使用指针时将2D数组用作多个1D数组 [英] Using 2D arrays as multiple 1D arrays when using pointers

问题描述

我使用下面的程序通过指针访问2D数组.

I used the below program to access 2D arrays using pointers.

    #include<stdio.h>

void main()
{

  int num[3][2]={ {00,01},{10,11},{20,21} }; 
  int i,j;
  printf("-----------------------------------");
  /* Treating 2d array as 1d array of each row */
  for(i=0;i<3;i++)
  {
        printf("\nThe Address(&num[%d]) is %u \n \n<= WHICH IS THE SAME AS =>\n \n(num[%d]) %u \t",i,&num[i],i,num[i]);
        printf("[*(num+%d)] %u \t [(num+%d)] %u ",i,*(num+i),i,(num+i)); 
        printf("\n The Value is %d \n",*num[i]);
        printf("\n -----------------------------------");
  }
}

这是输出:

    -----------------------------------
The Address(&num[0]) is 2140353424 

<= WHICH IS THE SAME AS =>

(num[0]) 2140353424     [*(num+0)] 2140353424    [(num+0)] 2140353424 
 The Value is 0 

 -----------------------------------
The Address(&num[1]) is 2140353432 

<= WHICH IS THE SAME AS =>

(num[1]) 2140353432     [*(num+1)] 2140353432    [(num+1)] 2140353432 
 The Value is 10 

 -----------------------------------
The Address(&num[2]) is 2140353440 

<= WHICH IS THE SAME AS =>

(num[2]) 2140353440     [*(num+2)] 2140353440    [(num+2)] 2140353440 
 The Value is 20 

 -----------------------------------

我了解程序中会发生什么，并且我知道 *(num + i)用于访问每一行的地址.

I understood what happens in the program and I know that *(num+i) is used to access each row's address.

但是，为什么 *(num + i)和(num + i)都指向相同的地址?

But, why both *(num+i) and (num+i) point to the same address?

要访问该特定行(第i行)的值，我们使用 **(num + i)，这很有意义，因为 *(num + i)指向第一个1D数组的行的地址，我们可以使用另一个间接运算符取消引用该指针.

To access the value at that particular row (i'th row) we use **(num+i) which makes sense because *(num+i) points to the 1st 1D array's row's address and we can use another indirection operator to dereference that pointer.

但是 *(num + i)和(num + i)为何都指向相同的地址?这个编译器依赖吗?还是一些未定义的行为?

But how come both *(num+i) and (num+i) point to the same address? Is this compiler dependent? Or some undefined behavior?

请提供尽可能多的信息.

Please provide as much information as possible.

推荐答案

多维数组连续存储在内存中.这是数组的内存布局:

Multi-dimensional arrays are stored contiguously in memory. Here's a memory layout of your array:

 [0][0] [0][1] [1][0] [1][1] [2][0] [2][1]
+------+------+------+------+------+------+
|  00  |  01  |  10  |  11  |  20  |  21  |
+------+------+------+------+------+------+
|_____________|_____________|_____________|
^             ^             ^
num[0]        num[1]        num[2]

在我的机器上，int的大小为4个字节.
num 的类型为 int(*)[2] -它是2个整数数组的指针.语句 num + 1 是一种指针算法，导致向 num 数组开始的地址添加8个字节(即两个整数的大小).
取消引用后- *(num + 1)仍然是一个指针，它指向与 num + 1 相同的地址.您可以检查以下简短代码以查看发生了什么:

On my machine int has size of 4 bytes.
num has type int (*)[2] - it is a pointer to array of 2 ints. Statement num+1 is a pointer arithmetic and results in adding 8 bytes (that's the size of two ints) to the address where num array begins.
After dereference - *(num+1) it is still a pointer and it points to the same address as num+1. You can check this short code to see what is happening:

#include <stdio.h>

int main(void)
{
    int num[3][2]={ {00,01},{10,11},{20,21} }; 
    int (*fp)[2] = num+1;
    int* fpp = *(num+1);
    printf("%d\n", (void*)fpp == (void*)fp);
    return 0;
}

输出为1-那些指针拥有相同的内存地址.

The output is 1 - those pointers hold the same memory address.

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