c中的双指针和二维数组 [英] double pointers and 2d arrays in c

问题描述

我正在尝试使用双指针访问二维数组

int x[2][2] = {{10, 20},{30, 40}};国际*xp;国际**xpp;printf ("%d %d
%d %d
", x[0][0], x[0][1], x[1][0], x[1][1]);printf ("
");xp = *x;printf ("%d %d
%d %d
", *xp, *(xp + 1), *(xp + 2), *(xp + 3));printf ("
");xpp = (int**)x;printf ("%d
", **xpp);

我得到的是:

<块引用>

10 2030 4010 2030 40分段故障

问题:我应该如何使用 xpp 访问数组?

解决方案

而不是 ...

int x[2][2] = {{10, 20},{30, 40}};国际**xpp;xpp = (int**)x;

... 意识到表达式中的 x 将第一个元素的地址转换为指针.x的第一个元素是x[0]，它的类型是int [2]，所以需要的类型是int(*)[2] 或 指向int 数组 2.

int (*p)[2];p = x;printf ("%p
", (void *) p);printf ("%p
", (void *) *p);printf ("%d
", **p);printf ("%d %d %d %d
", p[0][0], p[0][1], p[1][0], p[1][1]);

输出

0xffffcbd0(示例)0xffffcbd0(样本)1010 20 30 40

<小时>

提示:避免强制转换 - 它通常隐藏弱编程.

I'm trying to access a 2D array using double pointer

int x[2][2] = {{10, 20},{30, 40}};
int *xp;
int **xpp;

printf ("%d  %d
%d  %d
", x[0][0], x[0][1], x[1][0], x[1][1]);
printf ("
");

xp = *x;
printf ("%d  %d
%d  %d
", *xp, *(xp + 1), *(xp + 2), *(xp + 3));
printf ("
");

xpp = (int**)x;
printf ("%d
", **xpp);

What I get is:

10  20
30  40

10  20
30  40

Segmentation fault

Question: How should I access the array using xpp?

解决方案

Rather than ...

int x[2][2] = {{10, 20},{30, 40}};
int **xpp;
xpp = (int**)x;

... realize that x in the expression converts to a pointer the address of the first element. The first element of x is x[0] which has a type of int [2], so the type needed is int (*)[2] or pointer to array 2 of int.

int (*p)[2];
p = x;

printf ("%p
", (void *) p);
printf ("%p
", (void *) *p);
printf ("%d
", **p);
printf ("%d %d %d %d
", p[0][0], p[0][1], p[1][0], p[1][1]);

Output

0xffffcbd0 (sample)
0xffffcbd0 (sample)
10
10 20 30 40

---

Tip: avoid casting - it often hides weak programming.

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