Systèmes informatiques : Une perspective pour les programmeurs (édition mondiale) : L'évolution de la mémoire : Adresse virtuelle versus adresse physique

Les processeurs modernes utilisent un type d'adressage connu sous le nom de adressage virtuel. Mémoire virtuelle (MV) est une abstraction de la mémoire principale qui fournit à chaque processus un espace d'adressage linéaire privé et contigu espace d'adressage linéaire.

1. Évolution de l'adressage

Dans l'adressage physique (Figure 9.1), le processeur envoie une adresse physique (AP) directement à la DRAM. Dans adressage virtuel (Figure 9.2), les processeurs génèrent une adresse virtuelle (AV), qui est traduite par l' Unité de gestion de la mémoire (UGM) en une AP avant d'accéder à la mémoire.

2. Hiérarchie et mise en cache

La DRAM agit comme une mémoire cache DRAM pour le stockage disque. En raison de la latence du disque, les systèmes utilisent une politique d'écriture différée. Les traductions sont accélérées via la TLB en utilisant l' index de la TLB (TLBI) et protégée par des bits tels que PROT_ECRITURE. De grands espaces d'adressage ($N = 2^n$) permettent des segments et services du système d'exploitation pour les périphériques d'entrée/sortie.

TERMINALbash — 80x24

> Ready. Click "Run" to execute.

QUESTION 1

[Fill in the blanks] Complete the following table for a Virtual Address Space where $N = 2^n$ and the largest address is $N-1$. Use units: $K=2^{10}, M=2^{20}, G=2^{30}, T=2^{40}, P=2^{50}, E=2^{60}$.

n=16: N=64K, Max=0xFFFF; n=32: N=4G, Max=0xFFFFFFFF; n=48: N=256T, Max=0xFFFFFFFFFFFF; n=64: N=16E, Max=0xFFFFFFFFFFFFFFFF

n=16: N=16K; n=32: N=32M; n=48: N=48G; n=64: N=64T

QUESTION 2

Why does the DRAM cache use a Write-back policy instead of Write-through?

To maintain Stack Discipline.

Due to the massive latency penalty of accessing disk storage.

To prevent internal fragmentation.

Because the L3 unified cache is 16-way associative.

QUESTION 3

What happens if a process attempts to write to a page where the PTE has PROT_WRITE set to 0?

A General protection fault occurs.

The MMU performs Swapping automatically.

Deferred coalescing is triggered.

The Reference bit (A bit) is cleared.

QUESTION 4

Which component is responsible for translating the Virtual Address to a Physical Address?

The L3 unified cache (8 MB, 16-way).

The Memory Management Unit (MMU).

The Rio Package.

The Buddy System Allocator.

QUESTION 5

What is the physical address (PA) if the MMU finds a mapping for a VA where VPN=0x0d and VPO=0x14, given PPN=0x0d?

PA = 0x354: PPN=0x0d, PPO=0x14

$PA = 0x140d$

$PA = 0x0001$

PA = 0x354: VPN=0x0d, VPO=0x14

Case Study: Advanced Heap Management & VM Translation

Optimizing Allocators and Address Translation Efficiency

You are designing a high-performance allocator for an Intel Core i7 system with a k-level page table hierarchy. The allocator uses a single global variable (heap_listp) and must minimize external fragmentation using Buddy System Allocation. You also need to modify the block structure to improve Peak memory utilization.

1. Implement the 'find_fit' function for a simple implicit list allocator using first-fit logic. Your solution must account for Internal fragmentation and return NULL if no block is found.

Solution:
Implementation: static void *find_fit(size_t asize) { void *bp; /* Iterate through the heap starting from heap_listp */ for (bp = heap_listp; GET_SIZE(HDRP(bp)) > 0; bp = NEXT_BLKP(bp)) { /* If block is not allocated and size is sufficient */ if (!GET_ALLOC(HDRP(bp)) && (asize <= GET_SIZE(HDRP(bp)))) { return bp; } } return NULL; /* No fit found */ } This function searches the linear heap space, balancing search speed with memory utilization.

2. Explain how to modify an allocator to remove boundary tags (footers) from allocated blocks. How does this affect Peak memory utilization and the mm_free(void *bp) logic?

Solution:
To remove footers in allocated blocks, we store an 'extra' bit in the header of the *next* block to indicate whether the *current* block is allocated. This allows us to reduce the minimum block size by one word, directly increasing Peak memory utilization by reducing internal fragmentation in small objects. However, mm_free and coalesce(bp) must now inspect the header of the next block to determine if the previous block can be merged, maintaining constant-time coalescing for free blocks (which still require footers).

3. Describe the 'Buddy System Allocation' process for a block of size 32 bytes at address xxx...x10000. How does it handle recursive splitting?

Solution:
The Buddy System finds a block of size $2^j$ where $k \le j \le m$. To allocate 32 bytes ($2^5$), if only a $2^7$ block is available, it splits into two $2^6$ buddies. One $2^6$ block is then split again into two $2^5$ buddies. One buddy at address xxx...x10000 is allocated, while its buddy (address xxx...x11000) remains free. This minimizes external fragmentation and allows for fast buddy-merging using bitwise address operations.