为什么在将指针分配给2D数组时需要指定行? [英] Why does one need to specify the row while assigning a pointer to a 2D Array?

问题描述

当未提及2D数组的行时，编译器会指出来自不兼容指针类型的赋值"，我一直认为没有括号的数组表示第一个元素的地址，在这种情况下，元素twodstring [0]的地址[0]

The compiler states "assignment from incompatible pointer type" when the row of the 2D array is not mentioned, I always thought an array without brackets means the address of the first element, in this case address of the element twodstring[0][0]

在提到该行时，编译器未声明错误，我想知道为什么会这样吗?

Compiler does not state an error when the row is mentioned, I was wondering why is this the case?

#include<stdio.h>

int main()
{

  char onedstring[]={"1D Array"};
  char twodstring[][5]={"2D","Array"};
  char *p1,*p2;

  p1=onedstring;
  p2=twodstring;
  p2=twodstring[1];

}

推荐答案

二维数组

char a[M][N];

可以使用typedef通过以下方式声明

can be declared using a typedef the following way

typedef char T[N];

T a[M];

因此可以像这样声明指向数组a的第一个元素的指针

So a pointer to the first element of the array a can be declared like

T *p = a;

其中，T是类型char[N]的别名.现在进行反向替换，我们可以写

where T is an alias for the type char[N]. Now making the reverse substitution we can write

char ( *p )[N] = a;

也就是说，二维数组的元素是一维数组.

That is elements of a two-dimensional array are one-dimensional arrays.

此声明

char ( *p )[N] = a;

等同于

char ( *p )[N] = &a[0];

其中a[0]具有类型char[N].因此，指针指向数组的第一个行".

where a[0] has the type char[N]. So the pointer points to the first "row" of the array.

取消引用指针，您将获得类型为char[N]的对象.

Dereferencing the pointer you will get an object of the type char[N].

请注意，可以像这样声明二维数组

Pay attention to that a two-dimensional array can be declared like

char ( a[M] )[N];

因此，将一维数组声明符a[M]替换为指针，您将得到

So substituting the one-dimensional array declarator a[M] for pointer you will get

char ( a[M] )[N];
char (  *p  )[N] = a;

如果您要声明这样的指针

If you will declare a pointer like this

char *p1;

那么你可以写个例子

p1 = a[1];

该表达式中的

a[1]是类型为char[N]的一维数组.使用该表达式作为初始化程序，该数组将转换为指向其第一个类型为char *的元素的指针.

in this expression a[1] is a one-dimensional array of the type char[N]. Using the expression as an initializer the array is converted to pointer to its first element that has the type char *.

所以这个表达式语句

p1 = a[1];

等同于

p1 = &a[1][0];

取消引用此指针，您将获得类型为char的对象.

Dereferencing this pointer you will get an object of the type char.

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