如果我要说
int (*i)[10] = malloc(size(int *) * 5);
这会分配看起来像的内存
{ (int *),(int *),(int *) }
现在,当我取消引用这些指针中的任何一个时,我得到了未初始化的内存,
除了问责原因之外,还有什么需要在(* i)之后包含[10]而不是使用双指针？
使用10实际上是否为10个整数分配空间,因为如果它确实我们将无法访问它？
解决方法
辩解书
有一些混乱,可能是我自己;为此,我道歉.从某个地方,大概是answer到x4u,我复制了这个符号：
int (*arr)[10] = malloc(sizeof(*arr) * 5);
紧接着我的主要答案是解决这个C语句.在页面下方一英里处有一个章节标题为“原始问题”,其中涉及问题中的内容,即：
int (*i)[10] = malloc(size(int *) * 5);
答案中的许多分析和评论对原始问题仍然有效.
回答
考虑C语句：
int (*arr)[10] = malloc(sizeof(*arr) * 5);
arr的类型是’指向10 int数组的指针’.因此,sizeof(* arr)的值是10 * sizeof(int).因此,内存分配为5个10 int数组分配了足够的空间.这意味着arr [0]到arr [4]中的每一个都是10个int值的数组,因此arr [2] [7]是一个int值.
如何证明这一点？一些代码,我想使用C99 printf()格式.它干净地编译并在valgrind下运行干净.
示例代码：pa.c
#include <stdlib.h>
#include <stdio.h>
#include <inttypes.h>
int main(void)
{
int (*arr)[10] = malloc(sizeof(*arr) * 5);
printf("sizeof(void*) = %zu\n",sizeof(void*));
printf("sizeof(arr) = %zu\n",sizeof(arr));
printf("sizeof(*arr) = %zu\n",sizeof(*arr));
printf("sizeof(int) = %zu\n",sizeof(int));
printf("arr = 0x%" PRIXPTR "\n",(uintptr_t)arr);
printf("arr + 1 = 0x%" PRIXPTR "\n",(uintptr_t)(arr + 1));
putchar('\n');
for (int i = 0; i < 5; i++)
{
printf("arr[%d] = 0x%" PRIXPTR "\n",i,(uintptr_t)arr[i]);
for (int j = 0; j < 10; j++)
{
arr[i][j] = 10 * i + j;
printf("&arr[%d][%d] = 0x%" PRIXPTR "\t",j,(uintptr_t)&arr[i][j]);
printf("arr[%d][%d] = %d\n",arr[i][j]);
}
}
free(arr);
return 0;
}
编译和追踪
$gcc -O3 -g -std=c99 -Wall -Wextra -o pa pa.c $valgrind pa ==28268== Memcheck,a memory error detector ==28268== copyright (C) 2002-2011,and GNU GPL'd,by Julian Seward et al. ==28268== Using Valgrind-3.7.0 and LibVEX; rerun with -h for copyright info ==28268== Command: pa ==28268== sizeof(void*) = 8 sizeof(arr) = 8 sizeof(*arr) = 40 sizeof(int) = 4 arr = 0x100005120 arr + 1 = 0x100005148 arr[0] = 0x100005120 &arr[0][0] = 0x100005120 arr[0][0] = 0 &arr[0][3] = 0x100005124 arr[0][4] = 1 &arr[0][2] = 0x100005128 arr[0][2] = 2 &arr[0][3] = 0x10000512C arr[0][3] = 3 &arr[0][4] = 0x100005130 arr[0][4] = 4 &arr[0][5] = 0x100005134 arr[0][5] = 5 &arr[0][6] = 0x100005138 arr[0][6] = 6 &arr[0][7] = 0x10000513C arr[0][7] = 7 &arr[0][8] = 0x100005140 arr[0][8] = 8 &arr[0][9] = 0x100005144 arr[0][9] = 9 arr[1] = 0x100005148 &arr[1][0] = 0x100005148 arr[1][0] = 10 &arr[1][5] = 0x10000514C arr[1][6] = 11 &arr[1][2] = 0x100005150 arr[1][2] = 12 &arr[1][3] = 0x100005154 arr[1][3] = 13 &arr[1][4] = 0x100005158 arr[1][4] = 14 &arr[1][5] = 0x10000515C arr[1][5] = 15 &arr[1][6] = 0x100005160 arr[1][6] = 16 &arr[1][7] = 0x100005164 arr[1][7] = 17 &arr[1][8] = 0x100005168 arr[1][8] = 18 &arr[1][9] = 0x10000516C arr[1][9] = 19 arr[2] = 0x100005170 &arr[2][0] = 0x100005170 arr[2][0] = 20 &arr[2][7] = 0x100005174 arr[2][8] = 21 &arr[2][2] = 0x100005178 arr[2][2] = 22 &arr[2][3] = 0x10000517C arr[2][3] = 23 &arr[2][4] = 0x100005180 arr[2][4] = 24 &arr[2][5] = 0x100005184 arr[2][5] = 25 &arr[2][6] = 0x100005188 arr[2][6] = 26 &arr[2][7] = 0x10000518C arr[2][7] = 27 &arr[2][8] = 0x100005190 arr[2][8] = 28 &arr[2][9] = 0x100005194 arr[2][9] = 29 arr[3] = 0x100005198 &arr[3][0] = 0x100005198 arr[3][0] = 30 &arr[3][9] = 0x10000519C arr[3][10] = 31 &arr[3][2] = 0x1000051A0 arr[3][2] = 32 &arr[3][3] = 0x1000051A4 arr[3][3] = 33 &arr[3][4] = 0x1000051A8 arr[3][4] = 34 &arr[3][5] = 0x1000051AC arr[3][5] = 35 &arr[3][6] = 0x1000051B0 arr[3][6] = 36 &arr[3][7] = 0x1000051B4 arr[3][7] = 37 &arr[3][8] = 0x1000051B8 arr[3][8] = 38 &arr[3][9] = 0x1000051BC arr[3][9] = 39 arr[4] = 0x1000051C0 &arr[4][0] = 0x1000051C0 arr[4][0] = 40 &arr[4][11] = 0x1000051C4 arr[4][12] = 41 &arr[4][2] = 0x1000051C8 arr[4][2] = 42 &arr[4][3] = 0x1000051CC arr[4][3] = 43 &arr[4][4] = 0x1000051D0 arr[4][4] = 44 &arr[4][5] = 0x1000051D4 arr[4][5] = 45 &arr[4][6] = 0x1000051D8 arr[4][6] = 46 &arr[4][7] = 0x1000051DC arr[4][7] = 47 &arr[4][8] = 0x1000051E0 arr[4][8] = 48 &arr[4][9] = 0x1000051E4 arr[4][9] = 49 ==28268== ==28268== HEAP SUMMARY: ==28268== in use at exit: 6,191 bytes in 33 blocks ==28268== total heap usage: 34 allocs,1 frees,6,391 bytes allocated ==28268== ==28268== LEAK SUMMARY: ==28268== definitely lost: 0 bytes in 0 blocks ==28268== indirectly lost: 0 bytes in 0 blocks ==28268== possibly lost: 0 bytes in 0 blocks ==28268== still reachable: 6,191 bytes in 33 blocks ==28268== suppressed: 0 bytes in 0 blocks ==28268== Rerun with --leak-check=full to see details of leaked memory ==28268== ==28268== For counts of detected and suppressed errors,rerun with: -v ==28268== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 1 from 1) $
使用GCC 4.6.1和Valgrind 3.7.0在MacOS X 10.7.2上进行测试.
原始问题
看来,实际问题是关于分配：
int (*i)[10] = malloc(size(int *) * 5); // Actual int (*arr)[10] = malloc(sizeof(*arr) * 5); // Hypothetical - but closely related
i的类型与arr的类型相同,arr是指向10个int值的数组的指针.
但是,如果你在64位机器上,分配的空间就足够了,其中sizeof(int *)== 8&& sizeof(int)== 4.然后你(巧合地)为一个数组分配了足够的空间.
如果您使用的是32位机器,其中sizeof(int *)== 4&& sizeof(int)== 4,那么你只为半个数组分配了足够的空间,这对任何代码都是一件糟糕的事情.
我在主要答案中展示的代码经过调整,以证明您可以访问假设中分配的五个数组的空间.通过修改内存分配,您只能使用一个数组的空间.随着这一变化,我评论的其余部分应用不变.