I try to build an std::string in the form of "start:Pdc1;Pdc2;Pdc3;"
With following code I can build the repeated "Pdc" and the incremental string "123" but I'm unable to combine the two strings.
#include <string>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <iterator>
#include <numeric>
int main()
{
std::ostringstream ss;
std::string hdr("start:");
std::fill_n(std::ostream_iterator<std::string>(ss), 3, "Pdc;");
hdr.append(ss.str());
std::string v("abc");
std::iota(v.begin(), v.end(), '1');
std::cout << hdr << std::endl;
std::cout << v << std::endl;
std::cout << "Expected output: start:Pdc1;Pdc2;Pdc3;" << std::endl;
return 0;
}
How can I build this string? Preferable without a while or for loop.
The expected output is: start:Pdc1;Pdc2;Pdc3;
std::strings can be concatenated via their operator+ (or +=) and integers can be converted via std::to_string:
std::string res("start:");
for (int i=0;i<3;++i){
res += "Pdc" + std::to_string(i+1) + ";";
}
std::cout << res << "\n";
If you like you can use an algorithm instead of the handwritten loop, but it will still be a loop (your code has 2 loops, but only 1 is needed).
Code to generate your expected string, though with a small for loop.
#include <iostream>
#include <string>
#include <sstream>
std::string cmd(const std::size_t N)
{
std::ostringstream os;
os << "start:";
for(std::size_t n = 1; n <= N; ++n) os << "Pdc" << n << ";";
return os.str();
}
int main()
{
std::cout << cmd(3ul);
return 0;
}
Related
I need the positions of characters in a string.
The String contains:
"username":"secret", "password":"also secret", "id":"secret too", "token":"secret"
and I need the positions of the quotation marks from the token that are bold: "token":"secret".
I have experimented with the code from http://www.cplusplus.com/reference/string/string/find
but everything didn't work. Can anyone help me?
Here is what i have tried but it only gives out a 0:
#include <iostream>
#include <string>
int main() {
std::string buffer("\"username\":\"secret\", \"password\":\"also secret\", \"id\":\"secret too\", \"token\":\"secret\"");
size_t found = buffer.find('"');
if (found == std::string::npos)std::cout << "something went wrong\n";
if (found != std::string::npos)
std::cout << "first " << '"' << " found at: " << found << '\n';
for (int j = 0; j <= 17; ++j) {
found = buffer.find('"');
found + 1, 6;
if (found != std::string::npos)
std::cout << "second " << '"' << " found at : " << found << '\n';
}
return 0;
There are so many possible solutions. So, it is hard to answer.
What basically needs to be done, is to iterate through the string, position by position, then check if the character is the searched one, and then do something with the result.
A first simple implementation could be:
#include <iostream>
#include <string>
const std::string buffer("\"username\":\"secret\", \"password\":\"also secret\", \"id\":\"secret too\", \"token\":\"secret\"");
int main() {
for (size_t position{}, counter{}; position < buffer.length(); ++position) {
if (buffer[position] == '\"') {
++counter;
std::cout << "Character \" number " << counter << " found at position " << position << '\n';
}
}
return 0;
}
But then, your question was about the usage of std::string.find(). In your implementation, you start always the search at the beginning of the std::string. And because of that, you will always find the same " at position 0.
Solution: After you have found the first match, use the resulting pos (incremented by one) as the second parameter to the std::string.find() function. Then you will start the search after the first found " and hence find the next one. And all this can be done in a normal for-loop.
See below the next easy example:
#include <iostream>
#include <string>
const std::string buffer("\"username\":\"secret\", \"password\":\"also secret\", \"id\":\"secret too\", \"token\":\"secret\"");
int main() {
for (size_t position{}, counter{}; std::string::npos != (position = buffer.find("\"", position)); ++position, ++counter) {
std::cout << "Character \" number " << counter << " found at position " << position << '\n';
}
return 0;
}
There are more solutions, depending on what you really want to do. You coud extract all keywords and data with a simple regex.
Something like this:
#include <iostream>
#include <string>
#include <regex>
#include <vector>
const std::regex re{ R"(\"([ a-zA-Z0-9]+)\")" };
const std::string buffer("\"username\":\"secret\", \"password\":\"also secret\", \"id\":\"secret too\", \"token\":\"secret\"");
int main() {
std::vector part(std::sregex_token_iterator(buffer.begin(), buffer.end(), re, 1), {});
std::cout << part[7] << '\n';
return 0;
}
Or, you can split everything into tokens and values. Like this:
#include <iostream>
#include <string>
#include <regex>
#include <vector>
#include <map>
#include <iomanip>
const std::regex re1{ "," };
const std::regex re2{ R"(\"([^\"]+)\")" };
const std::string buffer("\"username\":\"secret\", \"password\":\"also secret\", \"id\":\"secret too\", \"token\":\"secret\"");
int main() {
std::vector<std::string> block(std::sregex_token_iterator(buffer.begin(), buffer.end(), re1, -1), {});
std::map<std::string, std::string> entry{};
for (const auto& b : block) {
std::vector blockPart(std::sregex_token_iterator(b.begin(), b.end(), re2, 1), {});
entry[blockPart[0]] = blockPart[1];
}
for (const auto& [token, value] : entry)
std::cout << std::setw(20) << token << " --> " << value << '\n';
return 0;
}
But if you have a complex given format, like JSON, there are so many special cases that the only meaningful approach is to use an existing library.
for (vector<int>::const_iterator i = vec.begin(); i != vec.end(); ++i)
{
int number = *i;
char* c;
itoa(number, c, 10);
result += c;
}
std::cout << result << std::endl;
I'm trying to convert each int in "vec" to a char and adding it to a string but I just get a compiler error. what am I doing wrong?
You can use the std::to_string available in C++11:
#include <iostream>
#include <vector>
#include <string>
int main()
{
std::vector<int> vec;
for (int i = 0; i < 100; i++)
{
vec.push_back(i);
}
std::string result;
for (std::vector<int>::const_iterator i = vec.begin(); i != vec.end(); ++i)
{
result += std::to_string(*i);
}
std::cout << result << std::endl;
}
Combining sounds like a job for std::accumulate.
#include <iostream>
#include <numeric>
#include <string>
#include <vector>
auto main() -> int
{
const std::vector<int> vec{ 1, 2, 3 };
const std::string result = std::accumulate(vec.begin(), vec.end(), std::string(),
[](const std::string& s, const int value)
{
return s + std::to_string(value);
});
std::cout << result << std::endl;
}
There's an error in your code:
You call itoa() with a non-initialized char pointer. That's bad, because itoa() needs a valid buffer! Besides, itoa() is not part of the c++ standard.
Better do leave char* behind and do it with more modern C++ features, especially std::stringstream, which is a string builder, which is powerful in conversion (and also often faster than the string += operator). It builds a string by pushing elements to it with the << operator (these can be string literals, strings, numbers of all kinds), it can be extended via external operators for own data types, and it also has a lot of formatting options (e.g. number of digits, hex etc), and returns its built string with the method str().
#include <iostream>
#include <vector>
#include <string>
#include <sstream>
int main()
{
std::vector<int> vec;
for (int i = 0;i<100;i++)
vec.push_back(i);
std::stringstream ss;
for (auto& number : vec)
ss << number;
std::cout << ss.str() << std::endl;
}
How can I convert this code from C into C++ ?
char out[61]; //null terminator
for (i = 0; i < 20; i++) {
snprintf(out+i*3, 4, "%02x ", obuf[i])
}
I can't find any alternative for snprintf.
Use stringstream class from <sstream>.
E.g.:
#include <iostream>
#include <iomanip>
#include <sstream>
#include <string>
using namespace std;
int main()
{
stringstream ss;
for (int i = 0; i < 20; i++) {
ss << setw(3) << i;
}
cout << "Resulting string: " << endl;
cout << ss.str() << endl;
printf("Resulting char*: \n%s\n", ss.str().c_str() );
return 0;
}
This code is valid C++11, if you have #include <cstdio> and type std::snprintf (or using namespace std;).
No need to "fix" what isn't broken.
You can use Boost.Format.
#include <boost/format.hpp>
#include <string>
std::string out;
for (size_t i=0; i<20; ++i)
out += (boost::format("%02x") % int(obuf[i])).str();
You can convert this code from C to C++ easily with standard library's std::stringstream and iomanip I/O stream manipulators:
#include <sstream>
#include <iomanip>
...
std::ostringstream stream;
stream << std::setfill('0') << std::hex;
for (const auto byte : obuf)
stream << std::setw(2) << byte;
const auto out = stream.str();
lexical_cast throws an exception in the following case. Is there a way to use lexical_cast and convert the string to integer.
#include <iostream>
#include "boost/lexical_cast.hpp"
#include <string>
int main()
{
std::string src = "124is";
int iNumber = boost::lexical_cast<int>(src);
std::cout << "After conversion " << iNumber << std::endl;
}
I understand, I can use atoi instead of boost::lexical_cast.
If I'm understanding your requirements correctly it seems as though removing the non-numeric elements from the string first before the lexical_cast will solve your problem. The approach I outline here makes use of the isdigit function which will return true if the given char is a digit from 0 to 9.
#include <iostream>
#include "boost/lexical_cast.hpp"
#include <string>
#include <algorithm>
#include <cctype> //for isdigit
struct is_not_digit{
bool operator()(char a) { return !isdigit(a); }
};
int main()
{
std::string src = "124is";
src.erase(std::remove_if(src.begin(),src.end(),is_not_digit()),src.end());
int iNumber = boost::lexical_cast<int>(src);
std::cout << "After conversion " << iNumber << std::endl;
}
The boost/lexical_cast uses stringstream to convert from string to other types,so you must make sure the string can be converted completely！ or， it will throw the bad_lexical_cast exception,This is an example:
#include <boost/lexical_cast.hpp>
#include <iostream>
#include <string>
#define ERROR_LEXICAL_CAST 1
int main()
{
using boost::lexical_cast;
int a = 0;
double b = 0.0;
std::string s = "";
int e = 0;
try
{
// ----- string --> int
a = lexical_cast<int>("123");//good
b = lexical_cast<double>("123.12");//good
// -----double to string good
s = lexical_cast<std::string>("123456.7");
// ----- bad
e = lexical_cast<int>("abc");
}
catch(boost::bad_lexical_cast& e)
{
// bad lexical cast: source type value could not be interpreted as target
std::cout << e.what() << std::endl;
return ERROR_LEXICAL_CAST;
}
std::cout << a << std::endl; // cout：123
std::cout << b << std::endl; //cout：123.12
std::cout << s << std::endl; //cout：123456.7
return 0;
}
Hi i want to save many different csv files from a function with a naming convention based on a different double value. I do this with a for loop and pass a string value to save each .csv file differently. Below is an example of what I'm trying to do the desired result would be
1.1_file.csv
1.2_file.csv
but instead i get
1.1_file.csv
1.11.2_file.csv
Here is a working sample code, what can i do to fix this
#include <sstream>
#include <iomanip>
#include <cmath>
#include <iostream>
#include <vector>
int main(){
std::string file = "_file.csv";
std::string s;
std::ostringstream os;
double x;
for(int i = 0; i < 10; i++){
x = 0.1 + 0.1 *i;
os << std::fixed << std::setprecision(1);
os << x;
s = os.str();
std::cout<<s+file<<std::endl;
s.clear();
}
return 0;
}
The ostringstream doesn't reset on each iteration of the loop, so you are just adding x to it every iteration; put it inside the scope of the for to make os be a different clean object on each iteration, or reset the contents with os.str("").
Also, the variable s is unnecessary; you can just do
std::cout << os.str() + file << std::endl;
And you don't need s and you eliminate the overhead of making a copy of the string.
Your ostringstream is getting appended for each iteration of the loop. You should clear it and reuse it as shown below (courtesy: How to reuse an ostringstream? on how to reuse an ostringstream)
#include <sstream>
#include <iomanip>
#include <cmath>
#include <iostream>
#include <vector>
int main() {
std::string file = "_file.csv";
std::string s;
double x;
std::ostringstream os;
for (int i = 0; i < 10; i++) {
x = 0.1 + 0.1 * i;
os << std::fixed << std::setprecision(1);
os << x;
s = os.str();
std::cout << s + file << std::endl;
os.clear();
os.str("");
}
return 0;
}