C:在动态分配期间通过引用传递 [英] C: Passing by reference during dynamic allocaion
问题描述
我的代码如下分配和初始化数组(floatalloc2是用于分配2D数组的函数):
My code is to allocate and initialize the array as follows (floatalloc2 is a function used to allocate the 2D array):
#include <stdio.h>
#include <stdlib.h>
void alloc_arr(int adjwfd, int nx, int nz, float **G, float **L1, float **L2)
{
G = floatalloc2(nx, nz);
switch (adjwfd){
case 1:
L1 = floatalloc2(nx, nz);
break;
case 2:
L2 = floatalloc2(nx, nz);
break;
}
}
void init_arr(int adjwfd, int nx, int nz, float **G, float **L1, float **L2)
{
memset(G[0],0,nx*nz*sizeof(float));
switch (adjwfd){
case 1:
memset(L1[0],0,nx*nz*sizeof(float));
break;
case 2:
memset(L2[0],0,nx*nz*sizeof(float));
break;
}
}
int main(int argc, char *argv[])
{
int adjwfd_P=1, nx=10, nz=10;
float **glob=NULL, **local1=NULL, **local2=NULL;
alloc_arr(adjwfd_P, nx, nz, glob, local1, local2);
init_arr(adjwfd_P, nx, nz, glob, local1, local2);
exit(0);
}
它通过编译.但是当我运行这段代码时，它说错了:
It passes the compilation. But when I run this code, it goes wrong says:
YOUR APPLICATION TERMINATED WITH THE EXIT STRING: Segmentation fault (signal 11)
但是，我发现如果按如下方式更改alloc_arr，它将成功运行:
However, I found that if I change the alloc_arr as follows, it runs successfully:
#include <stdio.h>
#include <stdlib.h>
void alloc_arr(int adjwfd, int nx, int nz, float ***G, float ***L1, float ***L2)
{
*G = floatalloc2(nx, nz);
switch (adjwfd){
case 1:
*L1 = floatalloc2(nx, nz);
break;
case 2:
*L2 = floatalloc2(nx, nz);
break;
}
}
void init_arr(int adjwfd, int nx, int nz, float **G, float **L1, float **L2)
{
memset(G[0],0,nx*nz*sizeof(float));
switch (adjwfd){
case 1:
memset(L1[0],0,nx*nz*sizeof(float));
break;
case 2:
memset(L2[0],0,nx*nz*sizeof(float));
break;
}
}
int main(int argc, char *argv[])
{
int adjwfd_P=1, nx=10, nz=10;
float **glob=NULL, **local1=NULL, **local2=NULL;
alloc_arr(adjwfd_P, nx, nz, &glob, &local1, &local2);
init_arr(adjwfd_P, nx, nz, glob, local1, local2);
exit(0);
}
我的问题是为什么我必须使用2D数组的地址，并且在alloc_arr中仅为分配部分定义3D数组，而在其他功能(例如init_arr)中，我只能传递原始2D数组插入函数?
My question is why I have to take the address of the 2D array, and in the alloc_arr define 3D array only for the allocation part, whereas in other functions such as init_arr, I can just pass the original 2D array into the function?
推荐答案
为什么我必须获取2D数组的地址，并在alloc_arr中仅为分配部分定义3D数组，......
why I have to take the address of the 2D array, and in the alloc_arr define 3D array only for the allocation part,......
在这种情况下:
alloc_arr(adjwfd_P, nx, nz, glob, local1, local2);
您实际上是将最后三个参数作为NULL传递给alloc_arr()函数参数G，L1和L2，因为glob，local1和local2用alloc_arr()
You are actually passing the last three argument as NULL to alloc_arr() function parameter G, L1 and L2 as the glob, local1 and local2 are initialized with NULL and in alloc_arr()
void alloc_arr(int adjwfd, int nx, int nz, float **G, float **L1, float **L2)
{
G = floatalloc2(nx, nz);
.....
L1 = floatalloc2(nx, nz);
.....
L2 = floatalloc2(nx, nz);
.....
正在将内存分配给G函数的局部变量的G，L1和L2.在main()中，返回alloc_arr()函数后，glob，local1和local2是仍然为空指针.这些空指针传递给init_arr()，当尝试在init_arr()中访问它们时，您的程序给出了分段错误.
the memory is getting allocated to G, L1 and L2 which are local variables of alloc_arr() function. In the main(), the glob, local1 and local2 are still null pointers after alloc_arr() function returned. These null pointers passed to init_arr() and when trying to access them in init_arr() your program is giving segmentation fault.
在这种情况下
alloc_arr(adjwfd_P, nx, nz, &glob, &local1, &local2);
您正在将指针glob，local1和local2的地址作为alloc_arr()函数参数G，L1和L2的参数传递，这意味着G，L1和L2将分别保存glob，local1和local2指针的地址.
you are passing the address of pointers glob, local1 and local2 as argument to alloc_arr() function parameters G, L1 and L2, which means G, L1 and L2 will hold address of glob, local1 and local2 pointers respectively.
void alloc_arr(int adjwfd, int nx, int nz, float ***G, float ***L1, float ***L2)
{
*G = floatalloc2(nx, nz);
.....
*L1 = floatalloc2(nx, nz);
.....
*L2 = floatalloc2(nx, nz);
.....
取消引用指针G将得到glob指针，即*G为glob.同样，取消引用*L1和*L2将分别给出指针local1和local2.因此，当为*G分配内存时，*L1和*L2实际上将为glob，local1和local2指针分配内存.
Dereferencing the pointer G will give glob pointer i.e. *G is glob. Similarly, dereferencing *L1 and *L2 will give pointer local1 and local2 respectively. So, when allocating memory to *G, *L1 and *L2 will actually allocate memory to glob, local1 and local2 pointers.
在其他函数(如init_arr)中，我可以将原始2D数组传递给该函数吗?
whereas in other functions such as init_arr, I can just pass the original 2D array into the function?
看看这个:
alloc_arr(adjwfd_P, nx, nz, &glob, &local1, &local2);
init_arr(adjwfd_P, nx, nz, glob, local1, local2);
alloc_arr()将为glob，local1和local2分配内存，这意味着从alloc_arr()返回(假设一切正常)后，指针glob，local1和local2指向有效的内存位置.在init_arr()中，您只是将这些内存位置作为参数传递给init_arr()函数参数G，L1和L2.在init_arr()中，访问G，L1和L2表示它正在访问这些内存位置.
the alloc_arr() will allocate memory to glob, local1 and local2 which means after returning from alloc_arr() (assuming everything works as expected) the pointers glob, local1 and local2 are pointing to valid memory locations. In the init_arr() you are just passing those memory locations as argument to the init_arr() function parameters G, L1 and L2. In the init_arr(), accessing G, L1 and L2 means it is accessing those memory locations.
其他:
遵循良好的编程习惯，请确保在程序退出之前释放已分配的内存.
Follow good programming practice, make sure to free the allocated memory before program exits.
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