[英]Accelerated C++ exercise 8-5 solution not clear
I am stuck at solving Accelerated C++ exercise 8-5 and I don't want to miss a single exercise in this book. 我坚持用8-5解决Accelerated C ++练习,我不想错过本书中的一个练习。
Accelerated C++ Exercise 8-5 is as follows: 加速C ++练习8-5如下:
Reimplement the
gen_sentence
andxref
functions from Chapter 7 to use output iterators rather than putting their entire output in one data structure.重新实现第7章中的
gen_sentence
和xref
函数以使用输出迭代器,而不是将它们的整个输出放在一个数据结构中。 Test these new versions by writing programs that attach the output iterator directly to the standard output, and by storing the results inlist <string>
andmap<string, vector<int> >
, respectively.通过编写将输出迭代器直接附加到标准输出的程序,并将结果分别存储在
list <string>
和map<string, vector<int> >
来测试这些新版本。
To understand scope of this question and current knowledge in this part of the book - this exercise is part of chapter about generic function templates and iterator usage in templates. 要理解这个问题的范围和本书这一部分的当前知识 - 本练习是关于通用函数模板和模板中迭代器用法的章节的一部分。 Previous exercise was to implement simple versions of
<algorithm>
library functions, such as equal, find, copy, remove_copy_if
etc. 以前的练习是实现
<algorithm>
库函数的简单版本,例如equal, find, copy, remove_copy_if
等。
If I understand correctly, I need to modify xref
function so it: 如果我理解正确,我需要修改
xref
功能,所以:
map<string, vector<int> >
map<string, vector<int> >
I tried to pass map iterator as back_inserter()
, .begin()
, .end()
to this function, but was not able to compile it. 我试图将map迭代器作为
back_inserter()
.begin()
.end()
传递给这个函数,但是无法编译它。 Answer here explains why. 这里的答案解释了原因
xref function as in Chapter 7: xref函数与第7章一样:
// 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;
}
Split implementation: 拆分实施:
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;
}
Please help to understand what am i missing. 请帮助理解我错过了什么。
I found more details on the exercise, here: https://stackoverflow.com/questions/5608092/accelerated-c-exercise-8-5-wording-help : 我在这里找到了有关练习的更多细节: 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 )
USAGE:
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 )
USAGE:
用法:
std::ostream_iterator<string> out_str (std::cout, " "); xref( cin, out_str, find_url ) ;
Frankly, I have to come to the conclusion that that question is ill-posed, specifically where they specified the new interface for xref
: xref should result in a map. 坦率地说,我必须得出这样的结论:这个问题是不适定的,特别是在他们指定
xref
的新接口的情况下:xref应该产生一个映射。 However, using output iterators would imply using std::inserter(map, map.end())
in this case. 但是,在这种情况下,使用输出迭代器意味着使用
std::inserter(map, map.end())
。 While you can write a compiling version of the code, this will not do what you expect since map::insert
will simply ignore any insertions with duplicated keys. 虽然您可以编写代码的编译版本,但这不会达到预期的效果,因为
map::insert
将忽略任何带有重复键的插入。
If the goal of xref is only to link the words to the line number of their first appearance this would still be ok, but I have a feeling that the author of the exercise simply missed this subtler point :) 如果外部参照的目标只是将单词链接到他们第一次出现的行号,这仍然可以,但我有一种感觉,练习的作者只是错过了这个更微妙的点:)
Here is the code anyways (note that I invented a silly implementation for split
, because it was both missing and required): 以下是代码(请注意,我发明了一个用于
split
的愚蠢实现,因为它既缺少又需要):
#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;
}
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