GDB的step over功能（下一个）似乎不工作 [英] GDB step over function(next) doesn't seem to work

问题描述

我试图调试我在C ++中编写的程序。这是代码：

I'm trying to debug a program I wrote in C++. Here is the code:

void a() { }
void b() { a(); }
int main() { b(); return 0; }

我编译它使用： g ++ -g3 -O0 -o cards。 exe cards.cpp 。
这是我的gdb会话的输出：

I compiled it using: g++ -g3 -O0 -o cards.exe cards.cpp. Here is the output of my gdb session:

(gdb) b main 
Breakpoint 1 at 0x401421: file cards.cpp, line 10. 
(gdb) r 
Starting program: C:\workspace\Cards\src/cards.exe  
[New thread 1624.0xa28]
Breakpoint 1, main () at cards.cpp:10 
10  int main() 
(gdb) n 
12      b(); 
(gdb) n 
b () at cards.cpp:5 5   
void b() 
(gdb) n 
7       a();
(gdb) quit 
The program is running.  Exit anyway? (y or n)

我的问题是，为什么发送下一个命令到gdb仍然进入功能？我使用g ++ 4.2.1-sjlj和gdb 6.8。

My question is, why does sending a next command to gdb still step into a function? I'm using g++ 4.2.1-sjlj and gdb 6.8.

推荐答案

步骤和下一个命令每次工作一个源代码行，所以当一切都在一行上单个下一个直接到 main（）结束。

The step and next commands work one source line at a time, so when everything is all on one line a single next takes me right to the end of main().

3   int main() { b(); return 0; }
(gdb) n
0x00001faa in start ()

格式化较少密集我仍然看不到你看到的结果。我把函数调用放在单独的行上，让gdb一次一个地跳过它们。这是我得到的：

With the code formatted less densely I still do not see the results you see. I put the function calls on separate lines to get gdb to step over them one at a time. Here's what I get then:

jkugelman$ cat cards.cpp
void a() {
}

void b() {
    a();
}

int main() {
    b();
    return 0;
}
jkugelman$ g++ -g3 -O0 -o cards cards.cpp
jkugelman$ gdb ./cards
GNU gdb 6.3.50-20050815 (Apple version gdb-960) (Sun May 18 18:38:33 UTC 2008)
<snip>
Reading symbols for shared libraries .... done

(gdb) b main
Breakpoint 1 at 0x1ff2: file cards.cpp, line 9.
(gdb) r
Starting program: /Users/jkugelman/Development/StackOverflow/cards 
Reading symbols for shared libraries +++. done

Breakpoint 1, main () at cards.cpp:9
9       b();
(gdb) n
10      return 0;
(gdb) n
11  }
(gdb) n
0x00001faa in start ()

我没有答案，但我只想分享gdb在我的iMac上的行为。在这两种情况下，gdb都将 b（）的调用作为一个指令处理，从未输入函数调用。

I don't have an answer, but I just wanted to share that gdb behaves as expected on my iMac. In either case gdb treated the call to b() as one instruction and never entered the function call.

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