.str() << "\n";
}
}
Output:
Count: 2
+
+
Issues with your code:
regex_match() returns false.
You don't have any capturing groups in the regular expression. So even if regex_match() returned true, it wouldn't capture anything.
The number of captures in a regex_match can be determined strictly by looking at the regular expression. So my re will capture exactly one group.
But we want to apply this regular expression on our string multiple times, because we want to find all of the matches. The tool for that is regex_iterator.
We also needed to escape the - in the regular expression. The minus has a special meaning within character classes.
I found the regex_iterator, which solved the problem. Here's the working code:
std::regex re("[-+*/()]");
std::smatch m;
std::cout << "My expression is " << expression << std::endl;
if(std::regex_search(expression, re)) {
std::cout << "Found a match!" << std::endl;
}
try {
std::sregex_iterator next(expression.begin(), expression.end(), re);
std::sregex_iterator end;
while (next != end) {
std::smatch match = *next;
std::cout << match.str() << "\n";
next++;
}
} catch (std::regex_error& e) {
// Syntax error in the regular expression
}
I am pretty sure this is no hard task, but I don’t get what causes the problems, and I would like to really understand this, since I often have some pointer/array/cast-related problems:
I store the bounding box values in a double*
// this is the calss-variable
double *_boundingBox;
// this is where I put some data in it
double boundingBox[6];
boundingBox[0] =
.
.
.
boundingBox[6] = ....;
// set pointer to boundingbox
_boundingBox = &boundingBox;
and in an other class I use this
double* getBoundingBoxInfo()
{
return _boundingBox;
}
to get my bounding box-data, which I would like to input in a QLabel as QString
double boundingBox[6];
boundingBox[0] = *_drawer->getBoundingBoxInfo();
std::string stringX = "x start: " << boundingBox[0] << "\tx end: " << boundingBox[3];
QLabel *labelX = new QLabel(QString(stringX.c_str()));
The current compile-error is
error: invalid operands of types ‘const char [10]’ and ‘double’ to binary ‘operator<<’
Could someone please tell me how this should work? Am I using double*, double[] and string the way they are supposed to be used?
You cannot stream data into a std::string as-is. A solution is to use std::ostringstream:
std::ostringstream out;
out << "x start: " << boundingBox[0] << "\tx end: " << boundingBox[3];
std::string stringX = out.str();
The compilation error you're getting is for "x start: " << boundingBox[0].
The type of "x start: " is const char*, and the type of boundingBox[0] is double.
But there is no definition for operator<<(const char*,double).
You can get this to work by using an ostringstream object:
ostringstream oss;
oss << "x start: " << boundingBox[0] << "\tx end: " << boundingBox[3];
std::string stringX = oss.str();
As a side-note, when you set _boundingBox = &boundingBox, you don't need the &, because boundingBox is an array, so in essence, boundingBox == &boundingBox.
The reason for this (in case you were wondering) is that arrays don't have an l-value, and you cannot change the value of an array (for example, you cannot do boundingBox = ...).
QString provides everything http://qt-project.org/doc/qt-4.8/qstring.html#arg-20
so just use
QString("some text for double value: %1").arg(yourdouble, <additional parameters>)
and in your case:
... new QLabel(QString("x start: %1\tx end: %2").arg(boundingBox[0]).arg(boundingBox[3]));
Hello I am new to regular expressions and from what I understood from the c++ reference website it is possible to get match results.
My question is: how do I retrieve these results? What is the difference between smatch and cmatch? For example, I have a string consisting of date and time and this is the regular expression I wrote:
"(1[0-2]|0?[1-9])([:][0-5][0-9])?(am|pm)"
Now when I do a regex_search with the string and the above expression, I can find whether there is a time in the string or not. But I want to store that time in a structure so I can separate hours and minutes. I am using Visual studio 2010 c++.
If you use e.g. std::regex_search then it fills in a std::match_result where you can use the operator[] to get the matched strings.
Edit: Example program:
#include <iostream>
#include <string>
#include <regex>
void test_regex_search(const std::string& input)
{
std::regex rgx("((1[0-2])|(0?[1-9])):([0-5][0-9])((am)|(pm))");
std::smatch match;
if (std::regex_search(input.begin(), input.end(), match, rgx))
{
std::cout << "Match\n";
//for (auto m : match)
// std::cout << " submatch " << m << '\n';
std::cout << "match[1] = " << match[1] << '\n';
std::cout << "match[4] = " << match[4] << '\n';
std::cout << "match[5] = " << match[5] << '\n';
}
else
std::cout << "No match\n";
}
int main()
{
const std::string time1 = "9:45pm";
const std::string time2 = "11:53am";
test_regex_search(time1);
test_regex_search(time2);
}
Output from the program:
Match
match[1] = 9
match[4] = 45
match[5] = pm
Match
match[1] = 11
match[4] = 53
match[5] = am
Just use named groups.
(?<hour>(1[0-2]|0?[1-9]))([:](?<minute>[0-5][0-9]))?(am|pm)
Ok, vs2010 doesn't support named groups. You already using unnamed capture groups. Go through them.
I have an application in which I need to combine strings within a variable like so:
int int_arr[4];
int_arr[1] = 123;
int_arr[2] = 456;
int_arr[3] = 789;
int_arr[4] = 10;
std::string _string = "Text " + int_arr[1] + " Text " + int_arr[2] + " Text " + int_arr[3] + " Text " + int_arr[4];
It gives me the compile error
Error C2210: '+' Operator cannot add pointers" on the second string of the expression.
As far as I can tell I am combining string literals and integers, not pointers.
Is there another concatenation operator that I should be using? Or is the expression just completely wrong and should figure out another way to implement this?
BTW I am using Visual Studio 2010
Neither C nor C++ allow concatenation of const char * and int. Even C++'s std::string, doesn't concatenate integers. Use streams instead:
std::stringstream ss;
ss << "Text " << int_arr[1] << " Text " << int_arr[2] << " Text " << int_arr[3] << " Text " << int_arr[4];
std::string _string = ss.str();
You can do this in Java since it uses the toString() method automatically on each part.
If you want to do it the same way in C++, you'll have to explicitly convert those integer to strings in order for this to work.
Something like:
#include <iostream>
#include <sstream>
std::string intToStr (int i) {
std::ostringstream s;
s << i;
return s.str();
}
int main (void) {
int var = 7;
std::string s = "Var is '" + intToStr(var) + "'";
std::cout << s << std::endl;
return 0;
}
Of course, you can just use:
std::ostringstream os;
os << "Var is '" << var << "'";
std::string s = os.str();
which is a lot easier.
A string literal becomes a pointer in this context. Not a std::string. (Well, to be pedantically correct, string literals are character arrays, but the name of an array has an implicit conversion to a pointer. One predefined form of the + operator takes a pointer left-argument and an integral right argument, which is the best match, so the implicit conversion takes place here. No user-defined conversion can ever take precedence over this built-in conversion, according to the C++ overloading rules.).
You should study a good C++ book, we have a list here on SO.
A string literal is an expression returning a pointer const char*.
std::stringstream _string_stream;
_string_stream << "Text " << int_arr[1] << " Text " << int_arr[2] << " Text " << int_arr[3] << " Text " << int_arr[4];
std::string _string = _string_stream.str();