GDB超越功能(下一个)似乎不起作用 [英] GDB step over function (next) doesn't seem to work

问题描述

我正在尝试调试用C ++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; }

我使用g++ -g3 -O0 -o cards.exe cards.cpp对其进行了编译.

I compiled it using: g++ -g3 -O0 -o cards.exe cards.cpp.

这是我的GDB会话的输出:

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)

为什么向GDB发送下一条命令仍然会进入功能?

Why does sending a next command to GDB still step into a function?

我正在使用g ++ 4.2.1-sjlj和GDB 6.8.

I'm using g++ 4.2.1-sjlj and GDB 6.8.

推荐答案

step和next命令一次运行一个源代码行，因此当所有内容全部都在一行上时，单个next就可以使我正确到main()的末尾.

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 ()

代码格式化得不太密集，我仍然看不到您看到的结果.我将函数调用放在单独的行上，以使gdb一次跨过它们.这就是我得到的:

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，但我只想分享一下gdb在我的iMac上的预期行为.无论哪种情况，gdb都将对b()的调用视为一条指令，而从未输入函数调用.

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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