Question

我有以下代码

serial.flushInput()
serial.readline() #do not save this value - throw away

while (serial.inWaiting()==0): #wait until there is data
    pass

valueRead = serial.readline() #keep this value

现在我要分配堆内存

这是我的尝试

struct USER{

   int human_id_number;

   char first_name_letter;

   int minutes_since_sneezing;

} *administrator;

不确定这是否正确...

Answer 1

struct USER {
   int human_id_number;
   char first_name_letter;
   int minutes_since_sneezing;
} *administrator;

这不仅是一个结构声明，它还是一个变量声明...它与：

struct USER {
   int human_id_number;
   char first_name_letter;
   int minutes_since_sneezing;
};

struct USER *administrator;

因此，当您随后使用sizeof(administrator)时，将得到“ 指针的大小” ...这很可能不是想要。

您可能想做更多类似的事情：

struct USER {
   int human_id_number;
   char first_name_letter;
   int minutes_since_sneezing;
};

int main(void) {
    struct USER *administrator;

    administrator = malloc(sizeof(*administrator));
    /* - or - */
    administrator = malloc(sizeof(struct USER));

    /* check that some memory was actually allocated */
    if (administrator == NULL) {
        fprintf(stderr, "Error: malloc() returned NULL...\n");
        return 1;
    }

    /* ... */

    /* don't forget to free! */
    free(administrator)

    return 0;
}

sizeof(*administrator)和sizeof(struct USER)都将为您提供“ USER结构的大小”，因此malloc()的结果将是指向足够的内存来存储结构的数据。

Answer 2

struct USER{
    int human_id_number;
    char first_name_letter;
    int minutes_since_sneezing;
} *administrator;

这将管理员定义为指针变量。但是，从其他代码开始

administrator *newStruct = (administor*)malloc(sizeof(administrator));

您似乎想将其用作类型。为此，您可以使用typedef。

typedef struct USER{
    int human_id_number;
    char first_name_letter;
    int minutes_since_sneezing;
} administrator;

然后使用

administrator *newStruct = (administrator *)malloc(sizeof(administrator));

如何为结构分配堆内存？

2 个答案: