指向数组类型的指针,c

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*) = %zun",sizeof(void*));
    printf("sizeof(arr)   = %zun",sizeof(arr));
    printf("sizeof(*arr)  = %zun",sizeof(*arr));
    printf("sizeof(int)   = %zun",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] = %dn",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.