增加指针的LSB位，但保留其余位(硬编码的环形缓冲区大小)

Question

我尝试编写一个非常紧凑的环形缓冲区。缓冲区包含2^X值(X: 1-7)，我需要的是增加X位(LSB)，但保留其余变量。我有一个解决方案，但我不知道这是否有效(以前从未修改过指针地址)，而且它不是那么紧凑。你知道如何改进这一点吗？

        // hardcoded ringbuffersize for better ram usage
        // this frees us out, start, end, size variables
        // 7 == 128, 6 == 64
        #define LIGHTWEIGHT_RING_BUFFER_BITS 7
        #define LIGHTWEIGHT_RING_BUFFER_SIZE (1<<LIGHTWEIGHT_RING_BUFFER_BITS)
        // thatswhy we cant use 256 buffer size

        #define LIGHTWEIGHT_RING_BUFFER_DISABLED LIGHTWEIGHT_RING_BUFFER_SIZE+1
        // save new data
        *Buffer->In = Data;

        // save LSB bits
        uint8_t pointermask = Buffer->In;
        // discard all LSB bits
        Buffer->In >>= LIGHTWEIGHT_RING_BUFFER_BITS;
        Buffer->In <<= LIGHTWEIGHT_RING_BUFFER_BITS;
        // save the LSB bits + 1
        Buffer->In |= ++pointermask & (LIGHTWEIGHT_RING_BUFFER_SIZE - 1)

        Buffer->Count++;

Answer 1

如果您要问的是如何以换行的方式递增指针，答案非常简单。您需要将uint64_t类型转换为您平台上的指针大小。如果所有低位都为零，则第二条语句包装指针。

p ++;
p -= ((uint64_t) p) & (LIGHTWEIGHT_RING_BUFFER_SIZE - 1) ?
    0 : LIGHTWEIGHT_RING_BUFFER_SIZE;

或者，再次将类型转换替换为适当的类型转换。

p = (uint8_t *) (((uint64_t) p & ~(LIGHTWEIGHT_RING_BUFFER_SIZE - 1)) |
                 (((uint64_t) p + 1) & (LIGHTWEIGHT_RING_BUFFER_SIZE - 1)));