我在cuda gdb，我可以使用((@global float *)array)[0]
但是如何在gdb中使用常量内存呢？
我尝试((@parameter float *)const_array)
我这样声明const_array：__constant__ float const_array[1 << 14]
我试过使用1 << 5，但问题是一样的。
我似乎没有任何问题。为了打印设备内存，必须在设备代码的断点处停止。
示例：
$ cat t1973.cu
const int cs = 1 << 14;
__constant__ int cdata[cs];
__global__ void k(int *gdata){
gdata[0] = cdata[0];
}
int main(){
int *hdata = new int[cs];
for (int i = 0; i < cs; i++) hdata[i] = i+1;
cudaMemcpyToSymbol(cdata, hdata, cs*sizeof(cdata[0]));
int *gdata;
cudaMalloc(&gdata, sizeof(gdata[0]));
cudaMemset(gdata, 0, sizeof(gdata[0]));
k<<<1,1>>>(gdata);
cudaDeviceSynchronize();
}
$ nvcc -o t1973 t1973.cu -g -G -arch=sm_70
$ cuda-gdb ./t1973
sh: python3: command not found
Unable to determine python3 interpreter version. Python integration disabled.
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Reading symbols from ./t1973...
(cuda-gdb) b 5
Breakpoint 1 at 0x403b0c: file t1973.cu, line 6.
(cuda-gdb) run
Starting program: /home/user2/misc/t1973
[Thread debugging using libthread_db enabled]
Using host libthread_db library "/lib64/libthread_db.so.1".
[Detaching after fork from child process 22872]
[New Thread 0x7fffef475700 (LWP 22879)]
[New Thread 0x7fffeec74700 (LWP 22880)]
[Switching focus to CUDA kernel 0, grid 1, block (0,0,0), thread (0,0,0), device 0, sm 0, warp 0, lane 0]
Thread 1 "t1973" hit Breakpoint 1, k<<<(1,1,1),(1,1,1)>>> (
gdata=0x7fffcdc00000) at t1973.cu:5
5 gdata[0] = cdata[0];
(cuda-gdb) print gdata[0]
$1 = 0
(cuda-gdb) print cdata[0]
$2 = 1
(cuda-gdb) s
6 }
(cuda-gdb) print gdata[0]
$3 = 1
(cuda-gdb) print cdata[0]
$4 = 1
(cuda-gdb) print cdata[1]
$5 = 2
(cuda-gdb)
尝试将__constant__放入.cuh中，然后用作经典的C全局变量。