c++ - 将 uint64_t 转换为 unsigned char serialized[8]

Question

我正在尝试将 uint64_t 值转换为 unsigned char 数组。目前我的函数正在使用 uint32_t。但是，一旦我尝试转换 64 位值，结果就不同了。

有人看错了吗？

#include <iostream>
#include <inttypes.h>

using namespace std;

void serialize_uint64(const uint64_t integer, unsigned char * const serializedInteger);
uint64_t deserialize_uint64(unsigned char * const serializedInteger);

int main (int argc, char *argv[])
{
    uint64_t test64 = 0xFF/*FFFFFFFFFFFFFF;*/;

    unsigned char serialized64[8];
    memset(serialized64, 0x00, sizeof(serialized64));

    serialize_uint64(test64, serialized64);

    printf("%llu = %llu\n", test64, deserialize_uint64(serialized64));

    return 0;
}

void serialize_uint64(const uint64_t integer, unsigned char * const serializedInteger)
{
    serializedInteger[0] = integer >> 56;
    serializedInteger[1] = integer >> 48;
    serializedInteger[2] = integer >> 40;
    serializedInteger[3] = integer >> 32;
    serializedInteger[4] = integer >> 24;
    serializedInteger[5] = integer >> 16;
    serializedInteger[6] = integer >> 8;
    serializedInteger[7] = integer;
}

uint64_t deserialize_uint64(unsigned char * const serializedInteger)
{
    uint64_t converted = 0;

    converted = (converted << 8) | serializedInteger[0];
    converted = (converted << 8) | serializedInteger[1];
    converted = (converted << 8) | serializedInteger[2];
    converted = (converted << 8) | serializedInteger[3];
    converted = (converted << 8) | serializedInteger[4];
    converted = (converted << 8) | serializedInteger[5];
    converted = (converted << 8) | serializedInteger[6];
    converted = (converted << 8) | serializedInteger[7];
    converted = (converted << 8) | serializedInteger[8];

    return converted; // 4294967295
}

Answer 1

你有一个 off-by-one error ，你解码 9 个字节。这:

converted = (converted << 8) | serializedInteger[8];

不应该在那里。请注意索引是 8，即它访问了不存在的第 9 个字节。