Question

假设我们使用test.cpp编译了以下文件g++ --std=c++11 -g test.cpp -o test.exe：

// test.cpp
// g++ --std=c++11 -g test.cpp -o test.exe
// gdb -ex "b test.cpp:37" -ex "r" -ex "p obj1" -ex "p obj2" --args test.exe
#include <iostream>
#include <vector>

class Tester {
public:
  std::string id = "";
  std::vector<int> important;
  std::vector<int> unimportant;
};

int main()
{
  Tester obj1;
  Tester obj2;

  obj1.id = "OBJ1";
  obj1.important.push_back(1);
  obj1.important.push_back(2);
  obj1.important.push_back(3);
  obj1.important.push_back(4);
  for (int i=0; i<25; i++) {
    obj1.unimportant.push_back(-10000000);
  }

  obj2.id = "OBJ2";
  obj2.important.push_back(5);
  obj2.important.push_back(6);
  obj2.important.push_back(7);
  obj2.important.push_back(8);
  for (int i=0; i<25; i++) {
    obj2.unimportant.push_back(-10020000);
  }

  std::cout << "Before exit (for breakpoint): " << obj1.important.size() << ", " << obj2.important.size() << std::endl;

  return 0;
}

如果这个程序在gdb中运行，通过上面给出的命令行，我得到了这个：

$ gdb -ex "b test.cpp:37" -ex "r" -ex "p obj1" -ex "p obj2" --args test.exe 
GNU gdb (Ubuntu 7.7.1-0ubuntu5~14.04.2) 7.7.1
...
Starting program: /tmp/test.exe 

Breakpoint 1, main () at test.cpp:37
37    std::cout << "Before exit (for breakpoint): " << obj1.important.size() << ", " << obj2.important.size() << std::endl;
$1 = {id = "OBJ1", important = std::vector of length 4, capacity 4 = {1, 2, 3, 4}, 
  unimportant = std::vector of length 25, capacity 32 = {-10000000, -10000000, -10000000, -10000000, -10000000, -10000000, 
    -10000000, -10000000, -10000000, -10000000, -10000000, -10000000, -10000000, -10000000, -10000000, -10000000, -10000000, 
    -10000000, -10000000, -10000000, -10000000, -10000000, -10000000, -10000000, -10000000}}
$2 = {id = "OBJ2", important = std::vector of length 4, capacity 4 = {5, 6, 7, 8}, 
  unimportant = std::vector of length 25, capacity 32 = {-10020000, -10020000, -10020000, -10020000, -10020000, -10020000, 
    -10020000, -10020000, -10020000, -10020000, -10020000, -10020000, -10020000, -10020000, -10020000, -10020000, -10020000, 
    -10020000, -10020000, -10020000, -10020000, -10020000, -10020000, -10020000, -10020000}}

在这种情况下，我并不真正关心上面的矢量属性unimportant，并且因为它有很多元素，所以它的打印输出使得阅读比它应该更加困难。但是，我对该对象的所有其他属性感兴趣，因此p单独划分它们可能需要我很多工作来简单地在gdb命令脚本中枚举它们（如果对象有很多属性）。

那么，有没有办法抑制对象中只有一个属性（在本例中为gdb）的.unimportant中的打印输出，甚至更好的类？而不是unimportant = std::vector of length 25, capacity 32 = {...}，我会对打印输出中unimportant = std::vector of length 25, capacity 32 = { __noprint__ }之类的内容感到满意......

Answer 1

有没有办法在gdb中仅抑制单个属性的打印输出

没有定义自定义漂亮的打印机：没有。

您可以通过为您的班级定义自定义漂亮的打印机，使print命令执行您想要的任何操作。文档here。

抑制gdb中特定C ++对象属性的打印输出？

1 个答案: