In the third post of the memory series we briefly explained locality and why it is an important principle to keep in mind while developing a memory-intensive program. This new post is going inflatable obstacle course to be more concrete and explains what actually happens behind the scene in a very simple example.

This post is a follow-up to a recent interview with a (brilliant) candidate. As a subsidiary question, we comprar castillo hinchable baratos presented him with the following two structure definitions:

struct foo_t { int len; char *data; }; struct bar_t { int len; char data[]; };

The question was: what is the difference between aufblasbarer hindernisparcours these two structures, what are the pros and the cons of both of them? For the remaining of the article we will suppose we are working on an x86_64 architecture.

By coincidence, an intern asked more or less at inflatable tent the same time why we were using bar_t -like structures in our custom database engine.

Pointer vs Array

So, the first expected remark is that here, we have the difference between a pointer ( foo_t ) and an array ( bar_t ).

In C, a pointer is a variable that stores the memory address of the data it references. In our case it is a 64 bits variable. So foo_t is composed of len that is a 4 bytes integer and data that is a 8 bytes integer. Because 8 bytes integer have to be aligned on 8 bytes , this has two consequences:

the whole structure foo_t has an alignement requirement of 8 bytes

has an alignement requirement of 8 bytes the variable data in foo_t has an alignement requirement of 8 bytes

This causes, len to be aligned on 8 bytes, and data to be placed on the next 8 bytes-aligned place, that is 4 bytes after len, creating a hole of 4 bytes between the two variables.

The actual memory that stores the data is placed elsewhere, probably in a separate allocation.

The bar_t structure ends with a variable-length array of characters. In C an array is a type that contains data. Developers are often confused by the difference between an array and a pointer because the language has an implicit cast from the array type to the pointer one, just by taking the address of the beginning of the array (which has the good property of making arrays and pointers accessible through the same syntax data[i] ).

void foo(void) { int buf[1024]; int *pbuf = buf; typeof(buf) == int[1024]; sizeof(buf) == sizeof(int) * 1024; typeof(pbuf) == int *; sizeof(pbuf) == 8; }

Another confusion comes from the fact the [] syntax can be used in function argument types and in that case both syntaxes are equivalent (however, the [] make it clear that we are talking about an array of values and not about a single pointed object). This is indeed so confusing that clang offers a dedicated warning for this ( -Wsizeof-array-argument ).

static void bar(int *pbuf, int buf[], int bufsiz[1024], int bufssiz[static 1024]) { sizeof(pbuf) == 8; sizeof(buf) == 8; sizeof(bufsiz) == 8; sizeof(bufssiz) == 8; } void foo(void) { int buf[1024]; bar(buf, buf, buf, buf); }

So, in the case of our bar_t structure, our data array just means that we want to able to access the memory that follows len as a sequence of char (that is 1 byte). An array has the alignment requirement of each individual object that it contains, so our data array is aligned on 1 byte which means it directly follows len in memory. bar_t has no intrinsic overhead: its size is the size of its length plus the size of the stored data.

This means our two structures have the following layouts in memory:

Allocations

In term of allocation, we already said that the data of the foo_t structure can point anywhere in memory. Most of the time, it comes from a separate allocation while the whole bar_t structure can only result from a single allocation. This causes foo_t to be more flexible: you can reallocate the data memory without affecting the pointer to the foo_t structure. On the other hand, the data array of the bar_t cannot be reallocated without reallocating the whole bar_t structure, potentially changing its pointer.

This causes the bar_t to be hard to embed by value in another structure (it will only be possible to put it at the end of the container structure) while foo_t can be placed anywhere. Both structures can be pointed to by other structures/variables without problem:

struct foo_container_t { struct foo_t foo_1; struct foo_t foo_2; struct foo_t *foo_3; }; struct bar_container_t { struct bar_t *bar_1; struct bar_t *bar_2; struct bar_t bar_3; /* Only possible here */ };

Because foo_t is larger that bar_t , it will always consumes more memory than bar_t for the same length: foo_t consumes 16 + len bytes while bar_t consumes 4 + len bytes. Moreover, since foo_t will most of the time require two allocations, each of which has an overhead of at least 8 bytes (with the glibc ‘s ptmalloc2 ), this causes the actual memory usage to be 16 + 2 * 8 + len for foo_t and 4 + 8 + len for bar_t . For a few large vectors, this is not a problem, but if you want a lot of small ones, the overhead of foo_t will be at least noticeable.

You can still allocate foo_t with a single allocation, however this will make reallocation harder (since the pointed data vector is not the allocated one and thus is not directly reallocable). This will slightly reduce the overhead for foo_t and improves the access performances (see next section). However the improvements are worth only for small allocations, when bar_t is already optimal.

struct foo_t *foo_alloc(int len) { struct foo_t *foo; foo = malloc(sizeof(struct foo_t) + len); foo->len = len; foo->data =, the gain observed by using a single dereferencement may be completely canceled by the context from which the structure is used. A careful design of your data structures is always required to benefit from the cache of your CPU. What did we expect from our brilliant candidate? Everything. In order to obtain a job as technical leader, the candidate has to master all these concepts and the discussion that follows the initial question must demonstrate this. Still that shouldn’t be a problem to any future brilliant candidate. struct bar_t *bar_alloc(int len) { struct bar_t *bar; bar = malloc(sizeof(struct bar_t) + len); bar->len = len; return bar; }

Locality

In order to better understand the performance’s implications, here is a small code snippet that can apply to both foo_t and bar_t . This simply gets the posth element of the data or -1 if the position is out of range. For this, we read the len member before dereferencing the data array.

int get_at(const struct foo_t *foo, int pos) { return pos >= foo->len ? -1 : foo->data[pos]; }

This produces the following assembly code (using clang -O2 ):

pushq %rbp movq %rsp, %rbp movl $-1, %eax cmpl %esi, (%rdi) jle LBB0_2 movslq %esi, %rax movq 8(%rdi), %rcx movsbl (%rcx,%rax), %eax LBB0_2: popq %rbp ret pushq %rbp movq %rsp, %rbp movl $-1, %eax cmpl %esi, (%rdi) jle LBB0_2 movslq %esi, %rax movsbl 4(%rdi,%rax), %eax LBB0_2: popq %rbp ret

Both assembly codes start with the same sequence: the len field is loaded ( (%rdi) ) and compared to the argument ( %esi ). If the comparison shows the position is greater (or equal) than len , we immediately go to the end of the function and we return %eax in which -1 has been previously stored.

In order to read len the processor has filled a line of its cache with the bytes surrounding the structure. Since a line of cache on current x86_64 processors is 64 bytes-long, the nearby memory is also loaded in the CPU, which gives on average 30 bytes of memory following the length. This means that the line of cache looks like this:

At this point, in the case of foo_t we loaded the 4 bytes of the length and the 8 bytes of the pointer to data as well as a lot of garbage. In the case of a bar_t , we fetched the 4 bytes of len and a few tens of bytes of the actual data . In case foo_t used the single allocation mechanism described above, a few tens of bytes of data would also be loaded, but on average 12 bytes less than bar_t (because of the size of the pointer and the hole in the structure). 12 of 64 bytes is a substantial part of the line of cache.

The next step in the case of a structure foo_t , is to load the pointer data ( 8(%rdi) , which means the content 8 bytes after the address stored in %rdi ) and stores the address in %rcx , then to load the data placed at that pointer plus the pos ( (%rcx,%rax) ) and move the value to %eax (since this stores the return value of the function). That particular instruction loads another line of cache.

In the case of the bar_t , the data array is in the continuation of len , which means that the address from which the result must be loaded is the address of the structure plus 4 (the size of len ) plus the position to fetch ( 4(%rdi,%rax) ). As a consequence, only a single instruction is required to load the value, not two, and in case pos is small, the source of the move will fall in the same line of cache as the one that was already fetched.

So, what’s the gain of bar_t here? In all cases, accessing the content of foo_t requires an indirection which result in an additional instruction to load the address of the pointer. However, in most cases, this additional instruction will have a very limited cost since the loaded pointer is already in the same line of cache as len (most of the time at least): loaded memory from a L1d cache costs only 4 CPU cycles.

Then, in both cases, the actual value is fetched. If the value is already in cache, this will cost 4 CPU cycles, however if it must be loaded from the main memory, this will cost more that 100 CPU cycles. In the case of a foo_t , the data will (almost) never belong to the same line of cache as the structure, as a consequence it is highly probable that a new line will have to be fetched from the main memory. In case of bar_t , if the pos is small, we will hit the same line of cache as the already fetched one, if pos is large, we will have to fetch a new line of cache.

As a consequence, the execution of the get_at function on a foo_t will cost one load from main memory, a load from L1d, and a second load from main memory, which means more than 204 cycles. The same execution on a bar_t will cost from 104 to 200 cycles, depending on the value of pos . These estimations are the worst case scenarios, however if that structure has been used recently, it is highly probable that the associated memory is already loaded in one of the levels of cache, reducing the latency. But caches are small, thus if you really want to benefit from them, you will have to group the accesses to the same line of caches together.

Conclusion