英文:
How to iterate through a dynamic, rectangular matrix in C?
问题
Sure, here's the translated code portion you requested:
我需要在C语言中使用动态内存中的指针创建一个矩阵，填充它们以随机数，然后打印出来。
这是大学作业的一部分（我必须为矩阵编写一整套函数），但是就帖子的目的而言，我认为我成功地找到并隔离了有问题的部分。以下是代码：
```c
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <stdbool.h>
int main()
{
int rows = 2;
int cols = 3;
int **matrix = malloc(sizeof(int*) * rows); //声明和分配矩阵的动态内存
for(int i=0;i<rows;i++)
*(matrix+i) = malloc(sizeof(int) * cols);
//生成矩阵的随机数：
srand(time(NULL));
for(int i=0;i<rows;i++)
for(int j=0;j<cols;j++)
*(*(matrix+j)+i) = rand() % 20;
//打印矩阵：
printf("{\t");
for(int i=0;i<rows;i++)
{
int j;
if(i>0)
printf("\t");
printf("{\t");
for(j=0;j<cols;j++)
printf("%d\t",*(*(matrix + j) + i));
printf("}");
if(i<rows-1)
printf("\n");
}
printf("\t}\n\n");
//销毁矩阵
for(int i=0;i<rows;i++)
free(matrix[i]);
free(matrix);
matrix = NULL;
return 0;
}
调试器在这里停止：
*(*(matrix+j)+i) = rand() % 20;
显然，在尝试为第二行的第一列生成随机数时（即首次更改行时），调试器无法访问该位置的内存，然后程序崩溃。
我尝试更改行数和列数，发现只有在行数小于列数时才会崩溃，这很奇怪，因为当它们相等或更多时，它运行得很好。
我正在尝试使用2行和3列。期望的输出是：
{ { 5 17 3 }
{ 1 8 11 } }
实际发生的是：
Process returned -1073741819 (0xC0000005)
当调试器停止（生成随机数时）时，变量的值如下：
i = 0
j = 2
**(matrix + 2)（即 matrix[0][2]）= 无法访问地址为 0xabababababababab 的内存
起初，我怀疑在为矩阵分配内存时出现了问题，但正如我所说，当矩阵是正方形时，它运行得很好，所以我真的不知道问题出在哪里。我知道StackOverflow社区不喜欢有人发布“我不知道我做错了什么，请帮助”的问题，但我确实花了很长时间尝试，无法找到问题的根本原因。请有人能够为我解答一下？
<details>
<summary>英文:</summary>
I have to create a matrix with pointers in dynamic memory in C, fill it with random numbers then print it.
This is part of a larger assignment from college (I had to do a whole library of functions for matrixes) but for the purposes of the post, I think I managed to track down and isolate the problematic bits. Here's the code:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <stdbool.h>
int main()
{
int rows = 2;
int cols = 3;
int *matrix = malloc(sizeof(int) * rows); //declaring and allocating dynamic memory for the matrix
for(int i=0;i<rows;i++)
*(matrix+i) = malloc(sizeof(int) * cols);
//generating random numbers for the matrix:
srand(time(NULL));
for(int i=0;i<rows;i++)
for(int j=0;j<cols;j++)
*(*(matrix+j)+i) = rand() % 20;
//printing matrix:
printf("{\t");
for(int i=0;i<rows;i++)
{
int j;
if(i>0)
printf("\t");
printf("{\t");
for(j=0;j<cols;j++)
printf("%d\t",*(*(matrix + j) + i));
printf("}");
if(i<rows-1)
printf("\n");
}
printf("\t}\n\n");
//destroying the matrix
for(int i=0;i<rows;i++)
free(matrix[i]);
free(matrix);
matrix = NULL;
return 0;
}
The debugger stops here:
`*(*(matrix+j)+i) = rand() % 20;`
Apparently, when trying to generate a random number for the first column of the second row (i.e. when it first changes rows) the debugger "cannot access memory" for that position, then the program crashes.
I tried changing the amount of rows and columns and I found it only crashes when row size is less than column size, which is weird because when it's equal or more, it works just fine.
I'm trying with 2 rows by 3 columns. Expectation:
{ { 5 17 3 }
{ 1 8 11 } }
What actually happens:
`Process returned -1073741819 (0xC0000005)`
Values of the variables when the debugger stops (at generating random numbers):
i = 0
j = 2
**(matrix + 2) (aka matrix[0][2]) = Cannot access memory at address 0xabababababababab
At first I suspected I messed up when allocating memory for the matrix, but as I said when the matrix is square it does work fine, so I really don't know what's the issue here. I know the stackOverflow community doesn't like when someone posts "I don't know what I did wrong please help" kind of thing but I've really been a long time trying and couldn't get to the root of the problem.
Can someone please shed some light on this for me?
</details>
# 答案1
**得分**: 2
These nested for loops contain a bug
这些嵌套的for循环包含一个错误
```c
for(int i=0;i<rows;i++)
for(int j=0;j<cols;j++)
*(*(matrix+i)+j) = rand() % 20;
这个指针matrix指向一个指针数组，这些指针又指向"rows"。所以表达式matrix + i指向第i行。然而，你却使用了索引j。
也就是说，你需要写成
*(*(matrix+i)+j) = rand() % 20;
而不是
*(*(matrix+j)+i) = rand() % 20;
否则，当cols的值大于rows的值时，表达式matrix + j可能会访问分配的指针数组（行）之外的内存。
在printf的这个调用中也存在同样的问题
printf("%d\t",*(*(matrix + j) + i));
你需要写成
printf("%d\t",*(*(matrix + i) + j));
也就是说，你实际上需要访问一个元素matrix[i][j]。这个表达式可以重写为
( *( matrix + i ) )[j]
然后再重写为
*( *( matrix + i ) + j )
或者以另一种方式重写为
*( matrix[i] + j )
然后
*( *( matrix + i ) + j )
为了引起你对下标操作符的兴趣，研究一下这些等价表达式，用来访问一个声明为
T a[M][N];
的二维数组的元素。这些表达式分别是
a[i][j]
i[a][j]
j[a[i]]
j[i[a]]
*( *( a + i ) + j )
*( a[i] + j )
*( i[a] + j )
( *( a + i ) )[j]
j[*( a + i ) ]
英文:
These nested for loops contain a bug
for(int i=0;i<rows;i++)
for(int j=0;j<cols;j++)
*(*(matrix+j)+i) = rand() % 20;
The pointer matrix points to an array of pointers that in turn point to "rows". So the expression matrix + i points to the i-th "row". However instead you are using the index j.
That is you need to write
*(*(matrix+i)+j) = rand() % 20;
instead of
*(*(matrix+j)+i) = rand() % 20;
Otherwise when the value of cols is greater than the value of rows the expression matrix + j can access memory outside the allocated array of pointers (rows).
The same problem exists in this call of printf
printf("%d\t",*(*(matrix + j) + i));
where you have to write
printf("%d\t",*(*(matrix + i) + j));
That is you need actually to access an element matrix[i][j]. This expression may be rewritten like
( *( matrix + i ) )[j]
that in turn may be rewritten like
*( *( matrix + i ) + j )
Or in other way it may be rewritten like
*( matrix[i] + j )
and then
*( *( matrix + i ) + j )
To raise your interest to the subscript operator investigate these equivalent expressions to access elements of a two-dimensional array declared for example like
T a[M][N];
The expressions are
a[i][j]
i[a][j]
j[a[i]]
j[i[a]]
*( *( a + i ) + j )
*( a[i] + j )
*( i[a] + j )
( *( a + i ) )[j]
j[*( a + i ) ]
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