# CS:APP Ch6 Memory Hierarchy Ex. 6.10 解析

## Presumption
* float is 4 bytes
* x is loaded to a contiguous memory address start from address 0
* y starts from address 32,
* assume a block is 16 bytes (sufficient to hold 4 float variables)
* Cache:
* 2 sets
* Capacity = 32 bytes

### Compute s, e, b from `(S,E,B,m) = (2, 1, 16, 32)`
* Capacity `(C)`:
* C = S * E * B
* 64 bytes = 2 * E * 16bytes
* `E = 2`
* sets:
* S = 2^s^ = 2,
* so s = 1
* blocks:
* B = 2^b^ = 16,
* so b = 4
* tag:
* s + b + t = m,
* 1 + 4 + t = 32,
* so, t = 27

Memory layout:

addr. 32:

[0000...01  0  0000] 
 |_______| |_|  |__|
     t      s     b
     
is at set 0

## Code

float dotprod(float x[8], float y[8])
{
    float sum = 0.0;
    int i;
    
    for (i = 0; i < 8; i++)
        sum += x[i] * y [i];
    return sum;
}

## Before padding

| n | addr (DEC) | set | | addr (DEC) | set |
|------|------------|-----|------|------------|-----|
| x[0] | 0 | 0 | y[0] | 32 | 0 |
| x[1] | 4 | 0 | y[1] | 36 | 0 |
| x[2] | 8 | 0 | y[2] | 40 | 0 |
| x[3] | 12 | 0 | y[3] | 44 | 0 |
| x[4] | 16 | 1 | y[4] | 48 | 1 |
| x[5] | 20 | 1 | y[5] | 52 | 1 |
| x[6] | 24 | 1 | y[6] | 56 | 1 |
| x[7] | 28 | 1 | y[7] | 60 | 1 |

## After padding (`x[8]` - `x[11]`)

* :question: Why we need padding?
* Because we want the term
* `sum += x[i] * y [i];`
* `x[i] and y[i]` not in the same set

| n | addr (DEC) | set | | addr (DEC) | set |
|------|------------|-----|------|------------|-----|
| x[0] | 0 | 0 | y[0] | 48 | 1 |
| x[1] | 4 | 0 | y[1] | 52 | 1 |
| x[2] | 8 | 0 | y[2] | 56 | 1 |
| x[3] | 12 | 0 | y[3] | 60 | 1 |
| x[4] | 16 | 1 | y[4] | 64 | 0 |
| x[5] | 20 | 1 | y[5] | 68 | 0 |
| x[6] | 24 | 1 | y[6] | 72 | 0 |
| x[7] | 28 | 1 | y[7] | 76 | 0 |
| x[8] | 32 | 0
| x[9] | 36 | 0
| x[10] | 40 | 0
| x[11] | 44 | 0

* After padding, for example, if we do