将令牌存储到数组中以稍后作为参数传递答案

【问题标题】：Storing tokens into an array to later pass as parameters将令牌存储到数组中以稍后作为参数传递
【发布时间】：2015-09-21 01:06:29
【问题描述】：

我有一个已经标记化的文件，但我需要将每个标记存储在一个数组中以供以后用作参数。我该怎么做呢？

//         Read in File           //
FILE *fp;
char buffer[100];

fp = fopen(params, "r");

printf("Here is filename...");
printf("%s\n", params);

fseek(fp, 0, SEEK_END);
//byte_size = ftell(fp);
rewind(fp);

if (fgets(buffer,sizeof(buffer),fp) != NULL)
{
    char*p, *b;
    b = buffer;
    printf("parsing %s", buffer);
    while ((p = strsep(&b, ",")) != NULL)
    {
        printf("param: %s\n",p);
    }
}
fclose(fp);

【问题讨论】：

我试过 strcpy() 但是每次我把它放入一个数组中我都会得到一个不兼容的指针类型错误。我需要保存每个标记，以便以后在新函数中作为新变量传递。

标签： c arrays stringtokenizer

【解决方案1】：

使用链表并稍后将其转换为数组可能会很好，因为我们不知道有多少令牌。

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

char *strsep(char **stringp, const char *delim);

typedef struct node_tag {
    char *str;
    struct node_tag* next;
} list_node;

list_node* create_node(const char *str) {
    list_node* n = malloc(sizeof(list_node));
    if (n == NULL) exit(1);
    if (str == NULL) {
        n->str = NULL;
    } else {
        n->str = malloc(sizeof(char) * (strlen(str) + 1));
        if (n->str == NULL) exit(1);
        strcpy(n->str, str);
    }
    n->next = NULL;
    return n;
}

int main(void) {
    const char *params = "dummy";
    FILE *fp;
    char buffer[100];
    list_node *head = NULL;
    list_node **tail = &head;
    unsigned int count = 0;
    unsigned int i;
    char **array;
    fp = stdin;//fopen(params, "r");
    printf("Here is filename...");
    printf("%s\n", params);
    fseek(fp, 0, SEEK_END);
    //byte_size = ftell(fp);
    rewind(fp);

    if (fgets(buffer,sizeof(buffer),fp) != NULL)
    {
        char*p, *b;
        b = buffer;
        printf("parsing %s", buffer);
        while ((p = strsep(&b, ",")) != NULL)
        {
            printf("param: %s\n",p);
            *tail = create_node(p);
            tail = &(*tail)->next;
            count++;
        }


    }

    array = malloc(sizeof(char*) * count);
    if (array == NULL) return 1;
    for (i = 0; i < count && head != NULL; i++) {
        list_node *next = head->next;
        array[i] = head->str;
        // Don't free(head->str) because it is used
        free(head);
        head = next;
    }

    for (i = 0; i < count; i++) {
        printf("array[%u] = %s\n", i, array[i]);
    }
    for (i = 0; i < count; i++) free(array[i]);
    free(array);

    //fclose(fp);
    return 0;
}

【讨论】：

有 8 个标记，但长度可能不同。

【解决方案2】：

如果你知道令牌的数量，你可以简单地使用数组。

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

char *strsep(char **stringp, const char *delim);

#define ARRAY_SIZE 8

int main(void) {
    const char *params = "dummy";
    FILE *fp;
    char buffer[100];
    char *array[ARRAY_SIZE];
    int array_count = 0;
    int i;
    fp = stdin;//fopen(params, "r");
    printf("Here is filename...");
    printf("%s\n", params);
    fseek(fp, 0, SEEK_END);
    //byte_size = ftell(fp);
    rewind(fp);

    if (fgets(buffer,sizeof(buffer),fp) != NULL)
    {
        char*p, *b;
        b = buffer;
        printf("parsing %s", buffer);
        while ((p = strsep(&b, ",")) != NULL)
        {
            printf("param: %s\n",p);
            if (array_count < ARRAY_SIZE)
            {
                array[array_count] = malloc(sizeof(char) * (strlen(p) + 1));
                strcpy(array[array_count], p);
                array_count++;
            }
        }


    }

    for (i = 0; i < array_count; i++) {
        printf("array[%u] = %s\n", i, array[i]);
    }
    for (i = 0; i < array_count; i++) free(array[i]);

    //fclose(fp);
    return 0;
}

【讨论】：