这条语句之后：

A *list = (A*) malloc(sizeOfA * sizeof(A));

列表是一个指向一个内存块的起始位置，可容纳类型的 sizeOfA 元素结构A 。因此， *名单的类型为结构A ，同样，列表[我] 的类型结构A ，不是指针到结构A的（这将是列表+ I ）。

list is a pointer to the starting location of a memory block that can hold sizeOfA elements of type struct A. Thus, *list is of type struct A, and similarly, list[i] is of type struct A, not pointer to struct A (that would be list+i).

如果你想列表[我] 是一个指向结构A ，那么你的第二件$的C $ C.将很正确的，因为你有足够的空间分配的存储空间来保存 sizeOfA 指针结构A 。请注意，您只分配空间来存放指针，而不是实际的结构A 实例。尝试读取列表[I] - 方式＆gt;ç将导致不确定的行为

If you want list[i] to be a pointer to struct A, then your second piece of code would be the correct one, since you're allocating a memory location with enough space to hold sizeOfA pointers to struct A. Note that you are only allocating space to hold pointers, not actual struct A instances. Attempting to read list[i]->c will result in undefined behavior.