C++ 如何计算文本文件中的字符数
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how to count the characters in a text file
提问by user836910
im trying to count the characters inside a text file in c++, this is what i have so far, for some reason im getting 4. even thou i have 123456 characters in it. if i increase or decrease the characters i still get 4, please help and thanks in advance
我正在尝试用 C++ 计算文本文件中的字符，这是我到目前为止所拥有的，出于某种原因，我得到了 4。即使我有 123456 个字符。如果我增加或减少字符我仍然得到 4，请帮助并提前致谢
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
const char FileName[] = "text.txt";
int main ()
{
string line;
ifstream inMyStream (FileName);
int c;
if (inMyStream.is_open())
{
while( getline (inMyStream, line)){
cout<<line<<endl;
c++;
}
}
inMyStream.close();
system("pause");
return 0;
}
回答by BeemerGuy
You're counting the lines.
You should count the characters. change it to:
你在数线。
你应该计算字符。将其更改为：
while( getline ( inMyStream, line ) )
{
cout << line << endl;
c += line.length();
}
回答by sbabbi
There are probably hundreds of ways to do that. I believe the most efficient is:
可能有数百种方法可以做到这一点。我认为最有效的是：
inMyStream.seekg(0,std::ios_base::end);
std::ios_base::streampos end_pos = inMyStream.tellg();
return end_pos;
回答by Brenden
this is how i would approach the problem:
这就是我将如何解决这个问题：
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int main ()
{
string line;
int sum=0;
ifstream inData ;
inData.open("countletters.txt");
while(!inData.eof())
{
getline(inData,line);
int numofChars= line.length();
for (unsigned int n = 0; n<line.length();n++)
{
if (line.at(n) == ' ')
{
numofChars--;
}
}
sum=numofChars+sum;
}
cout << "Number of characters: "<< sum << endl;
return 0 ;
}
回答by shengy
First of all, you have to init a local var, this means:
int c = 0;instead of
int c;
首先，你必须初始化一个本地变量，这意味着：
int c = 0;而不是
int c;
I think the old and easy to understand way is to use the get()function till the end char EOF
我认为古老且易于理解的方法是使用该get()函数直到结束字符EOF
char current_char;
if (inMyStream.is_open())
{
while(inMyStream.get(current_char)){
if(current_char == EOF)
{
break;
}
c++;
}
}
Then cwill be the count of the characters
然后c将是字符数
回答by Subbu
I found out this simple method , hope this helps
我发现了这个简单的方法，希望对您有所帮助
while(1)
{
if(txtFile.peek() == -1)
break;
c = txtFile.get();
if(c != txtFile.eof())
noOfChars++;
}
回答by Richard J. Ross III
Just use good old C FILE pointers:
只需使用旧的 C FILE 指针：
int fileLen(std::string fileName)
{
FILE *f = fopen(fileName.c_str(), "rb");
if (f == NULL || ferror(f))
{
if (f)
fclose(f);
return -1;
}
fseek(f, 0, SEEK_END);
int len = fell(f);
fclose(f);
return len;
}
回答by Martin Melichar
This works for sure, it is designed to read character by character.
这肯定有效，它旨在逐个字符地读取。
It could be easily put into a class and you may apply function for every char, so you may check for '\n', ' ' and so on. Just have some members in your class, where they can be saved, so you may only return 0 and use methods to get what exactly you want.
它可以很容易地放入一个类中，您可以为每个字符应用函数，因此您可以检查 '\n'、' ' 等。只要在你的班级中有一些成员，他们可以被保存在那里，所以你可能只返回 0 并使用方法来获得你想要的东西。
#include <iostream>
#include <fstream>
#include <string>
unsigned long int count(std::string string)
{
char c;
unsigned long int cc = 0;
std::ifstream FILE;
FILE.open(string);
if (!FILE.fail())
{
while (1)
{
FILE.get(c);
if (FILE.eof()) break;
cc++; //or apply a function to work with this char..eg: analyze(c);
}
FILE.close();
}
else
{
std::cout << "Counter: Failed to open file: " << string << std::endl;
}
return cc;
};
int main()
{
std::cout << count("C:/test/ovecky.txt") << std::endl;
for (;;);
return 0;
}
