Question

鉴于此CUDA代码，我正在尝试执行位移操作，并且这些值的返回值为零。这不应该发生。有谁知道如何解决这个问题？我错过了CUDA标题包含吗？

代码

      __device__ unsigned int FI( unsigned int in_data, unsigned int subkey, 
     unsigned int *KLi1, unsigned int *KLi2, unsigned int *KOi1, unsigned int *KOi2,
     unsigned int *KOi3, unsigned int *KIi1, unsigned int *KIi2, unsigned int *KIi3) {

      unsigned int nine, seven;

      unsigned int S7[128]  = {};

      unsigned int S9[512] = {};

      nine = (in_data>>7);
      seven = (in_data&0x7F);

      /* Now run the various operations */
      nine = (unsigned int)(S9[nine] ^ seven);
      seven = (unsigned int)(S7[seven] ^ (nine & 0x7F));
      seven ^= (subkey>>9);
      nine ^= (subkey&0x1FF);
      nine = (unsigned int)(S9[nine] ^ seven);
      seven = (unsigned int)(S7[seven] ^ (nine & 0x7F));
      in_data = (unsigned int)((seven<<9) + nine);
      return( in_data );
      }

断点分析

这是一个代码片段的示例，它将unsigned int 7个位置向右移动。当我在指令处cuda -gdb我的exec和断点时，我发现移位后的值在不应该的时候保持为零。当我通常在cuda-gdb命令提示符中执行相同的操作时，我得到一个非零值。有任何建议或提示吗？

根据in_data的值，变量9和7应该是非空的。

    nine = (in_data>>7);
    seven = (in_data&0x7F);

    [Switching focus to CUDA kernel 0, grid 1, block (0,0,0), thread (1,0,0), device 0, sm 0, warp 0, lane 1]
    Breakpoint 1, FI (KLi1=0x3fffae0, KLi2=0x3fffb00, KOi1=0x3fffb20, KOi2=0x3fffb40, KOi3=0x3fffb60, 
    KIi1=0x3fffb80, KIi2=0x3fffba0, KIi3=0x3fffbc0, in_data=461, subkey=0) at kasumiOp.cu:61
     61   nine = (in_data>>7);
     (cuda-gdb) p in_data
     $1 = 461
     (cuda-gdb) step
     62   seven = (in_data&0x7F);
     (cuda-gdb) p nine
     $2 = 0
     (cuda-gdb) step
     65   nine = (unsigned int)(S9[nine] ^ seven);
     (cuda-gdb) p seven
     $3 = 0
     (cuda-gdb) p 461 >> 7
     $4 = 3
     (cuda-gdb) cuda thread
     thread (1,0,0)
     (cuda-gdb) p 561 & 0x7f
     $5 = 49
     (cuda-gdb) p 461 & 0x7f
     $6 = 77

所以，in_data是一个值。我将尝试一个简单的例子，看看我是否可以重现相同的。

Answer 1

由于提供的信息有限（无代码），我可能会猜测：

CUDA-GDB documentation州：

The GDB print command has been extended to decipher the location of any program variable and can be used to display the contents of any CUDA program variable including:
* data allocated via cudaMalloc()
* data that resides in various GPU memory regions, such as shared, local, and global memory
* special CUDA runtime variables, such as threadIdx

如果 in_data 引用特定的内存区域，则可能是您正在处理内存指针而不是实际数据。

虽然只是我的两分钱。

CUDA中的Shift运算符

1 个答案: