传递给设备函数的共享内存地址仍然是共享内存

Question

假设我有这个__device__函数：

__device__ unsigned char* dev_kernel(unsigned char* array_sh, int params){
    return array_sh + params;
}

在__global__内核中，我以这种方式使用它：

uarray = dev_kernel (uarray, params);

其中uarray是位于共享内存中的数组。

但是当我使用 cuda-gdb 查看内核中uarray的地址时__global__我得到：

(@generic unsigned char * @shared) 0x1000010 "z377*"

在内核__device__我得到：

(unsigned char * @generic) 0x1000010 <Error reading address 0x1000010: Operation not permitted>

尽管有错误，但程序运行正常（也许是cuda-gdb的一些限制）。

所以，我想知道：在__device__内核中，uarray是共享的？我正在将数组从全局内存更改为共享内存，时间几乎相同（使用共享内存，时间稍差）。

Answer 1

所以，我想知道：在__device__内核中，uarray共享了吗？

是的，

当您以这种方式将指向共享内存的指针传递给设备函数时，它仍然指向共享内存中的同一位置。

为了回答下面发布的问题让我感到困惑，我选择展示一个简单的例子：

$ cat t249.cu
#include <stdio.h>
#define SSIZE 256
__device__ unsigned char* dev_kernel(unsigned char* array_sh, int params){
    return array_sh + params;
}
__global__ void mykernel(){
  __shared__ unsigned char myshared[SSIZE];
  __shared__ unsigned char *u_array;
  for (int i = 0; i< SSIZE; i++)
    myshared[i] = (unsigned char) i;
  unsigned char *loc = dev_kernel(myshared, 5);
  u_array = loc;
  printf("val = %dn", *loc);
  printf("val = %dn", *u_array);
}
int main(){
  mykernel<<<1,1>>>();
  cudaDeviceSynchronize();
  return 0;
}
$ nvcc -arch=sm_20 -g -G -o t249 t249.cu
$ cuda-gdb ./t249
NVIDIA (R) CUDA Debugger
5.5 release
....
Reading symbols from /home/user2/misc/t249...done.
(cuda-gdb) break mykernel
Breakpoint 1 at 0x4025dc: file t249.cu, line 9.
(cuda-gdb) run
Starting program: /home/user2/misc/t249
[Thread debugging using libthread_db enabled]
Breakpoint 1, mykernel () at t249.cu:9
9       __global__ void mykernel(){
(cuda-gdb) break 14
Breakpoint 2 at 0x4025e1: file t249.cu, line 14.
(cuda-gdb) continue
Continuing.
[New Thread 0x7ffff725a700 (LWP 26184)]
[Context Create of context 0x67e360 on Device 0]
[Launch of CUDA Kernel 0 (mykernel<<<(1,1,1),(1,1,1)>>>) on Device 0]
[Switching focus to CUDA kernel 0, grid 1, block (0,0,0), thread (0,0,0), device 0, sm 2, warp 0, lane 0]
Breakpoint 1, mykernel<<<(1,1,1),(1,1,1)>>> () at t249.cu:12
12        for (int i = 0; i< SSIZE; i++)
(cuda-gdb) continue
Continuing.
Breakpoint 2, mykernel<<<(1,1,1),(1,1,1)>>> () at t249.cu:14
14        unsigned char *loc = dev_kernel(myshared, 5);
(cuda-gdb) print &(myshared[0])
$1 = (@shared unsigned char *) 0x8 ""
       ^
       |
     cuda-gdb is telling you that this pointer is defined in a __shared__ statement, and therefore it's storage is implicit and it is unmodifiable.
(cuda-gdb) print &(u_array)
$2 = (@generic unsigned char * @shared *) 0x0
       ^                          ^
       |                          u_array is stored in shared memory.
      u_array is a generic pointer, meaning it can point to anything.  
(cuda-gdb) step
dev_kernel(unsigned char * @generic, int) (array_sh=0x1000008 "", params=5)
    at t249.cu:6
6           return array_sh + params;
(cuda-gdb) print array_sh
$3 = (@generic unsigned char * @register) 0x1000008 ""
          ^                      ^
          |                    array_sh is stored in a register.
         array_sh is a generic pointer, it can point to anything.
(cuda-gdb) print u_array
No symbol "u_array" in current context.
 (note that I can't access u_array from inside the __device__ function, so I don't understand your comment there.)
(cuda-gdb) step
mykernel<<<(1,1,1),(1,1,1)>>> () at t249.cu:15
15        u_array = loc;
(cuda-gdb) step
16        printf("val = %dn", *loc);
(cuda-gdb) print u_array
$4 = (
    @generic unsigned char * @shared) 0x100000d ......
       ^                         ^
       |                       u_array is stored in shared memory
     u_array is a generic pointer, it can point to anything
(cuda-gdb)

虽然你没有提供它，但我假设你对u_array的定义与我的相似，基于你得到的 cuda-gdb 输出。

请注意，像 @shared 这样的指示器不会告诉您指针指向哪种内存，它们告诉您它是哪种指针（在 __shared__ 语句中隐式定义）或存储位置（在共享内存中）。

如果这不能解决您的问题，请提供一个完整的示例，以及完整的 cuda-gdb 会话输出，就像我一样。