为什么不能使用“print”访问向量中的值在gdb调试器？ [英] Why can't I access value in a vector using "print" in gdb debugger?

问题描述

我正在学习使用gdb调试器，但是当该向量包含从另一个向量复制的值时，无法访问向量的值。最终，下面的程序做了我想要它做的事情（将nums的值复制到一个新的向量copyOfNums），但是当我尝试检查copyOfNums [0]的值在copyNums ，则无法访问内存中的值（请参阅下面的gdb会话）。

下面的copy nums已经执行了for循环的for语句，因此copyNums [0]应该包含nums [0]指向的值的地址，引用一个新的int（与nums [0]指向的l值相同）？无论我希望能够访问的值，但正如你可以看到从以下gdb输出，它是不可访问的。

 断点1，main.cpp的copyNums（nums = @ 0x7fff5fbffad0）：10 
 
（gdb）print nums [0] 
 $ 1 =（int&）@ 0x100103910：2 
（gdb）print copyOfNums [0] 
 $ 2 =（int&）@ 0x0：不能访问内存地址0x0 
（gdb）
 
 #include< iostream> ; 
 #include< fstream> 
 #include< sstream> 
 #include< string> 
 #include< vector> 
 
 using namespace std; 
 
 vector< int> copyNums（vector< int# nums）{
 int sizeNums = nums.size（）; 
 vector< int> copyOfNums; 
 
 => for（int i = 0; i  copyOfNums.push_back（nums [i]）; 
} 
 
 return copyOfNums; 
} 
 
 vector< int> getSmallNums（）{
 
 vector< int> nums; 
 nums.push_back（2）; 
 nums.push_back（1）; 
 nums.push_back（8）; 
 
 return nums; 
} 
 
 int main（int argc，char * argv []）{
 vector< int& nums = getSmallNums（）; 
 vector< int> copyOfNums = copyNums（nums）; 
 
 
 return 0; 
} 
  

这里是返回sort后的gbd输出。看起来副本包含对内存中新位置的引用，其l值为2（如预期）。为什么我无法访问函数copyNums中的copyOfNums [0]？

 （gdb）print nums [0] 
 $ 3 =（int&）@ 0x1001000e0：2 
（gdb）print copyOfNums [0] 
 $ 4 =（int&）@ 0x100103920：2 
  
 
 
 
编辑：这是调试会话的更完整的图片
 （gdb）info locals 
i = 1 
 sizeNums = 3 
 copyOfNums = {
< std :: _ Vector_base< int，std :: allocator< int> >> = {
 _M_impl = {
< std :: allocator< int>> = {
< __ gnu_cxx :: new_allocator< int>> = {< No data fields>}，< No data fields>}，
 std :: _ Vector_base< int，std :: allocator< int& > :: _ Vector_impl：
 _M_start = 0x0，
 _M_finish = 0x0，
 _M_end_of_storage = 0x0 
} 
}，< No data fields>} 
（gdb）print copyOfNums [0] 
 $ 1 =（const int&）@ 0x0：无法访问地址0x0处的内存
  
 
 
解决方案
我不知道编译器能够优化多少，但在现实中，这个代码：
 矢量< int> copyNums（vector< int# nums）{
 int sizeNums = nums.size（）; 
 vector< int> copyOfNums; 
 
 for（int i = 0; i  copyOfNums.push_back（nums [i]）; 
} 
 
 return copyOfNums; 
} 
  
与此代码具有相同的结果：
 矢量< int> copyNums（vector< int# nums）{
 return nums; 
} 
  
 
 
如果编译器正在进行这样的优化，优化， copyofNums 可以完全优化。
 
I'm learning to use the gdb debugger but having trouble accessing the values of a vector when that vector contains values copied from another vector. Ultimately the program below does what I want it to do (copies the values of "nums" into a new vector "copyOfNums"), but when I try to examine the value at copyOfNums[0] after executing the first for loop in "copyNums", the value in memory can't be accessed (see gdb session below). 

Below copy nums has executed 1 loop of the for statement, so should copyNums[0] contain the address of the value pointed to by nums[0] or should it contain a reference to a new int (that is the same as the l-value pointed to by nums[0])? Regardless I would expect to be able to access the value, but as you can see from the following gdb output, it is inaccessible.
Breakpoint 1, copyNums (nums=@0x7fff5fbffad0) at main.cpp:10

(gdb) print nums[0]
$1 = (int &) @0x100103910: 2
(gdb) print copyOfNums[0]
$2 = (int &) @0x0: Cannot access memory at address 0x0
(gdb) 

#include <iostream>
#include <fstream>
#include <sstream>
#include <string>
#include <vector>

using namespace std;

vector<int> copyNums(vector<int> nums){
  int sizeNums = nums.size();
  vector<int> copyOfNums;

=>for (int i = 0; i < sizeNums; i++){
    copyOfNums.push_back(nums[i]);
  }

  return copyOfNums;
}

vector<int> getSmallNums(){

  vector<int> nums;
  nums.push_back(2);
  nums.push_back(1);
  nums.push_back(8);

  return nums;
}

int main (int argc, char *argv[]){
    vector<int> nums = getSmallNums();
    vector<int> copyOfNums = copyNums(nums);


    return 0;
}
Here is the gbd output after sort has returned. Looks like the copy contains a reference to a new location in memory, the l-value of which is 2 (as expected). Why was I unable to access copyOfNums[0] within the function copyNums?
(gdb) print nums[0]
$3 = (int &) @0x1001000e0: 2
(gdb) print copyOfNums[0]
$4 = (int &) @0x100103920: 2
EDIT: here is a more complete picture of the debugging session
(gdb) info locals
i = 1
sizeNums = 3
copyOfNums = {
  <std::_Vector_base<int,std::allocator<int> >> = {
    _M_impl = {
      <std::allocator<int>> = {
        <__gnu_cxx::new_allocator<int>> = {<No data fields>}, <No data fields>},
      members of std::_Vector_base<int,std::allocator<int> >::_Vector_impl:
      _M_start = 0x0,
      _M_finish = 0x0,
      _M_end_of_storage = 0x0
    }
  }, <No data fields>}
(gdb) print copyOfNums[0]
$1 = (const int &) @0x0: Cannot access memory at address 0x0

解决方案
I'm not sure how much a compiler will be able to optimize this away, but in reality, this code:
vector<int> copyNums(vector<int> nums){
  int sizeNums = nums.size();
  vector<int> copyOfNums;

  for (int i = 0; i < sizeNums; i++){
    copyOfNums.push_back(nums[i]);
  }

  return copyOfNums;
}
Has the same result as this code:
vector<int> copyNums(vector<int> nums){
  return nums;
}
If a compiler is doing an optimization like that, or a number of other possible optimizations, copyofNums could be completely optimized away.

                        
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