gdb搜索源目录

Question

我如何提到unix中的gdb以递归方式搜索单个目录中的源文件,例如,如果一个模块中有一些不同的建筑块.a是b,c,d的父目录,其中b,c,d是子目录.和源文件分布在b,c,b中.我只需要向gdb提及所有源文件都位于(父目录)中.哪个gdb将用作参考并在调试程序时递归搜索源文件.

Answer 1

您需要的是命令set substitute-path.

    (gdb) set substitute-path /usr/src/include /mnt/include

但是,仅在最新版本(6.6+)的gdb中可用.

Answer 2

或者您可以执行以下操作,以便prog在目录中使用source 调试程序srcdir:

gdb `find srcdir -type d -printf '-d %p '` prog

我认为这是对你的问题更直接的回答.如果您的可执行文件不包含编译目录和/或您没有6.6+版本的gdb,也很有用.

Answer 3

(gdb) help files
Specifying and examining files.

List of commands:

add-shared-symbol-files -- Load the symbols from shared objects in the dynamic linkers link map  
add-symbol-file -- Load symbols from FILE  
add-symbol-file-from-memory -- Load the symbols out of memory from a dynamically loaded object file  
cd -- Set working directory to DIR for debugger and program being debugged  
core-file -- Use FILE as core dump for examining memory and registers  
directory -- Add directory DIR to beginning of search path for source files
edit -- Edit specified file or function  
exec-file -- Use FILE as program for getting contents of pure memory  
file -- Use FILE as program to be debugged  
forward-search -- Search for regular expression (see regex(3)) from last line listed  
generate-core-file -- Save a core file with the current state of the debugged process  

(gdb) help directory  

Add directory DIR to beginning of search path for source files.  
Forget cached info on source file locations and line positions.  
DIR can also be $cwd for the current working directory, or $cdir for the  
directory in which the source file was compiled into object code.  
With no argument, reset the search path to $cdir:$cwd, the default.