CUDA中的移位運算符

鑑於此CUDA代碼，我試圖執行位移操作，並且這些值的返回值爲零。這不應該發生。有誰知道如何解決這個問題？我是否缺少CUDA頭文件？CUDA中的移位運算符

代碼

 __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); 
     }

斷點分析

這裏是轉移一個unsigned int 7位到右側的代碼片段的例子。當我在指令中使用cuda-gdb命令和斷點時，我發現移位後的值在它不應該爲零時保持爲零。當我通常在cuda-gdb命令提示符中執行相同的操作時，我得到一個非零值。任何建議或提示？

根據in_data的值，變量9和7應該是非空的。

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

因此，in_data是一個值。我會嘗試一個微不足道的例子，看看我是否可以重現相同。

來源

2014-01-23 Anirudh Kaushik

CUDA中的移位運算符應該與C的工作方式相同。您可以發佈重現問題的最小大小示例而不是調試會話嗎？你確定'in_data'實際上包含你期望的值嗎？ – JackOLantern

你在調試調試版嗎？ –

這不是一個可編譯和可執行的完整代碼。只有'__device__'函數，我們不知道'in_data'是如何設置的。 – JackOLantern

由於提供的信息有限（無碼）我可能會猜測：

的CUDA-GDB documentation狀態：

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是指一個特定的存儲區，那麼它可能是你'處理內存指針而不是實際數據。

雖然我只有兩分錢。

來源

2014-01-23 21:19:05

CUDA中的移位運算符

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