如何在gdb中列出类方法？ [英] How to list class methods in gdb?

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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