malloc 返回的内存地址是否总是可以通过指向另一种类型的指针互换？

Question

char arr[512] = {0};
int *ptr = (int *)arr; // WRONG
                       // A bus error can be caused by unaligned memory access

printf("%d\n", *ptr);

另一方面：

The block that malloc gives you is guaranteed to be aligned so that it can hold any type of data.

char *arr= malloc(512);
int *ptr = (int *)arr; // OK, arr is properly aligned for ptr

memset(arr, 0, 512);
printf("%d\n", *ptr);

这个假设是正确的还是我遗漏了什么？

Answer 1

C 标准保证 malloc 将 return 内存适当对齐最严格的 基本类型 （例如 uint64_t）。如果您有更严格的要求，则必须使用 aligned_alloc 或类似的东西。

7.22.3

The pointer returned if the allocation succeeds is suitably aligned so that it may be assigned to a pointer to any type of object with a fundamental alignment requirement and then used to access such an object or an array of such objects in the space allocated (until the space is explicitly deallocated)

关于aligned_alloc：

void *aligned_alloc(size_t alignment, size_t size);
The aligned_alloc function allocates space for an object whose alignment is specified by alignment, whose size is specified by size, and whose value is indeterminate.

就对齐而言，您的代码是正确的。我不是特别喜欢指针转换（char * 到 int *），但我认为它应该可以正常工作。

malloc 返回的内存地址是否总是可以通过指向另一种类型的指针互换？

Is the memory address returned by malloc always interchangeable by a pointer to another type?

c

alignment

malloc