C++ printf std::vector

How I can do something like this in C++:

void my_print(format_string) {
   vector<string> dat开发者_Python百科a;
   //Fills vector
   printf(format_string, data);
}

my_print("%1$s - %2$s - %3$s");
my_print("%3$s - %2$s);

I have not explained well before. The format string is entered by the application user.

In C# this works:

void my_print(format_string) {
 List<string> data = new List<string>();
 //Fills list
 Console.WriteLine(format_string, data.ToArray);
}

my_print("{0} - {1} - {2}");
my_print("{2} - {1}");

If you're going to use streams, you can also use ostream_iterator in conjunction with a looping construct like copy:

vector<string> data;
data.assign(10, "hello");

copy( &data[0], &data[3], ostream_iterator<string>(cout, " "));

Note that the second parameter to copy points to one past the end. Output:

hello hello hello

printf("%s - %s - %s", data[0].c_str(), data[1].c_str(), data[2].c_str() );

Note that you must convert to C-style strings - printf cannot do this for you.

Edit: In response to your revised question, I think you will have to parse the format string yourself, as you will have to validate it. printf() won't do the job.

The Boost Format Library might be helpful.

#include <boost/format.hpp>
#include <vector>
#include <string>
#include <iostream>
int main(int arc, char** argv){
   std::vector<std::string> array;
   array.push_back("Hello");
   array.push_back("word");
   array.push_back("Hello");
   array.push_back("again");
   boost::format f("%s, %s! %s %s! \n");
   f.exceptions( f.exceptions() &
     ~ ( boost::io::too_many_args_bit | boost::io::too_few_args_bit )  );

   for (std::vector<std::string>::iterator i=array.begin();i!=array.end();++i){
      f = f % (*i);
   }
   std::cout << f;
   return 0;
}

I have temporarly solved with this function:

string format_vector(string format, vector<string> &items)
{   
   int counter = 1;

   replace_string(format,"\\n","\n");
   replace_string(format,"\\t","\t");

   for(vector<string>::iterator it = items.begin(); it != items.end(); ++it) {
        ostringstream stm; stm << counter;
        replace_string(format, "%" + stm.str(), *it);
        counter++;
   }
    return format;
}

I think you're looking to do the following:

1. Convert your std::vector<std::string> into a va_list of char*s
2. Pass that va_list, along with the user-supplied format string to vprintf.

I still don't know how to do step 1. (What I do know is that most higher-level languages, such as Java, Scala, and Ruby have a simple, safe, direct conversion for that. C++ doesn't.)

Now suports only %...f format. You can you can expand the possibilities by changing regex expression

https://regex101.com/r/dBaolO/1

template<typename ... Args>
        static std::string string_format(const std::string& format, Args ... args)
        {
            int size_s = std::snprintf(nullptr, 0, format.c_str(), args ...) + 1; // Extra space for '\0'
            if (size_s <= 0) { throw std::runtime_error("Error during formatting."); }
            auto size = static_cast<size_t>(size_s);
            auto buf = std::make_unique<char[]>(size);
            std::snprintf(buf.get(), size, format.c_str(), args ...);
            return std::string(buf.get(), buf.get() + size - 1); // We don't want the '\0' inside
        }
        
        static std::string string_format(const std::string& format, std::vector<float> args)
        {
            std::string formatted(format);
            std::string pattern = R"(\%(?:\d+\.\d+)?[f|F])";
            std::regex rx(pattern);

            int args_index = 0;
            int mismatch_offset = 0;
            for (std::sregex_iterator i = std::sregex_iterator(format.begin(), format.end(), rx); i != std::sregex_iterator(); ++i, ++args_index) {
                std::smatch match = *i;

                auto string_value = string_format(match.str(), args.at(args_index));

                formatted.replace(match.position() + mismatch_offset, match.length(), string_value);

                mismatch_offset += string_value.length() - match.length();
            }
            
            return formatted;
        }

Call the ones you want

printf("%1$s - %2$s - %3$s", date[0].c_str(), data[1].c_str(), data[2].c_str());