# CS:APP 第 2 章重點提示和練習 (解答)

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## `2.58`

On **big-endian (PowerPC, SPARC, and Internet)**, the 32-bit value x01234567 is stored as four bytes 0x01, 0x23, 0x45, 0x67, while on **little-endian (Intel x86)**, it will be stored in reverse order:

從 LSB / MSB 來看資料排列:

#include <assert.h>
#include <stdio.h>

typedef unsigned char *byte_ptr;

int is_little_endian() {
    int test_num = 0xff;
    byte_ptr byte_start = (byte_ptr) &test_num;
    if (byte_start[0] == 0xff) return 1;
    return   0;
}

int main(int argc, char *argv[]) {
    assert(is_little_endian());
    return 0;
}

Reference:
* [Little Endian & Big Endian](http://www.bogotobogo.com/Embedded/Little_endian_big_endian_htons_htonl.php)

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## `2.66`

#include <assert.h>
#include <stdio.h>

/* Generate mask indicating leftmost 1 in x. Assume w=32
 * For example, 0xFF00 -> 0x8000, and 0x6000 -> 0x4000.
 * If x = 0, then return 0
 */
int leftmost_one(unsigned x) {
    /* first, generate a mask that all bits after leftmost one are one
     * e.g. 0xFF00 -> 0xFFFF, and 0x6000 -> 0x7FFF
     * If x = 0, get 0
     */
    x |= x >> 1;
    x |= x >> 2;
    x |= x >> 4;
    x |= x >> 8;
    x |= x >> 16;

    /* then, do  mask & (~mask >> 1), reserve leftmost bit one
     * that's we want
     */
    return x & (~x >> 1);
}

int main(int argc, char *argv[]) {
    assert(leftmost_one(0xFF00) == 0x8000);
    assert(leftmost_one(0x6000) == 0x4000);
    return 0;
}

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