是否可以将相同的指向 struct 的指针数组指向不同类型的 struct？

Question

I'm working with pointers to structs and have the following set up which has been working.

/* Initialized. */
struct base { ... };  
struct base **db;
db = malloc( base_max * sizeof *db );

/* When a new struct base is required. */
db[ i ] = malloc( sizeof( struct base ) );

Now, if possible, although not really essential, I'd like only db[0] to point to a different struct, struct mem {... } . Is this possible and what is the proper way to do so?

I figured I can set db[0] = malloc( sizeof( struct mem ) ) ; but db is already declared to point to a pointer pointing to a struct base ; and the memory allocation of db is based on that also.

I'm confused because I read pointers must have a type or pointer arithmetic won't work properly; but I also read you don't have to cast the pointers from malloc even though it returns void pointers.

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Regarding the duplicate question, although it discusses void type it doesn't answer my question of how to accomplish this or the risks of it. The comment by @4386427 appears to point to the real issue to be considered which isn't addressed in the other question or the one answer to this one. Thank you.

Answer 1

In majority of the architecture, the size of a pointer (to any type) is constant, and any pointer type will generally occupy the same amount of memory.

Unless you're on some really uncommon hardware/platform, you can assign db[0] with any pointer returned by malloc() using a different structure type used for sizing calculation altogether.

 db[0] = malloc( sizeof( struct mem ));

should do good, you just need to worry about memory leak, as you'll be overwriting the previous memory location returned by malloc() .

However, since you're storing a pointer to memory of a differnt type, while using db[0] , ie, dereferencing it, you need to cast it to the proper pointer type (ie, struct mem* ) before you can use the pointer to access /operate based on the struct mem type.

That said,

I'm confused because I read pointers must have a type or pointer arithmetic won't work properly;

That's true

but I also read you don't have to cast the pointers from malloc even though it returns void pointers.

Also true, as when you assign the pointer to a variable of a certain type (other than void ), the conversion is implicit. You can use that variable to perform pointer arithmetic.