unsigned char 数组如何转换为long long数组

将char 数组转换为long long数组 转换后如图所示

unsigned char a[] = {

    0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12,


	0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 


	0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 


	0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, };

    long long llong[4] = {0};

    for (int i = 0; i < 4; i++)
    {
        for (int j = 0; j < 8; j++)
        {
            llong[i] |= (long long)a[j]<<(8*j);
        }
    }

楼上也可以，这是常规做法

可以使用循环左移的方法，每次左移8位，并用位或方法与char的数值进行拼接。

unsigned char a[] = {0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12,
							0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 
	0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 
	0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, };

long long* pLong = (long long*)a;
long long llong[4];
memcpy(&llong, pLong, sizeof(llong));

内存是连续的，用指针处理一下。

如果解决了你的问题，顺手点个采纳吧。

    unsigned char a[] = {

    0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x13,


	0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x14, 


	0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x15, 


	0x10, 0x12, 0x12, 0x12, 0x12, 0x12, 0x12, 0x16, };

    long long llong[4] = {0};

    for (int i = 0; i < 4; i++)
    {
        for (int j = 0; j < 8; j++)
        {
            llong[i] |= (long long)a[j + 8 * i] << (56 - 8 * j);
        }
    }

不好意思，昨天发急了没检查，有点毛病，缺了个位移；

端序也改过来了。

顺便把反转的也提上来吧。

    long long llong[4] = {0x1212121212121212, 0x1212121212121212,
                          0x1212121212121212, 0x1212121212121212};
    char ch[32];

    for (int i = 0; i < 4; i++)
    {
        for (int j = 0; j < 8; j++)
        {
            ch[j + 8 * i] = llong[i] >> (56 - 8 * j);
        }
    }