为函数中结构内部的指针分配内存

Allocating memory for pointers inside structures in functions

假设我声明了一个结构如下:

typedef struct {
    int *numbers;
    int size; // Size of numbers once treated as array
} sstruct;

然后我使用 sstruct *example;.

main() 中创建了一个指向结构的指针(以便稍后通过引用传递它)

然后我有一个函数,称之为allocpositions(),我在其中假装为*example中包含的*p指针分配内存位置。

如果我想将位置分配给 *example,将方向 &example 传递给函数就足够了,该函数将接收它作为 **a 然后做一些事情像 a = (sstruct **)malloc(N*sizeof(sstruct *)),但我不知道如何在函数内部直接分配给 *p

一旦分配,我是否仍然能够将 *p 中的元素引用为 allocpositions() 中的 example->p[index]

如有任何帮助,我将不胜感激!

编辑

说明我尝试实现的示例代码:

typedef struct {
    int *numbers;
    int size; // size of numbers once treated as array
} ssm;

main() {
    ssm *hello;
    f_alloc(&hello);
}

void f_alloc(ssm **a) {
    // Here I want to allocate memory for hello->p
    // Then I need to access the positions of hello->p that I just allocated
}

带注释的代码:

void f_alloc(ssm **a) {

    *a = malloc(sizeof(ssm)); // Need to allocate the structure and place in into the *a - i.e. hello in main
    (*a)->p = malloc(sizeof(int)); // Allocate memory for the integer pointer p (i.e. hello ->p;
}

编辑

我认为这就是您的要求:

void f_alloc(ssm **a, unsigned int length) {

    *a = malloc(sizeof(ssm)); // Need to allocate the structure and place in into the *a - i.e. hello in main
    (*a)->p = malloc(sizeof(int) * length); // Allocate memory for the integer pointer p (i.e. hello ->p;
    (*a)->v = length; // I am assuming that this should store the length - use better variable names !!!
}

然后一个函数到set/get

bool Set(ssm *s, unsigned int index, int value) {
   if (index >= s->v) {
       return false;
   }
   s->p[index] = value;
   return true;
}

bool Get(ssm *s, unsigned int index, int *value) {
   if (index >= s->v) {
       return false;
   }
   *value = s->p[index];
   return true;
}

我把免费位留给 reader。

编辑 2

因为心情好

void Resize(ssm**a, unsigned int new_length)
{
   (*a)->p = relloc((*a)->p, sizeof(int) * new_size);
   (*a)->v = new_length;
}
void Free(ssm *a)
{
   free(a->p);
   free(a);
}

你可以提高容错性来检查malloc/realloc是否有效