主

Question

我试图模仿GNU find命令。我将所有命令选项都设为linkedlist。我正在尝试处理argv中的所有选项并将它们存储在链表中。每个node一个选项，但似乎没有更新。我认为当函数返回时，它仍然指向最后一个node

主

int main(int argc, char *argv[]) {

    char *path_list[argc];


    s_option *p = process_parms(argc, path_list, argv);

    char **pp = path_list;

    printf("%s \n", path_list[0]); // prints Documents/
    printf("help is %d\n",p->help); // it doesn't work. is 0 should be 1



   return 0;
}

struct here

typedef struct _allopts {
    int help;
    int print;
    char f_type;
    int ls;
    char *user;  
    unsigned long user_id;
    char *name;
    struct _allopts *next;

} s_option;

过程参数方法

s_option *process_parms(const int len, char *spath[], char **pms) {

    int index = 0;

    s_option *op = malloc(sizeof(struct _allopts));


    for (int i = 1; (i < len); ++i, op = op->next) {

        op->next = malloc(sizeof(struct _allopts));


        if (op->next == NULL) {
            fprintf(stderr, "myfind: calloc(): %s\n", strerror(errno));
            exit(EXIT_FAILURE);
        }


        if (strcmp(pms[i], "-name") == 0) {
            size_t l = strlen(pms[i]);
            op->name = malloc(sizeof(char) * l + 1);
            strcpy(op->name, pms[i]);
            op->name[l + 1] = '\0';


            continue;
        } else if (strcmp(pms[i], "-help") == 0) {
            op->help = 1;

            continue;
        } else if (strcmp(pms[i], "-print") == 0) {
            op->help = 1;

            continue;

        } else if (strcmp(pms[i], "-ls") == 0) {
            op->ls = 1;

            continue;
        } else if (strcmp(pms[i], "-type") == 0) {
            char f = *(pms[++i]);
            if (f == 'f' || f == 'b' || f == 'c' ||
                f == 'd' || f == 's' || f == 'p' || f == 'l') {
                op->f_type = f;

                continue;
            } else {
                printf("myfind: Unknown argument to %s: %c\n", pms[i-1], *pms[i]);
                break;
            }


        } else



        if (strcmp(pms[i], "-user") == 0) {

            struct passwd *pd;
            if (pms[++i]) {
                op->user = strcpy(malloc(sizeof(strlen(pms[i]))), pms[i]);
                if ((pd = getpwnam(pms[i]))) {
                    op->user_id = pd->pw_uid;

                    continue;
                } else if (isdigit(pms[i][0])) {
                    sscanf(op->user, "%lu", &op->user_id);

                    continue;
                } else {
                    printf("myfind: `%s` is not a the name of a known user \n", pms[i]);
                    break;
                }


            } else {

                printf("myfind: missing argument to `%s`\n", pms[i]);
                break;

            }


        }



        /*Getting the path here.
        * */
        if (*(*(pms + 1)) != '-') {

            size_t l = strlen(pms[i] + 1);


            spath[index] = strcpy(malloc(sizeof(char) * l), pms[i]);
            (spath+index)[l + 1] = '\0';
            index++;


        } else {
            printf("myfind: Unknown predicate `%s`\n", pms[i]);
            //docs/ doc1/ doc2/ -ls -type m -user james
            break;

        }

    }

    spath[index] = NULL;


    return op;
}

我在这里错过了一些东西。我尝试使用带有printf的调试器，它非常好地更新了元素，但在主要内容它一直显示0以寻求帮助。谢谢你的帮助。

Answer 1

该函数返回最后一个节点，指针op：

指向该节点

return op;

因为指针op在这里递增：

for (int i = 1; (i < len); ++i, op = op->next) {    
                                ^^^^^^^^^^^^^

复制指向第一个节点的指针并改为返回：

s_option *op = malloc(sizeof(struct _allopts));
s_option *first = op;
//...
return first;

linkedlist元素在for循环中不能很好地更新

主

struct here

过程参数方法

1 个答案: