如何在gdb中列出类方法？

问题描述

我一直在谷歌搜索这个，检查通过gdb手册，但似乎找不到我想要做的解答。

I've been googling for this and checking through the gdb manual but can't seem to find an answer to what I'm trying to do.

有一种方法来获取gdb打印出一个给定类类型的所有方法的列表？打印命令只显示数据成员和字段，不显示任何方法。

Is there a way to get gdb to print out a listing of all the methods for a given class type? The print command only seems to show the data members and fields, none of the methods are displayed for it.

此外，为了进一步，打印所有正确的虚方法给一个base *指针？例如：

Additionally, to take it a step further, is there a way to print all the correct virtual methods given a base *pointer? Say like for example:

struct A
{
  virtual void foo() {}
};

struct B : public A
{
  void foo() {}
};

int main()
{
  A *b = new B;
}

如何获取gdb打印变量* b并显示正确的虚拟方法？

How can I get gdb to print variable *b and have it show the correct virtual method(s)?

感谢

推荐答案

ptype 。

假设我将这些行添加到您的示例中：

Suppose I add these lines to your example:

A alpha;
B beta;

现在在gdb中，我可以要求一个类类型（或一个实例）的描述：

Now in gdb I can ask for a description of a class type (or an instance of one):

(gdb) ptype alpha
type = class A {
  public:
    virtual void foo();
}

(gdb) ptype A
type = class A {
  public:
    virtual void foo();
}

(gdb) ptype beta
type = class B : public A {
  public:
    virtual void foo();
}

(gdb) ptype B
type = class B : public A {
  public:
    virtual void foo();
}

如果我使用指针尝试，我得到声明的类型：

If I try that with a pointer, I get the declared type:

(gdb) ptype b
type = class A {
  public:
    virtual void foo();
} *

如果我想要真正的类型，打印对象的变量：

(gdb) set print object on
(gdb) ptype b
type = /* real type = B * */
class A {
  public:
    virtual void foo();
} *

然后调用 ptype 再次看到 B 有（我不知道如何一步完成）。

and then call ptype again to see what B has (I don't know how to do it in one step).

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