在 C 中正确初始化递归结构类型 [英] Initialize recursive struct type in C properly

问题描述

我是 C 编程的新手，我有以下结构:

I am new to C programming and I have the following struct:

typedef struct _date {
    char d[10], t[5], s[3];
    struct _date *next;
} *date;

如何正确创建此实例?

我的解决方案:

date neuerTermin(char *d, char *t, char *s, date cal) {
    struct _date d = {*d, *t, *s, NULL};
    date d_ptr = malloc(sizeof *d);
    cal->next = d_ptr;

    return d;
}

但我收到一个错误:warning: 初始化从指针生成整数而无需强制转换

推荐答案

您可以这样做:

#include <stdio.h>

typedef struct _date {
    char d[11], t[6], s[4]; // +1 to size for null-terminator ('\0')
    struct _date *next;
} *date;


int main() {
    struct _date a = { "15.07.2017", "16:00", "Foo", NULL };
    date a_ptr = &a;

    printf("Description: %s\nDate: %s\nTime: %s\n", a_ptr->s, a_ptr->d, a_ptr->t);
    return 0;
}

上例中用大括号括起来的逗号分隔列表是 struct 初始化程序.

The brace-enclosed, comma-separated list in the example above is the struct initializer.

为了回应您的问题的编辑，如果您希望动态分配 struct _date 实例并在函数中初始化它们，请使用 malloc 如下:

To respond to the edits of your question, if you wish to dynamically allocate struct _date instances and initialize them in a function, then use malloc as follows:

date neuerTermin(const char* d, const char* t, const char* s) {
    date cal = (date)malloc(sizeof(struct _date));

    strncpy(cal->d, d, 10);
    strncpy(cal->t, t, 5);
    strncpy(cal->s, s, 3);
    cal->next = NULL;

    return cal;
}

在这种情况下，您必须逐个填充cal 指向的内存块.示例用法:

In this case you have to fill the memory block pointed by cal member-by-member. Sample usage:

date root = neuerTermin("15.07.2017", "16:00", "Foo");
root->next = neuerTermin("27.07.2017", "10:00", "Bar");
root->next->next = neuerTermin("01.08.2017", "12:30", "Baz");

重要提示:如果您使用 malloc 分配内存，您也必须使用 free 当您不再需要它时.

Important: if you used malloc to allocate memory, you have to deallocate it too with using free when you don't need it anymore.

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