动态分配向量C

Question

I'm trying to understand the concept of dynamically allocating a `struct´ in C, also I am interested in generic dynamically allocating, could you give me some help or information about this topic?

I've managed to understand the concept of pointers and I managed to to some kind of code but I am not sure if is right or if it's working as it should be...

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

int main(void)
{
    typedef struct {
        int id;
        char nume[20];
    } pers;

    pers *i = malloc(sizeof(pers)*100);

    i->id=22;
    i++;
    i->id=33;
}

My question is, have I declared this array properly, how do I print an element of this array, how do I refer to a specific element for eq. i[9] , and is there another way to do the same thing? I've heard about generic dynamically allocating memory, could someone give me an eq. of that?

Answer 1

When you do i++ you loose the original pointer. Instead you can use normal array indexing:

i[0].id = 22;
i[1].id = 33;

You need the original pointer when you later want to free the allocated memory.

Answer 2

以下所有都是同义词：

*(i + 9)

*(9 + i)

i[9]

9[i]

Answer 3

You refer to an element in the dynamically allocated array the same way you would refer to an element in an automatic array.

Using pers[i] will refer to the ith element. To access a field such as id you would use pers[i].id .

Answer 4

Normally you would do the typedef struct before (outside) the function but it works.

Your allocation is correct.

But i++ makes no sense here. You should keep the pointer i so that you still know where the allocated memory starts. What you probably want is another pointer into the array, and increase that:

pers* it = i;
it->id=22;
printf("%d", it->id);
it++;
it->id=33;
printf("%d", it->id);

Answer 5

You shouldn't change the pointer that malloc returned you. Treat it like a constant to avoid creating a memory leak (allocating and losing the pointer to the allocated memory). Instead of modifying i , iterate using a new index variable, let's call it j : To clear/initialize all array elements:

#include <string.h> /* for memset(). */
int j;
for (j=0; j<100; ++j) {
  i[j].id = 0;
  memset (i[j].nume, 0, 20); /* Or sizeof(i[0].nume) instead of 20. */
}

To print all array elements:

for (j=0; j<100; ++j) {
  printf ("id[%d] = %d\n", j, i[j].id);
  printf ("nume[%d] = '%s'\n", j, i[j].nume);
}

I would also suggest to rename i to person ; this makes the code so much more readable.