使用指针表示法在C中访问多维数组 [英] Accessing multi-dimensional arrays in C using pointer notation

问题描述

让 a 是一个由10个整数组成的一维数组，如下所示

Let a be an one dimensional array of 10 integers as shown below

int a[10];

以及上面声明的数组的表达式

and for above declared array the expression

a[i]

评估为 *(a + i)，并给出存储在 i th 索引中的值，例如

is evaluated as *(a+i) and gives the value stored at the ith index,e.g.

int a[5]={1,2,3,4,5};

此处 a [3] 的值为 *(a + 3)，等于 4 ，即存储在数组 a 的 3rd 索引处的元素的值. 但是在这种二维数组的情况下

here a[3] is evaluated as *(a+3) which is equal to 4 ,i.e. the value of element stored at 3rd index of array a. But in case of a 2D array like this one

int b[2][3]={ {1,2,3},{4,5,6} };

(这里我将2D数组b视为1D数组的数组，即b是具有2个元素的1D数组，每个元素本身都是3个int的1D数组) 如果我们使用单个下标运算符，数组名称为

(here I am thinking of 2D array b as array of 1D arrays,i.e. b is a 1D array having 2 elements,each element itself being an 1D array of 3 int) if we use single subscript operator with array name as

b[i]

i = 1的上述表达式给出了存储元素4的地址，所以

the above expression for i=1 gives address where the element 4 is stored,so

1. 当b是2D数组时，如何评估表达式 b [i] ?
2. 当b是2D数组时，如何评估表达式 b [i] [j] ?
3. 如何使用指针访问C中多维数组的元素.在这里，我想了解编译器如何在内部处理多维数组?
4. 如果 b 是2D数组， b 会如何评估?

1. How is the expression b[i] evaluated in case when b is 2D array?
2. How is the expression b[i][j] evaluated in case when b is 2D array?
3. How to access elements of a Multi-dimensional Array in C using pointers. Here I would like to know about how compiler treats a Multi-dimensional Array internally ?
4. What does b evaluate to if b is a 2D array?

推荐答案

在这里，我将2D数组b视为1D数组

here i am thinking of 2D array b as array of 1D arrays

这是解决此问题的正确方法，因为确实如此. C标准实际上并未将多维数组指定为某些特殊情况，但是由于数组的类型与其他数组相同，因此它们是可能的.这样我们就可以拥有一个数组数组.

That's the correct way to approach this, because that's indeed what it is. The C standard does not actually specify multi-dimensional arrays as some special case, but they are rather possible because an array is a type like any other. So we can have an array of arrays.

1. 当b是2D数组时，如何计算b [i]表达式?
   ...

1. How is the expression b[i] evaluated in case when b is 2D array?
   ...

1. 如果b是2D数组，b会得出什么结果?

对于任何数组，b在表达式中使用时，都会衰减为指向第一个元素的指针.因此，b[i]与*(b+i)等效，对于任何其他数组表达式.

As for any array, b when used in an expression, decays into a pointer to the first element. b[i] is therefore equivalent to *(b+i), as for any other array expression.

在这种情况下，b会衰减为类型为int(*)[3]的数组指针.

In this case b decays into an array pointer of type int(*)[3].

1. 当b是2D数组时，表达式b [i] [j]的值如何计算?

• b在表达式中使用，因此在该表达式中它衰减为指向第一个元素的指针，该元素是指向第一个数组的数组指针.
• b[i]导致将指针算术应用于数组指针，等效于*(b+i).这给出了数组编号i.
• 在此位置，我们有一个类型为int[3]的一维数组.由于此数组是另一个数组的一部分，因此它本身没有标识符.但是为了说明起见，让我们假设它得到一个临时名称"tmp".然后，我们将得到表达式tmp[j]，它一如既往地衰减为*(tmp+j)，从而导致int.
• b is used in an expression, so in this expression it decays into a pointer to the first element, which is an array pointer to the first array.
• b[i] causes pointer arithmetic to get applied to the array pointer, equivalent to *(b+i). This gives array number i.
• At this position we have a 1D array of type int[3]. Since this array is part of another array, it has no identifier by itself. But for the sake of illustration, lets pretend it gets a temporary name "tmp". We would then have the expression tmp[j], which as always decays into *(tmp+j), resulting in an int.

基本上整个表达式都可以视为*(*(b+i) + j).

Essentially the whole expression can be treated as *(*(b+i) + j).

1. 如何在C语言中使用指针访问多维数组的元素.这里我想了解编译器如何在内部处理多维数组?

如上所述，它将其视为数组数组.例如，您可以使用数组指针遍历2D数组:

As explained above, it treats it as an array of arrays. For example, you can iterate over a 2D array by using array pointers:

#include <stdio.h>

void print_array (int array[3])
{
  printf("%d %d %d\n", array[0], array[1], array[2]);
}

int main (void)
{
  int b[2][3]={ {1,2,3},{4,5,6} };
  const size_t b_size = sizeof b / sizeof *b;

  for(int(*ptr)[3] = b; ptr < b+b_size; ptr++)
  {
    print_array(*ptr);
  }
}

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