加速c++练习8-5解不清楚

// find all the lines that refer to each word in the input
map<string, vector<int> >
    xref(istream& in,
         vector<string> find_words(const string&) = split)
{
    string line;
    int line_number = 0;
    map<string, vector<int> > ret;
    // read the next line
    while (getline(in, line)) {
        ++line_number;
        // break the input line into words
        vector<string> words = find_words(line);
        // remember that each word occurs on the current line
        for (vector<string>::const_iterator it = words.begin();
             it != words.end(); ++it)
            ret[*it].push_back(line_number);
    }
    return ret;
}

vector<string> split(const string& s)
{
    vector<string> ret;
    typedef string::size_type string_size;
    string_size i = 0;
    // invariant: we have processed characters `['original value of `i', `i)'
    while (i != s.size()) {
        // ignore leading blanks
        // invariant: characters in range `['original `i', current `i)' are all spaces
        while (i != s.size() && isspace(s[i]))
            ++i;
        // find end of next word
        string_size j = i;
        // invariant: none of the characters in range `['original `j', current `j)' is a space
        while (j != s.size() && !isspace(s[j]))
            ++j;
        // if we found some nonwhitespace characters
        if (i != j) {
            // copy from `s' starting at `i' and taking `j' `-' `i' chars
            ret.push_back(s.substr(i, j - i));
            i = j;
        }
    }
    return ret;
}

我在这里找到了更多关于练习的细节:https://stackoverflow.com/questions/5608092/accelerated-c-exercise-8-5-wording-help:

template <class Out> 
void gen_sentence( const Grammar& g, string s, Out& out )

用法:

std::ostream_iterator<string> out_str (std::cout, " ");
gen_sentence(   g, "<sentence>", out_str   );

---

template <class Out, class In> 
void xref(   In& in, Out& out, vector<string> find_words( const string& ) = split   )

用法:

std::ostream_iterator<string> out_str (std::cout, " "); 
xref(   cin, out_str, find_url   ) ; 

坦率地说，我不得不得出结论，这个问题是不适定的，特别是他们为xref指定的新接口:xref应该导致映射。然而，在这种情况下，使用输出迭代器意味着使用std::inserter(map, map.end())。虽然您可以编写代码的编译版本，但这不会达到您的期望，因为map::insert将简单地忽略任何具有重复键的插入。

如果xref的目标只是将单词链接到它们第一次出现的行号，这仍然是可以的，但是我有一种感觉，练习的作者只是忽略了这个微妙的点:)

下面是代码(注意，我为split发明了一个愚蠢的实现，因为它既缺失又必需):

#include <map>
#include <vector>
#include <iostream>
#include <sstream>
#include <fstream>
#include <algorithm>
#include <iterator>
std::vector<std::string> split(const std::string& str)
{
    std::istringstream iss(str);
    std::vector<std::string> result;
    std::copy(std::istream_iterator<std::string>(iss),
              std::istream_iterator<std::string>(),
              std::back_inserter(result));
    return result;
}
// find all the lines that refer to each word in the input
template <typename OutIt>
OutIt xref(std::istream& in,
        OutIt out,
        std::vector<std::string> find_words(const std::string&) = split)
{
    std::string line;
    int line_number = 0;
    // read the next line
    while (getline(in, line)) {
        ++line_number;
        // break the input line into words
        std::vector<std::string> words = find_words(line);
        // remember that each word occurs on the current line
        for (std::vector<std::string>::const_iterator it = words.begin();
             it != words.end(); ++it)
            *out++ = std::make_pair(*it, line_number);
    }
    return out;
}
int main(int argc, const char *argv[])
{
    std::map<std::string, int> index;
    std::ifstream file("/tmp/test.cpp");
    xref(file, std::inserter(index, index.end()));
#if __GXX_EXPERIMENTAL_CXX0X__
    for(auto& entry: index)
        std::cout << entry.first << " first found on line " << entry.second << std::endl;
#else
    for(std::map<std::string, int>::const_iterator it = index.begin();
        it != index.end();
        ++it)
    {
        std::cout << it->first << " first found on line " << it->second << std::endl;
    }
#endif
    return 0;
}