C语言 3D 数组的动态内存分配
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Dynamic memory allocation for 3D array
提问by user239468
Possible Duplicates:
Malloc a 3-Dimensional array in C?
dynamic allocation/deallocation of 2D & 3D arrays
How can i allocate 3D arrays using malloc?
如何使用 malloc 分配 3D 数组？
采纳答案by Roman Dmitrienko
array = malloc(num_elem * num_elem * num_elem * sizeof(array_elem));
Why not? :)
为什么不？:)
回答by Il-Bhima
There are two different ways to allocate a 3D array. You can allocate it either as a 1D array of pointers to a (1D array of pointers to a 1D array). This can be done as follows:
有两种不同的方式来分配 3D 数组。您可以将它分配为指向（指向一维数组的指针的一维数组）的一维指针数组。这可以按如下方式完成：
int dim1, dim2, dim3;
int i,j,k;
double *** array = (double ***)malloc(dim1*sizeof(double**));
for (i = 0; i< dim1; i++) {
array[i] = (double **) malloc(dim2*sizeof(double *));
for (j = 0; j < dim2; j++) {
array[i][j] = (double *)malloc(dim3*sizeof(double));
}
}
Sometimes it is more appropriate to allocate the array as a contiguous chunk. You'll find that many existing libraries might require the array to exist in allocated memory. The disadvantage of this is that if your array is very very big you might not have such a large contiguous chunk available in memory.
有时将数组分配为连续的块更合适。您会发现许多现有的库可能需要数组存在于分配的内存中。这样做的缺点是，如果您的数组非常大，则内存中可能没有这么大的连续块可用。
const int dim1, dim2, dim3; /* Global variables, dimension*/
#define ARR(i,j,k) (array[dim2*dim3*i + dim3*j + k])
double * array = (double *)malloc(dim1*dim2*dim3*sizeof(double));
To access your array you just use the macro:
要访问您的数组，您只需使用宏：
ARR(1,0,3) = 4;
回答by Duleb
This would work
这会工作
int main()
{
int ***p,i,j;
p=(int ***) malloc(MAXX * sizeof(int **));
for(i=0;i<MAXX;i++)
{
p[i]=(int **)malloc(MAXY * sizeof(int *));
for(j=0;j<MAXY;j++)
p[i][j]=(int *)malloc(MAXZ * sizeof(int));
}
for(k=0;k<MAXZ;k++)
for(i=0;i<MAXX;i++)
for(j=0;j<MAXY;j++)
p[i][j][k]=<something>;
}
回答by AareP
@Poita_, ok, maybe you are right, but if somebody still wants to use 3-dimensional array allocated in one big chunk, here's how you add normal indexing to it:
@Poita_，好吧，也许你是对的，但如果有人仍然想使用分配在一个大块中的 3 维数组，这里是你如何向它添加普通索引：
void*** newarray(int icount, int jcount, int kcount, int type_size)
{
void*** iret = (void***)malloc(icount*sizeof(void***)+icount*jcount*sizeof(void**)+icount*jcount*kcount*type_size);
void** jret = (void**)(iret+icount);
char* kret = (char*)(jret+icount*jcount);
for(int i=0;i<icount;i++)
iret[i] = &jret[i*jcount];
for(int i=0;i<icount;i++)
for(int j=0;j<jcount;i++)
jret[i*jcount+j] = &kret[i*jcount*kcount*type_size+j*kcount*type_size];
return iret;
}
回答by John Bode
For a given type T (non-contiguous):
对于给定的类型 T（非连续）：
size_t dim0, dim1, dim2;
...
T ***arr = malloc(sizeof *arr * dim0); //type of *arr is T **
if (arr)
{
size_t i;
for (i = 0; i < dim0; i++)
{
arr[i] = malloc(sizeof *arr[i] * dim1); // type of *arr[i] is T *
if (arr[i])
{
size_t j;
for (j = 0; j < dim1; j++)
{
arr[i][j] = malloc(sizeof *arr[i][j] * dim2);
}
}
}
}
Unless you are working with a veryold (pre-C89) implementation, you do not need to cast the result of malloc(), and the practice is discouraged. If you forget to include stdlib.h or otherwise don't have a prototype for malloc()in scope, the compiler will type it to return int, and you'll get an "incompatible type for assignment"-type warning. If you cast the result, the warning is suppressed, and there's no guarantee that a conversion from a pointer to an intto a pointer again will be meaningful.
除非您正在使用非常旧的（C89 之前）实现，否则您不需要强制转换 的结果malloc()，并且不鼓励这种做法。如果您忘记包含 stdlib.h 或在其他情况下没有malloc()范围内的原型，编译器会将其键入为 return int，并且您将收到“不兼容的赋值类型”类型警告。如果您转换结果，警告将被抑制，并且不能保证从指针到 anint再到指针的转换将是有意义的。
