# Why malloc() 不需要轉型 (cast)
###### tags: `C`, `malloc`, `explicit`, `cast`

C99 2007 7.20.3 Memory management functions 提到:
> The order and contiguity of storage allocated by successive calls to the **calloc**, **malloc**, and **realloc** functions is unspecified. **The pointer returned if the allocation succeeds ==is suitably aligned== so that ==it may be assigned to a pointer to any type of object== and then used to access such an object or an array of such objects in the space allocated (until the space is explicitly deallocated)**.
>
> The **lifetime of an allocated object extends from the allocation until the deallocation.** Each such allocation shall yield a pointer to an object disjoint from any other object.
>
> **The pointer returned points to the start (lowest byte address) of the allocated space.** **If the space cannot be allocated, a null pointer is returned.**
>
> If the size of the space **requested is zero, the behavior is implementation-defined**: either a null pointer is returned, or the behavior is as if the size were some nonzero value, except that the returned pointer shall not be used to access an object.
* **C 語言 malloc 成功時是回傳開頭位址，並且會自動對齊，因此使用上根本不需要做轉型。**
* [細看 malloc 關於 alignment 的實作](https://hackmd.io/@sysprog/c-memory?type=view#data-alignment)
> 實際上到底 malloc 做了怎樣的 data alignment，繼續翻閱 The GNU C Library - Malloc Example，裡面特別提到:
> > In the GNU system, the address is always a multiple of eight on most systems, and a multiple of 16 on 64-bit systems.
>
> **在大多數系統下，malloc 會以 8 bytes 進行對齊；而在 64-bit 的系統下，malloc 則會以 16 bytes 進行對齊。**
> 對應的實驗，malloc 在 Linux x86_64 以 16 bytes 對齊：
>
> ```c
> for (int i = 0; i < 10000; ++i) {
> char *z;
> z = malloc(sizeof(char));
> }
> ```
> 結果用 gdb 測過之後，發現位址的結尾的確都是 0 結尾，表示真的是以 16-byte 做對齊。
* storage allocated 的生命週期是到 deallocation 為止，即 `free()`
* 若 malloc 失敗會回傳 null pointer
* malloc 一個 size = 0 的空間，此為未實作定義，不同編譯器有不同做法。