为什么在2D数组中a和* a指向相同的地址? [英] Why in a 2D array a and *a point to same address?

问题描述

我只是想学习2D数组的实现方式以及内存分配的方式.所以我在给定的c程序中有些疑问，为什么a和* a给出相同的地址.

I am just trying to learn how 2D arrays implemented and how memory allocation takes place. so I get some doubt in the given c program that why a and *a giving same address.

#include<stdio.h>
main()
{
    int i,j;
    int a[3][3]={1,2,3,4,5,6,7,8,9};


    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++)
        {
            printf("%d\t",*(*(a+i)+j));
        }
        printf("\n");
    }


    printf("%d\n",a);
    printf("%d\n",a[0]+1);
    printf("%d\n",a[0][0]);
    printf("%d\n",a+1);
    printf("%d\n",*a);
}

这是输出

推荐答案

一种了解2D数组的直观方法就是这样...

A visual way to understand the 2D array is like this...

a --> a[0] --> [1,2,3]
      a[1] --> [4,5,6]
      a[2] --> [7,8,9]

a [0]，a [1]，a [2]是逻辑表示，而不是实际的内存.

a[0], a[1], a[2] are logical representation and not actual memory.

so a[0] = a + 0*[bytes occupied by one row], 
   a[1] = a + 1*[bytes occupied by one row], 
   a[2] = a + 2*[bytes occupied by one row] 

因此*a => a[0] => a

请注意，即使地址相同，当我们调用a与a[0]或*a时，指针的类型"也会更改.通过(a + 1)和(a [0] +1)的打印输出可以很容易地看到这一点.

Note that even though the address is same the "type" of pointer changes when we invoke a vs a[0] or *a. This can be easily seen by your print output of (a+1) and (a[0]+1).

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