带有二维数组的指针 [英] Pointers with two dimensional array

问题描述

请考虑以下代码

#include <stdio.h>
#define ROW_SIZE 2
#define COL_SIZE 2

int main()
{
   int a[ROW_SIZE][COL_SIZE]={{1,2},{3,4}};

   // Base address:Pointer to the first element a 1D array
   printf("Base address of array:%p\n",a);

   //The value at the base address: should be the address of 1st 1D array
   printf("Value at the Base address:%p\n",*a);

   return 0;
}

获得的输出:

Sample Output:
Base address of array:0xbff77434
Value at the Base address:0xbff77434

不知何故，我无法理解二维数组基地址的概念和基地址处的值，而基地址又是一维数组的相同地址.请解释.

推荐答案

数组不是指针，在 C 中，多维数组只是数组的数组.在许多情况下，使用数组的名称衰减"为指向该数组第一个元素的指针.这就是您的两个打印语句中发生的情况.第一种情况:

Arrays aren't pointers, and in C, a multidimensional array is just an array of arrays. In many contexts, using the name of an array "decays" into a pointer to the first element of that array. That's what happens in both of your print statements. In the first case:

printf("Base address of array:%p\n",a);

a 成为指向数组第一个元素的指针 - 即指向数组第一行的指针.在你的情况下，这意味着你得到一个 int (*)[2] 类型的指针.

a becomes a pointer to the first element of the array - that is, a pointer to the first row of your array. In your case, that means you get a pointer of type int (*)[2].

第二种情况:

printf("Value at the Base address:%p\n",*a);

同样的衰减发生了，但随后你取消了该指针的引用.这意味着您取消引用 int (*)[2] 指向第一行的指针，再次为您留下数组(第一行).该数组本身衰减为指向它的第一个元素的指针，为您提供一个结果int *指针(指向第一行的第一个元素).

The same decaying happens, but then you dereference that pointer. That means you dereferenced that int (*)[2] pointer to the first row, leaving you with an array again (the first row). That array itself decays into a pointer to its first element, giving you a resulting int * pointer (to the first element of the first row).

在这两种情况下，地址是相同的，因为这就是数组在内存中的布局方式.如果我们说你的二维数组从地址 0 开始，它看起来像这样(假设是一个 4 字节的 int 类型):

In both cases the address is the same, since that's how the array is laid out in memory. If we said your 2D array started at address 0, it would look like this (assuming a 4 byte int type):

 Address       Value
    0            1
    4            2
    8            3
   12            4

第一行的地址和第一行的第一个元素的地址都是0.

The address of the first row and the address of the first element of the first row are both 0.

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