C++ printf std::vector

C++ printf std::vector

提问by Sebtm

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

我如何在 C++ 中做这样的事情：

void my_print(format_string) {
   vector<string> data;
   //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.

我之前没有解释清楚。格式字符串由应用程序用户输入。

In C# this works:

在 C# 中，这有效：

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}");

采纳答案by Sebtm

I have temporarly solved with this function:

我暂时解决了这个功能：

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;
}

回答by John Dibling

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

如果您打算使用流，您还可以ostream_iterator与循环结构结合使用，例如copy：

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

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

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

请注意，第二个参数 tocopy指向最后一个。输出：

hello hello hello

你好你好你好

回答by John Dibling

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.

请注意，您必须转换为 C 样式的字符串 - printf 无法为您执行此操作。

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.

编辑：针对您修改后的问题，我认为您必须自己解析格式字符串，因为您必须对其进行验证。printf() 不会完成这项工作。

回答by Ken Bloom

The Boost Format Librarymight be helpful.

该升压格式库可能会有所帮助。

#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;
}

回答by Ken Bloom

I think you're looking to do the following:

我认为您希望执行以下操作：

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

1. 将您的转换std::vector<std::string>为 a va_listof char*s
2. 将那个va_list以及用户提供的格式字符串传递给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.)

我仍然不知道如何执行第 1 步。（我所知道的是大多数高级语言，例如 Java、Scala 和 Ruby，都有一个简单、安全、直接的转换。C++ 没有。）

回答by wheaties

Call the ones you want

打电话给你想要的人

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