how to iterate into a smaller container (i.e. stride != 1)

问题

There is a question that is very similar in spirit here. Unfortunately that question didn't prompt much response - I thought I would ask a more specific question with the hope that an alternative method can be suggested.

I'm writing a binary file into std::cin (with tar --to-command=./myprog). The binary file happens to be a set of floats and I want to put the data into std::vector<float> - ideally the c++ way.

I can generate a std::vector<char> very nicely (thanks to this answer)

#include <fstream>
#include <iostream>
#include <iterator>
#include <algorithm>
#include <vector>

int
main  (int ac, char **av)
{
  std::istream& input = std::cin;
  std::vector<char> buffer;
  std::copy( 
        std::istreambuf_iterator<char>(input), 
           std::istreambuf_iterator<char>( ),
           std::back_inserter(buffer)); // copies all data into buffer
}

I now want to transform my std::vector<char> into a std::vector<float>, presumably with std::transform and a function that does the conversion (a char[2] to a float, say). I am struggling however, because my std::vector<float> will have half as many elements as std::vector<char>. If I could iterate with a stride of 2 then I think I would be fine, but from the previous question it seems that I cannot do that (at least not elegantly).

回答1:

I would write my own class that reads two chars and converts it to float.

struct FloatConverter
{
    // When the FloatConverter object is assigned to a float value
    // i.e. When put into the vector<float> this method will be called
    //      to convert the object into a float.
    operator float() { return 1.0; /* How you convert the 2 chars */ }

    friend std::istream& operator>>(std::istream& st, FloatConverter& fc)
    {
        // You were not exactly clear on what should be read in.
        // So I went pedantic and made sure we just read 2 characters.
        fc.data[0] = str.get();
        fc.data[1] = str.get();
        retun str;
    }
    char   data[2];
 };

Based on comments by GMan:

struct FloatConverterFromBinary
{
    // When the FloatConverterFromBinary object is assigned to a float value
    // i.e. When put into the vector<float> this method will be called
    //      to convert the object into a float.
    operator float() { return data }

    friend std::istream& operator>>(std::istream& st, FloatConverterFromBinary& fc)
    {
        // Use reinterpret_cast to emphasis how dangerous and unportable this is.
        str.read(reinterpret_cast<char*>(&fc.data), sizeof(float));
        retun str;
    }

    float  data;
};

Then use it like this:

int main  (int ac, char **av)
{
  std::istream& input = std::cin;
  std::vector<float> buffer;

  // Note: Because the FloatConverter does not drop whitespace while reading
  //       You can potentially use std::istream_iterator<>
  //
  std::copy( 
           std::istreambuf_iterator<FloatConverter>(input), 
           std::istreambuf_iterator<FloatConverter>( ),
           std::back_inserter(buffer));
}

回答2:

It seems to me that the best answer is to write a pair of your own iterators that parse the file the way that you want. You could change std::vector<char> to std::vector<float> and use the same streambuf iterators provided the input was formatted with at least one space between values.

回答3:

use boost range adaptors:

boost::copy(istream_range(input)|stride(2),back_inserter(buffer));

you might need to write your own istreambuf_iterator, which is trivial.

来源：https://stackoverflow.com/questions/5424093/how-to-iterate-into-a-smaller-container-i-e-stride-1

标签

c++

stl

iterator

stl-algorithm