释放calloc数组的数组在C负折射率 [英] Calloc for an array of array with negative index in C
问题描述
我有阵列的带负折射率的阵列。它是一种具有实际尺寸[dim_y + 40] [dim_x + 40]的数组，但用户使用阵列的像它的尺寸为[dim_y] [dim_x]。
首先我有全球和已定义的尺寸dim_x，dim_y，所以我有这个
I have an array of array with negative index. It is an array which has real dimensions [dim_y + 40][dim_x + 40] but the user uses the array like it has dimensions [dim_y][dim_x]. First i had global and already defined the dimensions dim_x, dim_y, so i had this
int map_boundaries[dim_y + 40][dim_x + 40];
int (*map)[dim_x+40] = (int(*)[dim_x+40])&map_boundaries[20][20];
这工作得很好。
现在我需要的尺寸dim_y和dim_x是可变的，有了这个，我的意思是，我想阵'地图'来没有固定的大小，但动态的，我需要阅读dim_y，dim_x从用户和阵'地图'是全球性的，所以我有
It worked fine. Now i need the dimensions dim_y and dim_x to be variable, and with this i mean that i want the array 'map' to not have fixed size but dynamic, i need to read the dim_y, dim_x from a user and the array 'map' to be global, so i have
int **map_boundaries;
和我使用释放calloc 在的main（）
map_boundaries = (int **)calloc(dim_y + 40,sizeof(int*));
for(i = 0; i < dim_y + 40; i++){
map_boundaries[i] = (int *)calloc(dim_x + 40,sizeof(int));
}
，但我不知道如何申报第二行
, but i don't know how to declare the second line
为了更好地理解边界的事情，我做了什么被张贴在第二评论在这里： HTTP： //everything2.com/title/Negative+array+indexing
For better understanding of the boundaries thing i did what is posted in the 2nd comment here: http://everything2.com/title/Negative+array+indexing
推荐答案
下面是我的code版本。在 INT **地图值是什么用户和图书馆使用访问数据。在矩阵结构封装了所有有关阵列的信息，包括在使用初始化地图的成员。在code设置大小，分配和初始化结构，然后初始化数组，然后使用它作为一个用户会，并作为库会。最后，分配的内存被释放。
Here's my version of the code. The int **map value is what both user and library use to access the data. The Matrix structure encapsulates all the information about the array, including a member that is use to initialize map. The code sets up the sizes, allocates and initializes the structure, then initializes the array, then uses it as a user would, and as the library would. Finally, the allocated memory is freed.
在实践中，你可能要拆分这件事成若干个独立的功能。
In practice, you'd probably want to split this up into a number of separate functions.
#include <inttypes.h>
#include <stdio.h>
#include <stdlib.h>
#include "stderr.h"
typedef struct Matrix
{
int *map_base;
int **map_rows;
int x_dim;
int y_dim;
int extra;
int **map;
} Matrix;
int main(void)
{
Matrix m;
int x_dim = 8;
int y_dim = 6;
int extra = 5;
/* Allocation */
m.extra = extra;
m.x_dim = x_dim;
m.y_dim = y_dim;
int x_cells = x_dim + 2 * extra;
int y_cells = y_dim + 2 * extra;
m.map_base = calloc(x_cells * y_cells, sizeof(**m.map));
if (m.map_base == 0)
err_syserr("Failed to allocate %zu bytes memory\n",
x_cells * y_cells * sizeof(**m.map));
m.map_rows = calloc(x_cells, sizeof(*m.map));
if (m.map_rows == 0)
err_syserr("Failed to allocate %zu bytes memory\n",
x_cells * sizeof(*m.map));
//printf("Map base: 0x%.8" PRIXPTR "\n", (uintptr_t)m.map_base);
//printf("Map rows: 0x%.8" PRIXPTR "\n", (uintptr_t)m.map_rows);
for (int i = 0; i < x_cells; i++)
{
m.map_rows[i] = &m.map_base[i * y_cells + extra];
//printf("Row[%2d] 0x%.8" PRIXPTR "\n", i, (uintptr_t)m.map_rows[i]);
}
m.map = &m.map_rows[extra];
int **map = m.map;
//printf("Map: 0x%.8" PRIXPTR "\n", (uintptr_t)map);
/* Initialization */
int x_min = -extra;
int y_min = -extra;
int x_max = x_dim + extra;
int y_max = y_dim + extra;
printf("Initialization:\n");
for (int i = x_min; i < x_max; i++)
{
for (int j = y_min; j < y_max; j++)
{
map[i][j] = i * 100 + j;
//printf("[%2d,%2d] = %4d\n", i, j, map[i][j]);
}
}
printf("User view:\n");
for (int i = 0; i < x_dim; i++)
{
for (int j = 0; j < y_dim; j++)
printf("[%2d,%2d] = %4d\n", i, j, map[i][j]);
}
printf("Library view:\n");
for (int i = x_min; i < x_max; i++)
{
for (int j = y_min; j < y_max; j++)
printf("[%2d,%2d] = %4d\n", i, j, map[i][j]);
}
/* Deallocation */
free(m.map_base);
free(m.map_rows);
return 0;
}
示例输出
当在Mac OS下 的valgrind 3.10.0运行点¯x10.9.5使用GCC 4.9.1，我得到了以下的输出。在stderr.h标题是 $ HOME / INC 并声明 err_syserr（） ，而 libjl.a 是 $ HOME / lib目录/ 64 并提供实施为 err_syserr（），即报告给定的错误消息，并通过标识的错误错误号并退出功能。我懒得写这个计划，这一次的code。在SO找到 err_setarg0（）在code在这个答案。）
Example output
When run under valgrind 3.10.0 on Mac OS X 10.9.5 with GCC 4.9.1, I got the following output. The "stderr.h" header is in $HOME/inc and declares err_syserr(), while libjl.a is in $HOME/lib/64 and provides the implementation for err_syserr(), a function that reports the given error message and the error identified by errno and exits. I'm too lazy to write the code in this program this time around. (Search for '[c] user:15168 err_syserr' in SO to find Protect a shared memory segment with semaphore does not work, which has essentially equivalent code. I didn't call err_setarg0() in the code in this answer.)
$ gcc -O3 -g -I/Users/jleffler/inc -std=c11 -Wall -Wextra -Wmissing-prototypes \
> -Wstrict-prototypes -Wold-style-definition -Werror 2da.c -o 2da \
> -L/Users/jleffler/lib/64 -ljl
$ valgrind ./2da
==26293== Memcheck, a memory error detector
==26293== Copyright (C) 2002-2013, and GNU GPL'd, by Julian Seward et al.
==26293== Using Valgrind-3.10.0.SVN and LibVEX; rerun with -h for copyright info
==26293== Command: 2da
==26293==
Initialization:
User view:
[ 0, 0] = 0
[ 0, 1] = 1
[ 0, 2] = 2
[ 0, 3] = 3
[ 0, 4] = 4
[ 0, 5] = 5
[ 1, 0] = 100
[ 1, 1] = 101
[ 1, 2] = 102
[ 1, 3] = 103
[ 1, 4] = 104
[ 1, 5] = 105
[ 2, 0] = 200
[ 2, 1] = 201
[ 2, 2] = 202
[ 2, 3] = 203
[ 2, 4] = 204
[ 2, 5] = 205
[ 3, 0] = 300
[ 3, 1] = 301
[ 3, 2] = 302
[ 3, 3] = 303
[ 3, 4] = 304
[ 3, 5] = 305
[ 4, 0] = 400
[ 4, 1] = 401
[ 4, 2] = 402
[ 4, 3] = 403
[ 4, 4] = 404
[ 4, 5] = 405
[ 5, 0] = 500
[ 5, 1] = 501
[ 5, 2] = 502
[ 5, 3] = 503
[ 5, 4] = 504
[ 5, 5] = 505
[ 6, 0] = 600
[ 6, 1] = 601
[ 6, 2] = 602
[ 6, 3] = 603
[ 6, 4] = 604
[ 6, 5] = 605
[ 7, 0] = 700
[ 7, 1] = 701
[ 7, 2] = 702
[ 7, 3] = 703
[ 7, 4] = 704
[ 7, 5] = 705
Library view:
[-5,-5] = -505
[-5,-4] = -504
[-5,-3] = -503
[-5,-2] = -502
[-5,-1] = -501
[-5, 0] = -500
[-5, 1] = -499
[-5, 2] = -498
[-5, 3] = -497
[-5, 4] = -496
[-5, 5] = -495
[-5, 6] = -494
[-5, 7] = -493
[-5, 8] = -492
[-5, 9] = -491
[-5,10] = -490
[-4,-5] = -405
[-4,-4] = -404
[-4,-3] = -403
[-4,-2] = -402
[-4,-1] = -401
[-4, 0] = -400
[-4, 1] = -399
[-4, 2] = -398
[-4, 3] = -397
[-4, 4] = -396
[-4, 5] = -395
[-4, 6] = -394
[-4, 7] = -393
[-4, 8] = -392
[-4, 9] = -391
[-4,10] = -390
[-3,-5] = -305
[-3,-4] = -304
[-3,-3] = -303
[-3,-2] = -302
[-3,-1] = -301
[-3, 0] = -300
[-3, 1] = -299
[-3, 2] = -298
[-3, 3] = -297
[-3, 4] = -296
[-3, 5] = -295
[-3, 6] = -294
[-3, 7] = -293
[-3, 8] = -292
[-3, 9] = -291
[-3,10] = -290
[-2,-5] = -205
[-2,-4] = -204
[-2,-3] = -203
[-2,-2] = -202
[-2,-1] = -201
[-2, 0] = -200
[-2, 1] = -199
[-2, 2] = -198
[-2, 3] = -197
[-2, 4] = -196
[-2, 5] = -195
[-2, 6] = -194
[-2, 7] = -193
[-2, 8] = -192
[-2, 9] = -191
[-2,10] = -190
[-1,-5] = -105
[-1,-4] = -104
[-1,-3] = -103
[-1,-2] = -102
[-1,-1] = -101
[-1, 0] = -100
[-1, 1] = -99
[-1, 2] = -98
[-1, 3] = -97
[-1, 4] = -96
[-1, 5] = -95
[-1, 6] = -94
[-1, 7] = -93
[-1, 8] = -92
[-1, 9] = -91
[-1,10] = -90
[ 0,-5] = -5
[ 0,-4] = -4
[ 0,-3] = -3
[ 0,-2] = -2
[ 0,-1] = -1
[ 0, 0] = 0
[ 0, 1] = 1
[ 0, 2] = 2
[ 0, 3] = 3
[ 0, 4] = 4
[ 0, 5] = 5
[ 0, 6] = 6
[ 0, 7] = 7
[ 0, 8] = 8
[ 0, 9] = 9
[ 0,10] = 10
[ 1,-5] = 95
[ 1,-4] = 96
[ 1,-3] = 97
[ 1,-2] = 98
[ 1,-1] = 99
[ 1, 0] = 100
[ 1, 1] = 101
[ 1, 2] = 102
[ 1, 3] = 103
[ 1, 4] = 104
[ 1, 5] = 105
[ 1, 6] = 106
[ 1, 7] = 107
[ 1, 8] = 108
[ 1, 9] = 109
[ 1,10] = 110
[ 2,-5] = 195
[ 2,-4] = 196
[ 2,-3] = 197
[ 2,-2] = 198
[ 2,-1] = 199
[ 2, 0] = 200
[ 2, 1] = 201
[ 2, 2] = 202
[ 2, 3] = 203
[ 2, 4] = 204
[ 2, 5] = 205
[ 2, 6] = 206
[ 2, 7] = 207
[ 2, 8] = 208
[ 2, 9] = 209
[ 2,10] = 210
[ 3,-5] = 295
[ 3,-4] = 296
[ 3,-3] = 297
[ 3,-2] = 298
[ 3,-1] = 299
[ 3, 0] = 300
[ 3, 1] = 301
[ 3, 2] = 302
[ 3, 3] = 303
[ 3, 4] = 304
[ 3, 5] = 305
[ 3, 6] = 306
[ 3, 7] = 307
[ 3, 8] = 308
[ 3, 9] = 309
[ 3,10] = 310
[ 4,-5] = 395
[ 4,-4] = 396
[ 4,-3] = 397
[ 4,-2] = 398
[ 4,-1] = 399
[ 4, 0] = 400
[ 4, 1] = 401
[ 4, 2] = 402
[ 4, 3] = 403
[ 4, 4] = 404
[ 4, 5] = 405
[ 4, 6] = 406
[ 4, 7] = 407
[ 4, 8] = 408
[ 4, 9] = 409
[ 4,10] = 410
[ 5,-5] = 495
[ 5,-4] = 496
[ 5,-3] = 497
[ 5,-2] = 498
[ 5,-1] = 499
[ 5, 0] = 500
[ 5, 1] = 501
[ 5, 2] = 502
[ 5, 3] = 503
[ 5, 4] = 504
[ 5, 5] = 505
[ 5, 6] = 506
[ 5, 7] = 507
[ 5, 8] = 508
[ 5, 9] = 509
[ 5,10] = 510
[ 6,-5] = 595
[ 6,-4] = 596
[ 6,-3] = 597
[ 6,-2] = 598
[ 6,-1] = 599
[ 6, 0] = 600
[ 6, 1] = 601
[ 6, 2] = 602
[ 6, 3] = 603
[ 6, 4] = 604
[ 6, 5] = 605
[ 6, 6] = 606
[ 6, 7] = 607
[ 6, 8] = 608
[ 6, 9] = 609
[ 6,10] = 610
[ 7,-5] = 695
[ 7,-4] = 696
[ 7,-3] = 697
[ 7,-2] = 698
[ 7,-1] = 699
[ 7, 0] = 700
[ 7, 1] = 701
[ 7, 2] = 702
[ 7, 3] = 703
[ 7, 4] = 704
[ 7, 5] = 705
[ 7, 6] = 706
[ 7, 7] = 707
[ 7, 8] = 708
[ 7, 9] = 709
[ 7,10] = 710
[ 8,-5] = 795
[ 8,-4] = 796
[ 8,-3] = 797
[ 8,-2] = 798
[ 8,-1] = 799
[ 8, 0] = 800
[ 8, 1] = 801
[ 8, 2] = 802
[ 8, 3] = 803
[ 8, 4] = 804
[ 8, 5] = 805
[ 8, 6] = 806
[ 8, 7] = 807
[ 8, 8] = 808
[ 8, 9] = 809
[ 8,10] = 810
[ 9,-5] = 895
[ 9,-4] = 896
[ 9,-3] = 897
[ 9,-2] = 898
[ 9,-1] = 899
[ 9, 0] = 900
[ 9, 1] = 901
[ 9, 2] = 902
[ 9, 3] = 903
[ 9, 4] = 904
[ 9, 5] = 905
[ 9, 6] = 906
[ 9, 7] = 907
[ 9, 8] = 908
[ 9, 9] = 909
[ 9,10] = 910
[10,-5] = 995
[10,-4] = 996
[10,-3] = 997
[10,-2] = 998
[10,-1] = 999
[10, 0] = 1000
[10, 1] = 1001
[10, 2] = 1002
[10, 3] = 1003
[10, 4] = 1004
[10, 5] = 1005
[10, 6] = 1006
[10, 7] = 1007
[10, 8] = 1008
[10, 9] = 1009
[10,10] = 1010
[11,-5] = 1095
[11,-4] = 1096
[11,-3] = 1097
[11,-2] = 1098
[11,-1] = 1099
[11, 0] = 1100
[11, 1] = 1101
[11, 2] = 1102
[11, 3] = 1103
[11, 4] = 1104
[11, 5] = 1105
[11, 6] = 1106
[11, 7] = 1107
[11, 8] = 1108
[11, 9] = 1109
[11,10] = 1110
[12,-5] = 1195
[12,-4] = 1196
[12,-3] = 1197
[12,-2] = 1198
[12,-1] = 1199
[12, 0] = 1200
[12, 1] = 1201
[12, 2] = 1202
[12, 3] = 1203
[12, 4] = 1204
[12, 5] = 1205
[12, 6] = 1206
[12, 7] = 1207
[12, 8] = 1208
[12, 9] = 1209
[12,10] = 1210
==26293==
==26293== HEAP SUMMARY:
==26293== in use at exit: 29,341 bytes in 374 blocks
==26293== total heap usage: 452 allocs, 78 frees, 36,581 bytes allocated
==26293==
==26293== LEAK SUMMARY:
==26293== definitely lost: 0 bytes in 0 blocks
==26293== indirectly lost: 0 bytes in 0 blocks
==26293== possibly lost: 0 bytes in 0 blocks
==26293== still reachable: 4,096 bytes in 1 blocks
==26293== suppressed: 25,245 bytes in 373 blocks
==26293== Rerun with --leak-check=full to see details of leaked memory
==26293==
==26293== For counts of detected and suppressed errors, rerun with: -v
==26293== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)
$
在'仍然到达块是标准输出缓冲区;如果我包括 FCLOSE（标准输出）; 从返回前的main（），它会去。 Mac OS X的运行时库分配大量内存太多，而是使用在SUP pressed信息标识。
The 'still reachable' block is the buffer for stdout; if I included fclose(stdout); before returning from main(), it would go. The Mac OS X runtime library allocates a lot of memory too, but that usage is identified in the suppressed information.
我做全initiallization如您在2线看到的。但它不是这就是问题的初始化;的问题是，在每帧映射的值被改变。所以，我只需要一个'地图'，它直接指向边界地图。 地图[0] [0] 是 boundaries_map [20] [20] 等。我想声明全局 **地图，然后写我的第二行到主，但我得到施放的错误。 （*地图）dim_x + 40] 这是一个 **地图是不是？
I do the whole initiallization as you see in 2 lines. But it is not the initialization that's the problem; the problem is that in every frame the values of map are changed. So I need just a 'map' which points directly to 'boundaries map'.
map[0][0]isboundaries_map[20][20]and so on. I tried to declare global**mapand then write my second line into the main but I get an error about cast.(*map)[dim_x+40]this is a**mapisn't it?
我不知道我完全理解你以后，这是一个原因，这种反应已经采取了很长一段时间。
I'm not sure I completely understand what you're after, which is one reason this response has taken a long time.
您不能是沃拉斯全局变量;他们只能在功能通过的malloc分配，无论是在堆栈局部变量或在堆（）等。原来的code我发现，您可以通过一个全局变量，双指针，可以从任何code进行访问，可以读取双指针。
You cannot have global variables that are VLAs; they can only be allocated in functions, either on the stack as local variables or on the heap via malloc() et al. The original code I showed allows you to define an array via a global variable, a double pointer, that can be accessed from any code that can read the double pointer.
下code，可考虑构建沃拉斯的练习。它采用的malloc（）以创建一个VLA，记录它是在一个指针VLA创建。这可以被传递到就像一个基于堆栈的VLA可以传递的功能。在code完全编译GCC 4.9.1下在Mac OS X 10.9.5：
The code below could be considered an exercise in constructing VLAs. It uses malloc() to create a VLA, recording where it was created in a pointer to a VLA. This can be passed to a function just like a stack-based VLA can be passed. The code compiles cleanly under GCC 4.9.1 on Mac OS X 10.9.5:
$ gcc -g -O3 -std=c99 -Wall -Wextra -Wmissing-prototypes -Wstrict-prototypes -Werror \
> 2dv.c -o 2dv
它运行干净，无泄漏，根据 的valgrind
#include <stdio.h>
#include <stdlib.h>
static void dump_vla_1(int x_size, int y_size, int vla[x_size][y_size]);
static void dump_vla_2(int x_min, int y_min, int x_size, int y_size,
int vla[x_size][y_size]);
static void dump_vla_3(int x_min, int x_max, int y_min, int y_max,
int x_size, int y_size, int vla[x_size][y_size]);
int main(void)
{
int extra = 3;
int x_base = 5;
int y_base = 4;
int x_size = x_base + 2 * extra;
int y_size = y_base + 2 * extra;
int (*vla)[x_size][y_size] = calloc(x_size * y_size, sizeof(int));
for (int i = 0; i < x_size; i++)
for (int j = 0; j < y_size; j++)
(*vla)[i][j] = (i + 1) * 100 + (j + 1);
for (int i = 0; i < x_size; i++)
{
for (int j = 0; j < y_size; j++)
printf(" %4d", (*vla)[i][j]);
putchar('\n');
}
dump_vla_1(x_size, y_size, *vla);
dump_vla_2(0, 0, x_size, y_size, *vla);
/* Harsh cast! */
int (*vla_2)[x_size][y_size] = (int (*)[x_size][y_size])&(*vla)[extra][extra];
dump_vla_2(-extra, -extra, x_size, y_size, *vla_2);
dump_vla_3(-extra, x_size - extra, -extra, y_size - extra, x_size, y_size, *vla_2);
dump_vla_3(0, x_base, 0, y_base, x_size, y_size, *vla_2);
free(vla);
return 0;
}
static void dump_vla_1(int x_size, int y_size, int vla[x_size][y_size])
{
printf("Matrix %dx%d\n", x_size, y_size);
for (int i = 0; i < x_size; i++)
{
for (int j = 0; j < y_size; j++)
printf(" %4d", vla[i][j]);
putchar('\n');
}
}
static void dump_vla_2(int x_min, int y_min, int x_size, int y_size,
int vla[x_size][y_size])
{
printf("Matrix %dx%d (%d..%d, %d..%d)\n",
x_size, y_size, x_min, x_min + x_size - 1, y_min, y_min + y_size - 1);
for (int i = x_min; i < x_min + x_size; i++)
{
for (int j = y_min; j < y_min + y_size; j++)
printf(" %4d", vla[i][j]);
putchar('\n');
}
}
static void dump_vla_3(int x_min, int x_max, int y_min, int y_max,
int x_size, int y_size, int vla[x_size][y_size])
{
printf("Matrix %dx%d (%d..%d, %d..%d)\n",
x_size, y_size, x_min, x_max, y_min, y_max);
for (int i = x_min; i < x_max; i++)
{
for (int j = y_min; j < y_max; j++)
printf(" %4d", vla[i][j]);
putchar('\n');
}
}
请注意使用的* VLA 和 * vla_2 来的VLA传递给函数。因为 VLA 和 vla_2 变量指向一个VLA， * VLA 的是的VLA的该功能的预期。
Note the use of *vla and *vla_2 to pass the VLA to the functions. Since the vla and vla_2 variables are pointers to a VLA, *vla is the VLA which the functions expect.
输出：
101 102 103 104 105 106 107 108 109 110
201 202 203 204 205 206 207 208 209 210
301 302 303 304 305 306 307 308 309 310
401 402 403 404 405 406 407 408 409 410
501 502 503 504 505 506 507 508 509 510
601 602 603 604 605 606 607 608 609 610
701 702 703 704 705 706 707 708 709 710
801 802 803 804 805 806 807 808 809 810
901 902 903 904 905 906 907 908 909 910
1001 1002 1003 1004 1005 1006 1007 1008 1009 1010
1101 1102 1103 1104 1105 1106 1107 1108 1109 1110
Matrix 11x10
101 102 103 104 105 106 107 108 109 110
201 202 203 204 205 206 207 208 209 210
301 302 303 304 305 306 307 308 309 310
401 402 403 404 405 406 407 408 409 410
501 502 503 504 505 506 507 508 509 510
601 602 603 604 605 606 607 608 609 610
701 702 703 704 705 706 707 708 709 710
801 802 803 804 805 806 807 808 809 810
901 902 903 904 905 906 907 908 909 910
1001 1002 1003 1004 1005 1006 1007 1008 1009 1010
1101 1102 1103 1104 1105 1106 1107 1108 1109 1110
Matrix 11x10 (0..10, 0..9)
101 102 103 104 105 106 107 108 109 110
201 202 203 204 205 206 207 208 209 210
301 302 303 304 305 306 307 308 309 310
401 402 403 404 405 406 407 408 409 410
501 502 503 504 505 506 507 508 509 510
601 602 603 604 605 606 607 608 609 610
701 702 703 704 705 706 707 708 709 710
801 802 803 804 805 806 807 808 809 810
901 902 903 904 905 906 907 908 909 910
1001 1002 1003 1004 1005 1006 1007 1008 1009 1010
1101 1102 1103 1104 1105 1106 1107 1108 1109 1110
Matrix 11x10 (-3..7, -3..6)
101 102 103 104 105 106 107 108 109 110
201 202 203 204 205 206 207 208 209 210
301 302 303 304 305 306 307 308 309 310
401 402 403 404 405 406 407 408 409 410
501 502 503 504 505 506 507 508 509 510
601 602 603 604 605 606 607 608 609 610
701 702 703 704 705 706 707 708 709 710
801 802 803 804 805 806 807 808 809 810
901 902 903 904 905 906 907 908 909 910
1001 1002 1003 1004 1005 1006 1007 1008 1009 1010
1101 1102 1103 1104 1105 1106 1107 1108 1109 1110
Matrix 11x10 (-3..8, -3..7)
101 102 103 104 105 106 107 108 109 110
201 202 203 204 205 206 207 208 209 210
301 302 303 304 305 306 307 308 309 310
401 402 403 404 405 406 407 408 409 410
501 502 503 504 505 506 507 508 509 510
601 602 603 604 605 606 607 608 609 610
701 702 703 704 705 706 707 708 709 710
801 802 803 804 805 806 807 808 809 810
901 902 903 904 905 906 907 908 909 910
1001 1002 1003 1004 1005 1006 1007 1008 1009 1010
1101 1102 1103 1104 1105 1106 1107 1108 1109 1110
Matrix 11x10 (0..5, 0..4)
404 405 406 407
504 505 506 507
604 605 606 607
704 705 706 707
804 805 806 807
这篇关于释放calloc数组的数组在C负折射率的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
