GDB是否知道在调试模式下编译的C#代码(.pdb / .mdb文件)

时间:2017-11-29 12:43:39

标签: c# debugging msbuild mono gdb

我正在调试在单声道运行时运行的C#应用​​程序。用w msbuild X.sln '/p:Configuration=Debug'或w / msbuild X.sln '/p:Configuration=Release'编译它会在gdb中给出相同的堆栈跟踪(这是相同的read()调用):

Thread 1 (Thread 0x7f37f8ebd720 (LWP 16269)):
#0  0x00007f37f840a82d in read () at ../sysdeps/unix/syscall-template.S:82
#1  0x00000000005f956d in console_read (handle=0x0, buffer=0x7f37f13d0db8, numbytes=1024, bytesread=0x7fff748848c4) at w32file-unix.c:1903
#2  0x00000000005fb87c in mono_w32file_read (handle=0x0, buffer=0x7f37f13d0db8, numbytes=1024, bytesread=0x7fff748848c4) at w32file-unix.c:3109
#3  0x00000000007561e0 in ves_icall_System_IO_MonoIO_Read (handle=0x0, dest=0x7f37f13d0d98, dest_offset=0, count=1024, error=0x7fff74884a10) at w32file.c:866
#4  0x00000000403066fa in ?? ()
#5  0x0000000000000000 in ?? ()


Thread 1 (Thread 0x7fa729652720 (LWP 610)):
#0  0x00007fa728b9f82d in read () at ../sysdeps/unix/syscall-template.S:82
#1  0x00000000005f956d in console_read (handle=0x0, buffer=0x7fa71d940580, numbytes=1024, bytesread=0x7fff6b7bd784) at w32file-unix.c:1903
#2  0x00000000005fb87c in mono_w32file_read (handle=0x0, buffer=0x7fa71d940580, numbytes=1024, bytesread=0x7fff6b7bd784) at w32file-unix.c:3109
#3  0x00000000007561e0 in ves_icall_System_IO_MonoIO_Read (handle=0x0, dest=0x7fa71d940560, dest_offset=0, count=1024, error=0x7fff6b7bd8d0) at w32file.c:866
#4  0x0000000041b396fa in ?? ()
#5  0x0000000000000000 in ?? ()

问题:在调试以调试模式编译的C#应用​​程序时,它对GDB有什么不同?如果它确实/没有 - 如何(看看G​​DB如何使用.pdb / .mdb文件操作会很有趣)?

0 个答案:

没有答案