C ++中的警告消息：delete [] a [英] Warning message in C++: delete [ ] a

问题描述

我的代码中有以下警告信息：



$ g ++ GES1.cpp

GES1.cpp：在函数'int main（int，char **）'中：

GES1.cpp：100：14：警告：删除数组'A'

删除[] A;



没有GaussElimination（....）函数的代码是：

Hi,
I am gettingfollowing warning message in my code:

$ g++ GES1.cpp
GES1.cpp: In function ‘int main(int, char**)’:
GES1.cpp:100:14: warning: deleting array ‘A’
delete [ ] A;

My code without the GaussElimination(....) function is:

#include <iostream>
#include <cmath>
#include <fstream>
#include <ctime>
#include <vector>

#define n  3

using namespace std;


void GaussianElimination(double **,double *b ,double *y);
int main(int argc, char * argv[])
{
  /* values
  Row0 = 0, 2, 1; b0= -8
  Row1 = 1, -2, -3; b1 = 0
  Row3 = -1, 1, 2; b2= 3

  Ans = -4, -5, 2 */
  //int **A = new {{0.0, 2.0, 1.0, -8.0}, {1.0, -2.0, -3.0, 0.0},{-1.0, 1.0, 2.0,3.0}};
  
  
  double *A[n];
  for (int i=0; i < n; i++)
    A[i] = new double [n];
  double *b = new double[n];
  double *y = new double[n];
  
  A[0][0] =0.0;A[0][1] =2.0; A[0][2] = 1.0; //A[0][3] = -8.0;
  A[1][0] =1.0;A[1][1] =-2.0; A[1][2] =-3.0; //A[1][3] = 0.0;
  A[2][0] =-1.0;A[0][1] =1.0; A[0][2] = 2.0; //A[0][3] = 3.0;
  b[0] = -8.0;
  b[1] = 0.0;
  b[2] = 3.0;

  
  GaussianElimination(A, b ,y);
  delete [] y;
  delete [ ]b;
  for(int i = 0; i < n; i++)
    delete [] A[i];
  delete [ ] A;  
}

有些人请指导我。



Zulfi。



我尝试过的事情：



对不起我无法理解，因为我使用的是同样的代码在另一个程序中，没有任何警告信息。



Zulfi。

Some body please guide me.

Zulfi.

What I have tried:

Sorry I can't understand because I am using the same code in another program without any warning message.

Zulfi.

推荐答案

g ++ GES1.cpp

GES1.cpp：在函数'int main（int，char **）'中：

GES1.cpp：100：14：警告：删除数组'A'

删除[] A;



我没有GaussElimination（....）函数的代码是：

g++ GES1.cpp
GES1.cpp: In function ‘int main(int, char**)’:
GES1.cpp:100:14: warning: deleting array ‘A’
delete [ ] A;

My code without the GaussElimination(....) function is:

#include <iostream>
#include <cmath>
#include <fstream>
#include <ctime>
#include <vector>

#define n  3

using namespace std;


void GaussianElimination(double **,double *b ,double *y);
int main(int argc, char * argv[])
{
  /* values
  Row0 = 0, 2, 1; b0= -8
  Row1 = 1, -2, -3; b1 = 0
  Row3 = -1, 1, 2; b2= 3

  Ans = -4, -5, 2 */
  //int **A = new {{0.0, 2.0, 1.0, -8.0}, {1.0, -2.0, -3.0, 0.0},{-1.0, 1.0, 2.0,3.0}};
  
  
  double *A[n];
  for (int i=0; i < n; i++)
    A[i] = new double [n];
  double *b = new double[n];
  double *y = new double[n];
  
  A[0][0] =0.0;A[0][1] =2.0; A[0][2] = 1.0; //A[0][3] = -8.0;
  A[1][0] =1.0;A[1][1] =-2.0; A[1][2] =-3.0; //A[1][3] = 0.0;
  A[2][0] =-1.0;A[0][1] =1.0; A[0][2] = 2.0; //A[0][3] = 3.0;
  b[0] = -8.0;
  b[1] = 0.0;
  b[2] = 3.0;

  
  GaussianElimination(A, b ,y);
  delete [] y;
  delete [ ]b;
  for(int i = 0; i < n; i++)
    delete [] A[i];
  delete [ ] A;  
}

有些人请指导我。



Zulfi。



我尝试过的事情：



对不起我无法理解，因为我使用的是同样的代码在另一个程序中没有任何警告信息。



Zulfi。

Some body please guide me.

Zulfi.

What I have tried:

Sorry I can't understand because I am using the same code in another program without any warning message.

Zulfi.

double *A[n];

A是一个你没有动态创建的变量，它只是一个变量。那么为什么你认为你可以删除A？



另见我对你的问题的评论。

"A" is a variable you did not created dynamically, it is simply a variable. So why you think you can delete "A"?

See also my comment to your question.

你误解了你的A [] - 它是一个指针数组，你不能删除。



最好的解决方案是：

You misunderstood your A[] - it is an array of pointers, which you cant delete.

The best solution is:

double A[n][n];

无需正确描述0x01AA等其他分配或删除。

No need for additional allocations or deletes like 0x01AA correctly described.

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