指向连续二维数组的指针

Question

我使用的是带有标志符-Wall -std=gnu99的gcc版本4.8.0。

我需要在C中使用malloc动态分配连续二维数组的内存;这一事实不容置疑。但是，为了便于使用，我仍然希望能够使用方便的x[r][c]符号访问数组。这是我勇敢的尝试创建一个指向连续二维数组的指针，并通过执行*array[r][c]:

对数组进行索引

#include <stdlib.h>
#include <stdio.h>
int main(void)
{
    size_t rows = 3, cols = 5;
    printf("sizeof(int) = %linn", sizeof(int));
    int (*array)[rows][cols] = malloc(sizeof(int) * rows * cols);
    printf("array starts at %pn", array);
    printf("sizeof(array) = %lin", sizeof(array));
    printf("sizeof(array[0][0]) = 0x%lxn", sizeof(array[0][0]));
    puts("");
    unsigned short r, c;
    for (r = 0; r <= rows - 1; r++) {
        for (c = 0; c <= cols - 1; c++) {
            printf("array[%i][%i] is at %pn", r, c, &(*array[r][c]));
        };
        puts("");
    };
}

虽然它编译时没有警告，但它有一些意想不到的输出:

sizeof(int) = 4
array starts at 0x16cc010
sizeof(array) = 8
sizeof(array[0][0]) = 0x14
array[0][0] is at 0x16cc010
array[0][1] is at 0x16cc024
array[0][2] is at 0x16cc038
array[0][3] is at 0x16cc04c
array[0][4] is at 0x16cc060
array[1][0] is at 0x16cc04c
array[1][1] is at 0x16cc060
array[1][2] is at 0x16cc074
array[1][3] is at 0x16cc088
array[1][4] is at 0x16cc09c
array[2][0] is at 0x16cc088
array[2][1] is at 0x16cc09c
array[2][2] is at 0x16cc0b0
array[2][3] is at 0x16cc0c4
array[2][4] is at 0x16cc0d8

我真的不明白这里发生了什么。元素是连续的，但是对于整型来说它们太大了，而数组又太小了，容纳不了所有的元素。

我的数组声明是我最不确定的部分-我正在阅读int (*array)[rows][cols]作为"array是指向整数数组的指针(stride cols)"，但显然这是不正确的。

我到底做错了什么?

Answer 1

int (*array)[cols] = malloc( sizeof *array * rows );
for (r = 0; r <= rows - 1; r++) {
    for (c = 0; c <= cols - 1; c++) {
        printf("array[%i][%i] is at %pn", r, c, &array[r][c]);

这允许您将array索引为array[i][j]，因为array[i]隐式地解引用指针。

要解除分配，只需要调用

free( array );

Answer 2

代码应为

#include <stdlib.h>
#include <stdio.h>
int main(void)
{
    size_t rows = 3, cols = 5;
    printf("sizeof(int) = %linn", sizeof(int));
    int (*array)[rows][cols] = malloc(sizeof(int) * rows * cols);
    printf("array starts at %pn", array);
    printf("sizeof(array) = %lin", sizeof(*array));  // Note the *
    printf("sizeof(array[0][0]) = 0x%lxn", sizeof((*array)[0][0]));
    puts("");
    unsigned short r, c;
    for (r = 0; r <= rows - 1; r++) {
        for (c = 0; c <= cols - 1; c++) {
            printf("array[%i][%i] is at %pn", r, c, &((*array)[r][c]));
        };
        puts("");
    };
}

由于数组是指向数组[x][y]的指针，因此必须首先对其解引用

 (*array)

访问项

 (*array)[x][y]

这样就得到

sizeof(int) = 4
array starts at 0x12de010
sizeof(array) = 60
sizeof(array[0][0]) = 0x4
array[0][0] is at 0x12de010
array[0][1] is at 0x12de014
array[0][2] is at 0x12de018
array[0][3] is at 0x12de01c
array[0][4] is at 0x12de020
array[1][0] is at 0x12de024
array[1][1] is at 0x12de028
array[1][2] is at 0x12de02c
array[1][3] is at 0x12de030
array[1][4] is at 0x12de034
array[2][0] is at 0x12de038
array[2][1] is at 0x12de03c
array[2][2] is at 0x12de040
array[2][3] is at 0x12de044
array[2][4] is at 0x12de048

在预期的值范围内