我遇到的问题是,我不确定如何重置字数。我创建了一个单词搜索,但是当我计算出10个不同单词的出现次数时,它与它计算的第一个单词保持相同的数字。我相信我遇到的问题是我使用for
循环
输出
boy appeared 3 times
Snape appeared 3 times
Dumbledore appeared 3 times
he appeared 3 times
her appeared 3 times
the appeared 3 times
it appeared 3 times
is appeared 3 times
will appeared 3 times
all appeared 3 times
应该是什么样子
boy appeared 3 times
Snape appeared 7 times
Dumbledore appeared 4 times
he appeared 27 times
her appeared 4 times
the appeared 13 times
it appeared 6 times
is appeared 12 times
will appeared 2 times
all appeared 3 times
通过阅读我的代码,我确信我已经让它变得更复杂了。我很感激我所做的任何建议和更正。
完整代码
#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
#include <vector>
// Main Function
int main()
{
// Declaration
std::string list, passage, word[10];
std::ifstream listFile("WordList.txt", std::ios::in);
std::ifstream passageFile("HarryPotterPassage.txt", std::ios::in);
std::vector<std::string> vec_wordList, vec_passage;
/* Read a file that contains a list of 10 words */
if (listFile.is_open())
{
// Store text file in a vector
while (listFile)
{
listFile >> list;
vec_wordList.push_back(list);
}
// Assign vector to individual strings
for (int i = 0; i < 10; i++)
{
word[i] = vec_wordList[i];
}
// Close file
listFile.close();
}
else
std::cout << "No file found.\n";
/* Read another file containing a paragraph */
if (passageFile.is_open())
{
while (passageFile)
{
// Store text file in a string
std::getline(passageFile, passage);
}
// Close file
passageFile.close();
}
else
std::cout << "No file found.\n";
//std::cout << passage << '\n';
/* Count the number of words from the first file
from the second file that contains the paragraph */
size_t count = 0;
std::string::size_type pos = 0;
for (int i = 0; i < 10; i++)
{
while ((pos = passage.find(word[i], pos)) != std::string::npos)
{
count++;
pos += word[i].size();
}
std::cout << word[i] << " appeared " << count << " many times\n";
}
system("pause");
return 0;
}
提前致谢。
答案 0 :(得分:1)
您使用单词[9]而不是单词[i],因此您可以得到最后一个单词的结果而不是每个单词的结果。 尝试:
for (int i = 0; i < 10; i++)
{
while ((pos = passage.find(word[i], pos)) != std::string::npos)
{
count++;
pos += word[i].size();
}
std::cout << word[i] << " appeared " << count << " many times\n";
}
答案 1 :(得分:0)
您需要在外循环的每次迭代开始时重置count
和pos
。
换句话说,改变这个:
size_t count = 0;
std::string::size_type pos = 0;
for (int i = 0; i < 10; i++)
{
...
}
对此:
for (int i = 0; i < 10; i++)
{
size_t count = 0;
std::string::size_type pos = 0;
...
}
顺便说一句,我也将10
更改为sizeof(word)/sizeof(*word)
。