如何使用 POSIX 函数计算 C 语言文件中的多个字符？

Question

我正在尝试编写一个程序，使用标准 POSIX 函数获取一个文件和一个字符串，程序计算字符串包含的文件中的所有字符。

例如，如果用户写：

count.exe x.txt abcd

程序计算文件中a,b,c,d每个字符的个数x.txt

示例消息：

Number of 'a' characters in 'x.txt' file is: 4
Number of 'b' characters in 'x.txt' file is: 9
Number of 'c' characters in 'x.txt' file is: 7
Number of 'd' characters in 'x.txt' file is: 0

目前得到的代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <errno.h>

#define BUFSIZE     1024


void exit_sys(const char* msg)
{
    perror(msg);
    exit(EXIT_FAILURE);
}

void exit_fail(const char* msg)
{
    fprintf(stderr, "%s\n", msg);
    exit(EXIT_FAILURE);
}

int get_count(char* p, size_t size, char c)
{
    int count = 0;
    size_t i;

    for (i = 0; i < size; ++i)
        if (p[i] == c)
            ++count;

    return count;
}


void run_count_characters_application(int argc, char** argv)
{
    int fd;
    char c;
    char buf[BUFSIZE];
    int n;
    int count;

    if (argc != 3)
        exit_fail("usage: ./mycounter file character");

    if (strlen(argv[2]) < 0)
        exit_fail("You have to give at least one character");

    c = argv[2][0];

    if ((fd = open(argv[1], O_RDONLY)) < 0)
        exit_sys("open");

    count = 0;


    while ((n = read(fd, buf, BUFSIZE)) > 0)
        count += get_count(buf, n, c);

    if (n < 0)
        exit_sys("read");


    printf("Count:%d\n", count);

    close(fd);
}

int main(int argc, char** argv)
{
    run_count_characters_application(argc, argv);

    return 0;
}

到目前为止我在这段代码中得到的问题是它只计算一个字符（只计算第一个字符），我想知道如何让它读取和计算我在命令中写的其他字符, 提前谢谢你:)

Answer 1

由于您在评论中要求示例：

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

void die(const char *reason)
{
        fprintf(stderr, "%s\n", reason);
        exit(EXIT_FAILURE);
}

int main(int argc, char **argv)
{
        if (argc != 3)
                die("usage: ./count file characters");

        FILE *f = fopen(argv[1], "r");
        if (f == NULL)
                die("unable to open file");

        unsigned int counter[UCHAR_MAX + 1] = { 0 };
        int c;
        while ((c = fgetc(f)) != EOF)
                counter[c]++;
        fclose(f);

        size_t len = strlen(argv[2]);
        for (unsigned int i = 0; i < len; i++) {
                char c = argv[2][i];
                unsigned int count = counter[c];
                printf("Number of '%c' characters in '%s' file is: %u\n", c, argv[1], count);
        }
}

$ cc count.c -o count
$ echo "bbaaafff" > test.txt
$ ./count test.txt afm
Number of 'a' characters in 'test.txt' file is: 3
Number of 'f' characters in 'test.txt' file is: 3
Number of 'm' characters in 'test.txt' file is: 0
$

函数fgetc()

return the character read as an unsigned char cast to an int or EOF on end of file or error.

unsigned char 可以存储 UCHAR_MAX + 1 个可能的值（通常为 0 到 255）所以我制作了一个数组，可以通过这些值（counter）进行索引来存储我们的很重要。将其视为来自其他语言的 "map"，将字符映射到它们的计数。

然后在最后，我遍历输入字符串中的字符并打印它们的计数。

如评论中所述，这仅适用于 ASCII 个字符。

如何使用 POSIX 函数计算 C 语言文件中的多个字符？

How to count multiple characters from a file in C language using POSIX functions?

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posix

counting