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我正在编译以下代码片段nvcc -g -G gdbfail.cu。

#include <cstdio>
#include <cinttypes>

__global__ void mykernel() {
    uint8_t* ptr = (uint8_t*) malloc(8);
    for (int i = 0; i < 8; i++) {
        ptr[i] = 7 - i;
    }

    for (int i = 0; i < 8; i++) { // PUT BREAKPOINT HERE
        printf("%" PRIx8 " ", ptr[i]);
    }
    printf("\n");
}

int main() {

    uint8_t* ptr = (uint8_t*) malloc(8);
    for (int i = 0; i < 8; i++) {
        ptr[i] = 7 - i;
    }

    for (int i = 0; i < 8; i++) { // PUT BREAKPOINT HERE
        printf("%" PRIx8 " ", ptr[i]);
    }
    printf("\n");

    mykernel<<<1,1>>>();
    cudaDeviceSynchronize();
}

当我运行cuda-gdb ./a.out并在第 10 行 () 放置断点时b 10，运行代码 ( r)，并尝试在位于的地址处打印值，ptr我得到了令人惊讶的结果

(cuda-gdb) x/8b ptr
0x7fffcddff920: 7       6       5       4       3       2       1       0
(cuda-gdb) x/8b 0x7fffcddff920
0x7fffcddff920: 0       0       0       0       0       0       0       0

当我在主机代码 ( b 23, r) 中做同样的事情时，我得到了预期的结果：

(cuda-gdb) x/8b ptr
0x5555556000a0: 7       6       5       4       3       2       1       0
(cuda-gdb) x/8b 0x5555556000a0
0x5555556000a0: 7       6       5       4       3       2       1       0

0x7fffcddff920为什么当 cuda-gdb 以数字 ( ) 而不是符号 ( )提供地址时，它不显示正确的内存值ptr？

4

1 回答 1

1

显然，并非所有gdb可用于主机代码的命令功能也可用于设备代码。在设备代码中使用时，支持的命令可能具有不同的语法或期望。这在cuda-gdb 文档中有所说明。

这些文档表明检查内存的方法是print命令，并指出“裸”地址/指针所需的一些额外解码语法。这是您的示例：

$ cuda-gdb ./t1869
NVIDIA (R) CUDA Debugger
11.4 release
Portions Copyright (C) 2007-2021 NVIDIA Corporation
GNU gdb (GDB) 10.1
Copyright (C) 2020 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.
Type "show copying" and "show warranty" for details.
This GDB was configured as "x86_64-pc-linux-gnu".
Type "show configuration" for configuration details.
For bug reporting instructions, please see:
<https://www.gnu.org/software/gdb/bugs/>.
Find the GDB manual and other documentation resources online at:
    <http://www.gnu.org/software/gdb/documentation/>.

For help, type "help".
Type "apropos word" to search for commands related to "word"...
Reading symbols from ./t1869...
(cuda-gdb) b 10
Breakpoint 1 at 0x403b05: file t1869.cu, line 14.
(cuda-gdb) r
Starting program: /home/user2/misc/t1869
[Thread debugging using libthread_db enabled]
Using host libthread_db library "/lib64/libthread_db.so.1".
7 6 5 4 3 2 1 0
[Detaching after fork from child process 25822]
[New Thread 0x7fffef475700 (LWP 25829)]
[New Thread 0x7fffeec74700 (LWP 25830)]
[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 "t1869" hit Breakpoint 1, mykernel<<<(1,1,1),(1,1,1)>>> () at t1869.cu:10
10          for (int i = 0; i < 8; i++) { // PUT BREAKPOINT HERE
(cuda-gdb) x/8b ptr
0x7fffbcdff920: 7       6       5       4       3       2       1       0
(cuda-gdb) p/x *(@global unsigned char *)0x7fffbcdff920@8
$1 = {0x7, 0x6, 0x5, 0x4, 0x3, 0x2, 0x1, 0x0}
(cuda-gdb)

请注意，上面的print命令需要一些帮助来解释您期望内存地址引用哪个“空间”（例如@shared，@global等）

如果我们给你的命令同样的“帮助”，我们会得到预期的结果：

(cuda-gdb) x/8b ptr
0x7fffbcdff920: 7       6       5       4       3       2       1       0
(cuda-gdb) x/8b (@global unsigned char *)0x7fffbcdff920
0x7fffbcdff920: 7       6       5       4       3       2       1       0
(cuda-gdb)
于 2021-08-14T18:18:33.633 回答