Question

所以我想创建一个程序来计算文件中每个字符的出现次数。例如：

4个字符0x67（g）的实例

字符0x68（h）的11个实例

等等

我不确定如何显示和计算实例。

有什么想法吗？

#include <stdio.h>
const char FILE_NAME[] = "input.txt";
#include <stdlib.h>

int main() {

    int             count = 0;  /* number of characters seen */
    FILE           *in_file;    /* input file */

   /* character or EOF flag from input */
    int             ch;

    in_file = fopen(FILE_NAME, "r");
    if (in_file == NULL) {
        printf("Cannot open %s\n", FILE_NAME);
        exit(8);
    }

    while (1) {
        ch = fgetc(in_file);
        if (ch == EOF)
            break;
        ++count;
    }
    printf("Number of characters in %s is %d\n",
                  FILE_NAME, count);

    fclose(in_file);
    return (0);

Answer 1

这就是我想出来的......

#include<stdio.h>
#include<stdlib.h>

int main() {
  /* a buffer to hold the count of characters 0,...,256; it is
   * initialized to zero on every element */
  int count[256] = { 0 };

  /* loop counter */
  int k;

  /* file handle --- in this case I am parsing this source code */
  FILE *fp = fopen("ccount.c", "r");

  /* a holder for each character (stored as int) */
  int c;

  /* for as long as we can get characters... */
  while((c=fgetc(fp))) {

    /* break if end of file */
    if(c == EOF) break;

    /* otherwise add one to the count of that particular character */
    count[c]+=1;
  }

  /* now print the results; only if the count is different from
   * zero */
  for(k=0; k<256; k++) {
    if(count[k] > 0) {
      printf("char %c: %d times\n", k, count[k]);
    }
  }
  /* close the file */
  fclose(fp);

  /* that's it */
  return 0;
}

我使用以下命令（OS X 10.7.4上的GCC 4.8.1）编译代码

gcc ccount.c -Wall -Wextra -pedantic -ansi

它编译时没有警告也没有错误;这是输出：

char 
: 40 times
char  : 190 times
char ": 6 times
char #: 2 times
char %: 2 times
char ': 1 times
char (: 11 times
char ): 11 times
char *: 23 times
char +: 3 times
char ,: 5 times
char -: 3 times
char .: 9 times
char /: 20 times
char 0: 5 times
char 1: 1 times
char 2: 3 times
char 5: 3 times
char 6: 3 times
char :: 1 times
char ;: 13 times
char <: 3 times
char =: 7 times
char >: 3 times
char E: 2 times
char F: 2 times
char I: 2 times
char L: 1 times
char O: 1 times
char [: 4 times
char \: 1 times
char ]: 4 times
char a: 29 times
char b: 4 times
char c: 36 times
char d: 15 times
char e: 49 times
char f: 25 times
char g: 4 times
char h: 22 times
char i: 36 times
char k: 9 times
char l: 19 times
char m: 5 times
char n: 35 times
char o: 38 times
char p: 9 times
char r: 34 times
char s: 22 times
char t: 49 times
char u: 16 times
char v: 1 times
char w: 4 times
char y: 2 times
char z: 3 times
char {: 5 times
char }: 5 times

Answer 2

你需要使用数组，请查看：

int charArray[256];
memset(charArray, 0, 256*sizeof(int)); // instead of memset, for 0 values you can you just {0}

while (1) {
        ch = fgetc(in_file);
        if (ch == EOF)
            break;
        charArray[ch]++;
    }
for (int i=0; i<256; i++)
    if (charArray[i] > 0)
        printf("Number of character %c is %d\n", (char)i, charArray[i]);

Answer 3

如果要检索字母字符数，则它可能如下所示：

int counts[26];
memset(&counts[0], 0, sizeof(counts));

while ( (ch = fgetc(in_file)) != EOF) {
    if (ch >= 'a' && ch <= 'z')
        ++count[ch - 'a'];
}

打印它们可以很简单：

for (char c = 'a'; c <= 'z', ++c)
    printf("Count of '%c' is %d\n", c, count[c - 'a']);

Answer 4

我找到了一个简单的程序here。它需要两个输入，首先是你要计算的字符和必须计算字符出现的文件名。

Answer 5

int strchro(char c, char *str) {

    char *pch;
    int found = 0;
    pch=strchr(str,c);
    while (pch!=NULL)
    {
        //printf("found at %d\n",pch-str+1);
        found++;
        pch=strchr(pch+1,c);
    }
    return found;
}

我刚才写的一个旧功能..希望这有帮助;）
更多信息：http://en.cppreference.com/w/c/string/byte/strchr

Answer 6

您可以使用256个整数的数组（在许多平台上char是一个8位值）。由于字符在文件中出现的次数不能为负，因此无符号类型是有意义的。

unsigned charCount[256] = { 0 };

数组中的每个插槽表示具有该值的字符出现在该文件中的次数。

while ((ch = fgetc(in_file)) != EOF)
{
    // increment the count of character ch
    charCount[ch]++;
}

打印出来时，某些字符不可打印或是空格（如果您正在阅读二进制文件，这尤其适用），您可以使用isprint和isspace函数找到{ {1}}标题。

ctype.h

C - 文件中的字符数

6 个答案: