c++ - 计算文件中字母的出现次数

Question

我正在尝试计算每个字母在文件中出现的次数。当我运行下面的代码时，它会计算“Z”两次。谁能解释一下为什么？

测试数据为:

abcdefghijklmnopqrstuvwxyz

ABCDEFGHIJKLMNOPQRSTUVWXYZ

#include <iostream>                 //Required if your program does any I/O
#include <iomanip>                  //Required for output formatting
#include <fstream>                  //Required for file I/O
#include <string>                   //Required if your program uses C++ strings
#include <cmath>                    //Required for complex math functions
#include <cctype>                   //Required for letter case conversion

using namespace std;                //Required for ANSI C++ 1998 standard.

int main ()
{
string reply;
string inputFileName;
ifstream inputFile;
char character;
int letterCount[127] = {};

cout << "Input file name: ";
getline(cin, inputFileName);

// Open the input file.
inputFile.open(inputFileName.c_str());      // Need .c_str() to convert a C++ string to a C-style string
// Check the file opened successfully.
if ( ! inputFile.is_open())
{
    cout << "Unable to open input file." << endl;
    cout << "Press enter to continue...";
    getline(cin, reply);
    exit(1);
}

while ( inputFile.peek() != EOF )
{
      inputFile >> character;
      //toupper(character);

      letterCount[static_cast<int>(character)]++;
}

for (int iteration = 0; iteration <= 127; iteration++)
{
    if ( letterCount[iteration] > 0 )
    {
         cout << static_cast<char>(iteration) << " " << letterCount[iteration] << endl;
    }
}

system("pause");
exit(0);
}

Answer 1

正如其他人所指出的，您在输入中有两个 Q。你有两个 Z 的原因是最后一个

inputFile >> character;

(可能当流中只剩下一个换行符时，因此不是 EOF)无法转换任何内容，在上一次迭代的全局“字符”中留下“Z”。之后尝试检查 inputFile.fail() 以查看:

while (inputFile.peek() != EOF)
{
    inputFile >> character;

    if (!inputFile.fail())
    {
        letterCount[static_cast<int>(character)]++;
    }
}

编写循环的惯用方法是:

while (inputFile >> character)
{
      letterCount[static_cast<int>(character)]++;
}