Malloc and alignment

October 17, 2023

As far as I know, every type – like int – can only be stored at memory addresses which start at a power of 2. Now my question is, how can malloc cope with this requirement? For instance, if I call

void * ptr;
ptr = malloc(8);

Then it is not obvious what I want to store in the given memory block of eight bytes. Is it therefore true that malloc always returns a starting address of the block which is located the largest power of 2 such that every standard type could be stored? If I misunderstand something, please feel free to correct me.

>Solution :

C 2018 7.22.3 specifies the behavior of the memory allocation routines and its paragraph 1 says “… 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…”

6.2.8 2 says:

… The alignment requirements of the following types shall be fundamental alignments:

all atomic, qualified or unqualified basic types;

all atomic, qualified or unqualified enumerated types;

all atomic, qualified or unqualified pointer types;

all array types whose element type has a fundamental alignment requirement;

all types specified in Clause 7 as complete object types;

all structure or union types all of whose elements have types with fundamental alignment requirements and none of whose elements have an alignment specifier specifying an alignment that is not a fundamental alignment.