c - 这个C函数是如何工作的

Question

#define BASE32_ONIONLEN 16
#define BASE32_ALPHABET "abcdefghijklmnopqrstuvwxyz234567"

void base32_onion(char *dst, unsigned char *src) { // base32-encode hash
    uint8_t byte = 0, // dst location
    offset = 0; // bit offset
    for(; byte < BASE32_ONIONLEN; offset += 5) {
        if(offset > 7) {
            offset -= 8;
            src++;
        }
        dst[byte++] = BASE32_ALPHABET[(htobe16(*(uint16_t*)src) >> (11-offset))
                                       & (uint16_t)0x001F];
    }
    dst[byte] = '\0';
}

我无法理解以 dst[byte++] 开头的部分。我是一名 python 程序员，我不太确定所有类型转换是如何工作的。 src 指向一个字节，对吗？并且 (uint16_t*) 将其转换为指向 2 字节值的指针？那么这是否会使 src 处的字节成为两个字节值的开始或结束？ >>(11-offset) 是什么？

Answer 1

uint16_t t1, t2, t3;
uint8_t t4, t5, t6;

t1 = *(uint16_t*)src;       // Whatever src is pointing to, treat the next
                            // 16 bits as an unsigned number
                            // Safer would be memcpy(&t1, src, sizeof(t1));
t2 = htobe(t1);             // Guessing this changes the value to have big-endian
                            // byte order
t3 = t2 >> (11-offset);     // Shift t2 to the right by (11-offset) bits
t4 = t3 & (uint16_t)0x001F; // t4 contains the lower 5 bits of t3
t5 = BASE32_ALPHABET[t4];   // t5 gets the t4-th byte of the base32 alphabet
t6 = byte++;                // t6 gets the value of byte, byte's value is then
                            // incremented, so t6 + 1 == byte
dest[t6] = t5;              // t5 is assigned to dest[t6]

将值向右移位 n 位相当于将该值除以 2ⁿ。