Question

我从文件中读取了一个文本行，我需要将其转换为类似于main函数参数的结构。例如，如果char数组是char* text="There are books in the library."并且我有以下结构定义：

struct mystruct{
   int argc;
   char** argv;
};

如果我有struct mystruct a，功能foo(a, text)我最终a.argc等于6，a.argv[0]等于那里，{等于的{1}} ，...。

C中是否有任何可用于此功能的函数或库？因为当我们执行C main函数时，这个转换是为输入参数自动完成的。

Answer 1

您可以使用strtok进行此拆分，如下所示：

struct split_result {
    int cnt;
    char *buf;
    char **strs;
};

int
split(const char *str, struct split_result *rst)
{
    int idx, str_num;
    char *buf, *sep, **strs;

    buf = strdup(str);
    if (buf == NULL) {
        perror("strdup");
        return -1;
    }

    str_num = 1;
    strs = malloc(str_num * sizeof(char *));
    if (strs == NULL) {
        perror("malloc");
        return -1;
    }

    sep = " \t";
    idx = 0;
    for (strs[idx] = strtok(buf, sep);
         strs[idx];
         strs[idx] = strtok(NULL, sep))
    {
        idx++;
        if (idx >= str_num) {
            str_num += 10;
            strs = realloc(strs, str_num * sizeof(char *));
            if (strs == NULL) {
                perror("realloc");
                return -1;
            }
        }
    }

    rst->cnt = idx;
    rst->strs = strs;
    rst->buf = buf;

    return 0;
}

这是一个测试程序及其输出：

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

struct split_result {
    int cnt;
    char *buf;
    char **strs;
};

int
split(const char *str, struct split_result *rst)
{
    int idx, str_num;
    char *buf, *sep, **strs;

    buf = strdup(str);
    if (buf == NULL) {
        perror("strdup");
        return -1;
    }

    str_num = 1;
    strs = malloc(str_num * sizeof(char *));
    if (strs == NULL) {
        perror("malloc");
        return -1;
    }

    sep = " \t";
    idx = 0;
    for (strs[idx] = strtok(buf, sep);
         strs[idx];
         strs[idx] = strtok(NULL, sep))
    {
        idx++;
        if (idx >= str_num) {
            str_num += 10;
            strs = realloc(strs, str_num * sizeof(char *));
            if (strs == NULL) {
                perror("realloc");
                return -1;
            }
        }
    }

    rst->cnt = idx;
    rst->strs = strs;
    rst->buf = buf;

    return 0;
}

int
main(void)
{
    int i, j;
    struct split_result rst;
    const char *msg[] = {
        "",
        " One",
        " One  Two",
        " One  Two Tree  ",
        "There are books in the library.",
        NULL,
    };

    for (i = 0; msg[i]; i++) {
        if (split(msg[i], &rst) < 0) {
            exit(EXIT_FAILURE);
        }

        printf("msg[%d] = >>%s<<\n", i, msg[i]);
        for (j = 0; j < rst.cnt; j++) {
            printf("cnt = %d: |%s|\n", j, rst.strs[j]);
        }

        free(rst.strs);
        free(rst.buf);
    }

    exit(EXIT_SUCCESS);
}

输出：

$ ./a.out 
msg[0] = >><<
msg[1] = >> One<<
cnt = 0: |One|
msg[2] = >> One  Two<<
cnt = 0: |One|
cnt = 1: |Two|
msg[3] = >> One  Two Tree  <<
cnt = 0: |One|
cnt = 1: |Two|
cnt = 2: |Tree|
msg[4] = >>There are books in the library.<<
cnt = 0: |There|
cnt = 1: |are|
cnt = 2: |books|
cnt = 3: |in|
cnt = 4: |the|
cnt = 5: |library.|

Answer 2

我假设文件有一个数字，指定参数的数量。如果是这种情况，那么创建一个循环并读取每个字符串，直到遇到分隔符（再次，假设你被告知有一个分隔符）。如果没有“数”参数，则在结构中添加unsigned类型的成员，以跟踪读取的单词数。你真的不需要创建一个split_function，但如果你决定创建一个split_function，它就不会受到伤害：）

将char数组转换为类似于int argc和char ** argv的结构

2 个答案: