问题描述
我一直在写我的电磁仿真课程一块code和我所遇到的一个问题。我决定做一点额外的通过扩大原有的计算，可达10 ^ 8个元素的真正大的网，所以现在我不得不使用malloc（）。
I have been writing a piece of code for my coursework in electromagnetic simulation and I have run into a problem. I decided to do a bit extra by expanding the original calculations to really large meshes of up to 10^8 elements, so now I have to use malloc().
到目前为止，一切都很好，但因为我preFER让我的code库中，然后用编译器的内嵌选项编译，我需要一种方法来传递函数之间的信息。所以，我开始使用结构来跟踪网格的参数，以及指针的信息的阵列。我定义的结构方式如下：
So far, so good, but since I prefer to keep my code in libraries and then compile with the inline option of the compiler, I needed a way to pass information between functions. So, I started using structs to keep track of the parameters of the mesh, as well as the pointer to the array of information. I defined the struct the following way:
typedef struct {
int height;
int width;
int bottom; //position of the bottom node
unsigned int*** dat_ptr;//the pointer to the array with all the data
} array_info;
凡三重指向一个unsigned int是指向一个二维数组。我必须这样做，这样一来，因为否则它是按值传递，我无法从函数内改变。
Where the triple pointer to an unsigned int is the pointer to a 2D array. I have to do it this way because otherwise it is passed by value and I cannot change it from within the function.
现在，当我尝试使用以下功能结构分配内存：
Now, when I try to allocate memory for the struct with the following function:
void create_array(array_info A)//the function accepts struct of type "array_info" as argument
{
int i;
unsigned int** array = malloc(sizeof(*array) * A.height);//creates an array of arrays
for(i = 0; i<A.height; ++i)
{
array[i] = malloc(sizeof(**array) * A.width);//creates an array for each row
}
*A.dat_ptr=array;//assigns the position of the array to the input pointer
}
我在执行操作时出现段故障。我不明白为什么的sizeof（* A.dat_ptr）是一样的sizeof（数组）。因此，在最坏的情况下，我应该得到的地方乱码的路线，而不是在分配行，对吧？
I get a segmentation fault upon executing the operation. I cannot see why: sizeof(*A.dat_ptr) is the same as sizeof(array). Thus, in the worst case I should be getting gibberish somewhere down the line, not in the assignment line, right?
推荐答案
您要么需要从函数返回的 array_info 结构（修订本）或（更通常）一个指针传递给 array_info 结构到功能，使您所做的更改会影响调用函数的值。
You either need to return the array_info structure (as amended) from the function or (more usually) pass a pointer to the array_info structure into the function so that the changes you make affect the value in the calling function.
typedef struct
{
int height;
int width;
int bottom;
unsigned int **dat_ptr; // Double pointer, not triple pointer
} array_info;
void create_array(array_info *A)
{
unsigned int **array = malloc(sizeof(*array) * A->height);
for (int i = 0; i < A->height; ++i)
array[i] = malloc(sizeof(**array) * A->width);
A->dat_ptr = array;
}
我假设你做的内存分配一些地方检查;逻辑的地方就是这个功能，虽然。从故障中的一部分的方式，通过恢复是繁琐的（但必要的，如果你打算从函数返回，而不是退出程序）。
I assume you do some checking on the memory allocations somewhere; the logical place is this function, though. Recovery from a failure part way through is fiddly (but necessary if you are going to return from the function rather than exit from the program).
void create_array(array_info *A)
{
unsigned int **array = malloc(sizeof(*array) * A->height);
if (array != 0)
{
for (int i = 0; i < A->height; ++i)
{
if ((array[i] = malloc(sizeof(**array) * A->width)) == 0)
{
for (int j = 0; j < i; j++)
free(array[j]);
free(array);
array = 0;
break;
}
}
}
A->dat_ptr = array;
}
调用函数知道，如果 dat_ptr 成员是从 create_array返回NULL功能失败（）。这可能是更好的提供成功/失败的返回值。
The calling function knows that the function failed if the dat_ptr member is null on return from create_array(). It might be better to provide a success/failure return value.
我使用的是C99，因此调用code可能是：
I'm using C99, so the calling code might be:
array_info array = { .height = 10, .width = 20, .dat_ptr = 0 };
create_array(&array);
if (array->dat_ptr == 0)
...error handling...
请注意，在 create_array的code（）可能需要检查空指针，为负或零宽度或高度。我不清除底元素应该包含什么，所以我把它初始化，这给了我使用指定的初始值一半的借口。你也可以写初始化很清楚，而无需使用指定的初始值。
Note that the code in create_array() might need to check for a null pointer, for negative or zero width or height. I'm not clear what the bottom element should contain, so I left it uninitialized, which gives me half an excuse for using designated initializers. You can also write the initializer quite clearly without using designated initializers.
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