# linux2021: mucrolores

## 測驗 $\beta - 1$

static inline uintptr_t align_up(uintptr_t sz, size_t alignment)
{
    uintptr_t mask = alignment - 1;
    if ((alignment & mask) == 0) { /* check if alignment is power of 2 */
        return ((sz+mask)&(~mask));        
    }
    return (((sz + mask) / alignment) * alignment);
}

這段程式碼中主要的部份就是以`((sz+mask)&(~mask))` 為主要核心

sz+mask
會讓所有小於等於下一個alignment且大於前一個alignment的
都變成大於等於下一個alignment
ex:

120 + 3 = 123 // next alignment: 120
121 + 3 = 124 // next alignment: 124
122 + 3 = 125 // next alignment: 124
123 + 3 = 126 // next alignment: 124
124 + 3 = 127 // next alignment: 124

之後直接取 `&(~mask)` 相當於直接把mask的bit都移除掉 類似
`k >> (bit_of_mask) << (bit_of_mask)` 的作用

達到最後所需要的效果

## 測驗 $\gamma - 1$

#include <stdio.h>
#include <unistd.h>

int main(void)
{
    for (int i = 0; i < NNN; i++) {
        fork();
        printf("-");                         
    }

    fflush(stdout);
    return 0;
}

因為在執行`fork()`的過程中會複製到很多東新到新的process, stdio中的buffer也是，在stdio中，執行`printf()`時，只要沒有遇到 `\n` 或是執行 `fflush(stdout)`，其實資料就會先暫時放在stdio的buffer中，所以呼叫 `fork()`的時候 就會把先前 `printf("-");` 也複製到新的process。

所以在迴圈執行的過程中可以簡化為下列的式子。

i = 0,  p = 2,    char per buf = 1,  buf_total = 2
i = 1,  p = 4,    char per buf = 2,  buf_total = 8
i = 2,  p = 8,    char per buf = 3,  buf_total = 24
i = 3,  p = 16,   char per buf = 4,  buf_total = 64
i = 4,  p = 32,   char per buf = 5,  buf_total = 160
i = 5,  p = 64,   char per buf = 6,  buf_total = 384
i = 6,  p = 128,  char per buf = 7,  buf_total = 896
i = 7,  p = 256,  char per buf = 8,  buf_total = 2048
i = 8,  p = 512,  char per buf = 9,  buf_total = 4608
i = 9,  p = 1024, char per buf = 10, buf_total = 10240
i = 10, p = 2048, char per buf = 11, buf_total = 22528
i = 11, p = 4096, char per buf = 12, buf_total = 49152

就會形成最後的結果