c++ - 将模板函数作为参数传递给 C++ 中的另一个函数时出现问题

Question

问题来源-Accelerated C++，问题8-5

我编写了一个小程序来检查字符串输入的行数，并计算单词在给定行中出现的次数。以下代码实现了这一点:

#include <map>
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <cctype>
#include <iterator>

using std::vector;         using std::string;
using std::cin;            using std::cout;
using std::endl;           using std::getline;
using std::istream;        using std::string;
using std::list;           using std::map;
using std::isspace;        using std::ostream_iterator;
using std::allocator;

inline void keep_window_open()
{
    cin.clear();
    cout << "Please enter EOF to exit\n";
    char ch;
    cin >> ch;
    return;
}

template <class Out>
void split(const string& s, Out os)
{
    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
            *os++ = (s.substr(i, j - i));
            i = j;
        }
    }
}


// find all the lines that refer to each word in the input
map<string, vector<int> > xref(istream& in)     // works
// now try to pass the template function as an argument to function - what do i put for templated type?
//map<string, vector<int> > xref(istream& in, void find_words(vector<string, typedef Out) = split)      #LINE 1#
{
    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;   // works            // #LINE 2#
        split(line, back_inserter(words));          // #LINE 3#
        //find_words(line, back_inserter(words));   // #LINE 4# attempting to use find_words as an argument to function

        // 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;
}

int main()
{
    cout << endl << "Enter lines of text, followed by EOF (^Z):" << endl;

    // call `xref' using `split' by default
    map<string, vector<int> > ret = xref(cin);

    // write the results
    for (map<string, vector<int> >::const_iterator it = ret.begin();
         it != ret.end(); ++it) {
        // write the word
        cout << it->first << " occurs on line(s): ";

        // followed by one or more line numbers
        vector<int>::const_iterator line_it = it->second.begin();
        cout << *line_it;   // write the first line number

        ++line_it;
        // write the rest of the line numbers, if any
        while (line_it != it->second.end()) {
            cout << ", " << *line_it;
            ++line_it;
        }
        // write a new line to separate each word from the next
        cout << endl;
    }
    keep_window_open();
    return 0;
}

如您所见，split 函数是一个模板函数，可以根据需要处理各种类型的输出迭代器。

当我尝试通过传入模板化的 split 函数作为参数来概括 xref 函数时，我的问题就来了。我似乎无法获得正确的类型。

所以我的问题是，你能否将一个模板函数作为参数传递给另一个函数，如果可以，是否必须在传递之前声明所有类型？还是编译器可以从主体中使用模板化函数的方式推断出类型？

为了演示我得到的错误，注释掉现有的 xref 函数 header ，并取消注释我试图开始工作的备用 header (就在以下注释行下方。)同时注释这些行标记第 2 行和第 3 行并取消注释第 4 行，它试图使用参数 find_words(默认为 split。)

感谢任何反馈!

Answer 1

以下是解决方法。 您可以使用具有静态函数的类

struct split
{
  template <class Out>
  static apply(const string& s, Out os) { 
    // include the body of your split function or call to an existing function
  }

};

现在，使外部参照通用

template <typename FIND_WORDS>
map<string, vector<int> > xref(istream& in, FIND_WORDS find_words = split()) 
{

  // replace #2 and #3 by
  find_words.apply(line, back_inserter(words));

};