C语言 C - 文件中的字符数
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C - Character Count from a File
提问by Sarah97
So I want to make a program which counts the occurrences of each character in a file. For example:
所以我想制作一个程序来计算文件中每个字符的出现次数。例如：
4 instances of character 0x67 (g)
字符 0x67 (g) 的 4 个实例
11 instances of character 0x68 (h)
字符 0x68 (h) 的 11 个实例
and so on
等等
I am not sure how to display and count instances.
我不确定如何显示和计算实例。
Any thoughts?
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);
回答by Escualo
This is what I came up with...
这是我想出来的……
#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;
}
I compile the code using the following command (GCC 4.8.1 on OS X 10.7.4)
我使用以下命令编译代码（OS X 10.7.4 上的 GCC 4.8.1）
gcc ccount.c -Wall -Wextra -pedantic -ansi
it compiles with no warnings and no errors; this is the output:
它编译时没有警告和错误；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
回答by I?ya Bursov
you need to use array, check out:
您需要使用数组，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]);
回答by Ajeet Khan
回答by LihO
If you want to retrieve counts of alphabet characters, then it could look like this:
如果要检索字母字符的计数，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'];
}
printing them out could be simple as:
打印出来可能很简单：
for (char c = 'a'; c <= 'z', ++c)
printf("Count of '%c' is %d\n", c, count[c - 'a']);
回答by Joe DF
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;
}
an old function i wrote a while ago.. hope this helps ;)
more info here: http://en.cppreference.com/w/c/string/byte/strchr
我前段时间写的一个旧函数..希望这有帮助;)
更多信息在这里：http: //en.cppreference.com/w/c/string/byte/strchr
回答by dreamlax
You could use an array of 256 integers (on many platforms charis an 8-bit value). Since the number of times a character appears in a file can't be negative, an unsigned type makes sense.
您可以使用 256 个整数的数组（在许多平台上char是 8 位值）。由于字符在文件中出现的次数不能为负数，因此无符号类型是有意义的。
unsigned charCount[256] = { 0 };
Each slot in the array represents the number of times the character with that value appears in that file.
数组中的每个槽表示具有该值的字符在该文件中出现的次数。
while ((ch = fgetc(in_file)) != EOF)
{
// increment the count of character ch
charCount[ch]++;
}
When printing them out, some characters are not printable or are whitespace (this is particularly applicable if you are reading a binary file), you can use the isprintand isspacefunctions found in the ctype.hheader.
打印出来时，有些字符是不可打印的或者是空白字符（这在读取二进制文件时特别适用），您可以使用标题中的isprint和isspace函数ctype.h。
for (int i = 0; i < 256; i++)
{
// only display characters with a count of at least 1
if (charCount[i] > 0)
{
if (!isprint(i) || isspace(i))
printf("%u instances of character %x\n", charCount[i], (unsigned) i);
else
printf("%u instances of character '%c'\n", charCount[i], i);
}
}
