Accessing a two-dimensional array with a pointer

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Preface

How to understand two-dimensional arrays ？ We understand two-dimensional arrays in a two-dimensional way , It is one-dimensional in storage .

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Access a binary array with a pointer

example ：int a[3] [4] = {1,3,5,7,9,11,13,15,17,19,21,23};

a[0] [0]	a[0] [1]	a[0] [2]	a[0] [3]
a[1] [0]	a[1] [1]	a[1] [2]	a[1] [3]
a[2] [0]	a[2] [1]	a[2] [2]	a[2] [3]

We imagine the storage mode of two-dimensional array ：3 That's ok 4 Column

a[0] [0]
a[0] [1]
a[0] [2]
a[0] [3]
a[1] [0]
a[1] [1]
a[1] [2]
a[1] [3]
a[2] [0]
a[2] [1]
a[2] [2]
a[2] [3]

We think the first four are one with a[0] A one bit array of array names , There are four int Elements

The middle four are named a[1] One dimensional array of

The last four are named a[2] One dimensional array of

Access different forms of two-dimensional arrays through pointers

*a and a It's two completely different concepts ：

a Look at this array and think that this array is a one-dimensional array , It adds itself for a while and turns to the next one-dimensional array, that is a+1

*a It refers to a one-dimensional array in which each element is an integer, so it is considered *a+0 It points to an integer , Plus one is the next integer

Why can't you use *p=a

int x,*p;
double y,*q;
p=&x;
q=&y;
p=q;// It's wrong. 
//p It refers to the address with four bytes as the storage unit 
//q It refers to the address with eight bytes as the storage unit 
// The two cannot be equivalent 

There are four ways to access two-dimensional array elements ：

1、 Access array elements by subscript

int a[3][4];
for (int i=0; i<3; i++)
    for (int j=0; j<4; j++)
        a[i][j] = i+j;

int a[3][4];
for (int i=0; i<3; i++)
    for (int j=0; j<4 ;j++)
        *(*(a+i)+j)=i+j;

2、 Access array elements through the first address of the array

For two-dimensional array elements a[i] [j], “【】” It's actually an index operator , Upcoming element a[i] [j] The storage address of is converted to a[i] + j.

3、 Access array elements through pointers （ From the perspective of a one-dimensional array “ To view ” Two dimensional array ”）

int a[3][4];
int *p=&a[0][0];
for (int i=0; i<3; i++)
    for(int j=0; j<4; j++)
        *(p++) = i+j;		// amount to *p=i+j;p++;

ad locum , The pointer p Is defined as ：“ Point to int The pointer to ”, in other words ：p The pointer “ Think ” It points to a one-dimensional array , Every element is int Type , share 12 Elements .

4、 Accessing arrays through pointers （ From the perspective of a two-dimensional array “ To view ” Array ）

int a[3][4]={
    1,3,5,7,9,11,13,15,17,19,21,23};
int (*p)[4];		
/*  Yes p Understanding of pointer ： *p The statement p The pointer to . p What pointer is it pointing to ？  Points to a containing 4 One dimensional array of integers  */
p=a;
for(int i=0; i<3;i++)
	for (int j=0;j < 4;j++)
        *(*(p+i)+j)=i+j;

After the pointer is defined like this ,== The pointer p And array a Exactly the same “ visual angle ” Look at two-dimensional arrays .== in other words ,a How can I use ,p It can also be used in the same way

// for example ： here *(*(p+i)+j) Replace with p[i][j],a[i][j],*(*(a+i)+j) They're all equivalent 

Point to a two-dimensional array of strings

char *str[3]={
    "Red","Green","Blue"};// Square brackets take precedence over "*" priority 
// intend str Is an array with three elements , The array stores pointers to characters 

example ： Student achievement

score The array holds 3 A student 4 Your grades

By calling average Function to calculate the average of all students' grades

By calling search The function displays the n A student 4 Grades in this subject .

Ideas ：

You need to define a two-dimensional array score, Each row stores a student's 4 Your grades . Yes 3 That's ok , It stores 3 Results of students .

65	70	70	60
80	87	90	81
90	99	100	98

When writing functions average When calculating the average of all students' grades , It can be used One dimensional array From the perspective of score Array , in other words ,“ Think ” This array has 12 Achievements , Direct summation , Just calculate the average score

When writing functions search According to the first n Of a student 4 When you learn a subject , Then use Two dimensional array From the perspective of score Array , in other words ,“ Think ” This array has 3 That's ok , Each row 4 Elements

void average(float *p,int n)
{
    
    // Look at students' grades in the way of one-dimensional array 
    float *p_end;
	float sum=0,aver;
    p_end=p+n;
    for ( ; p<p_end;p++)
        sum=sum+(*p);
    aver=sum/n;
    printf(" The average score is ：%5.1f\n",aver);
}

void search(float (*p)[4],int n)// It can be written.  float p[][4]
{
    
    // Look at students' grades in a two-dimensional array , Perspective and score identical 
    int i;
    printf(" The first %d A student's grade is : ",n);
    for (i=0;i<4;i++)
        printf("%5.1f",*(*(p+n)+i));  // Can be substituted with p[n][i]
}

int main()
{
    
    float score[3][4]={
    {
    65,67,70,60},{
    80,87,90,81},{
    90,99,100,98}};
    average(&score[0][0],12);		// Can be substituted with score[0],*(score+0)
    search(score,2);
    return 0;
}

reflection （ If you don't understand , Please step back and look ahead ）

(1) It is also passed to the function score Two dimensional array , Why? average Functions and search Different methods are used to transfer parameters of functions ？

average(&score[0][0],12);		// Can be substituted with score[0],*(score+0)
search(score,2);				// Show the second student's grades 

Two dimensional arrays as arguments to functions

1、 From the perspective of a one-dimensional array “ To view ” Two dimensional array

2、 From the perspective of a two-dimensional array “ To view ” One dimensional array

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Practice after class

Title Description ： Write a inverse function , The function of this function is ： Will a 3*4 The values in the two-dimensional array of are stored in reverse order .

Example of program running results

Please enter 3*4 Two dimensional array of ：

1 2 3 4

5 6 7 8

9 10 11 12

The result of storage in reverse order is ：

12 11 10 9

8 7 6 5

4 3 2 1

Tips ：

There are many ways to realize this program , Try two ways to achieve

answer

Method 1 ： Look at the program in a two-dimensional array ：

void inverse(int a[][N],int b[][N])
{
    
    int i,j;
    for (i=0;i<M;i++)
        for (j=0;j<N;j++)
            b[M-i-1][N-j-1]=a[i][j]
}

int main()
{
    
    int a[M][N],b[M][N];
    int i,j;
    printf(" Please enter  %d*%d  Two dimensional array of ：\n",M,N);
    for (i=0;i<M;i++)
        for (j=0;j<N;j++)
            scanf("%d",&a[i][j]);
    inverse(a,b);
    printf(" The result of storage in reverse order is ：\n");
    for (i=0;i<M;i++)
    {
    
        for (j=0;j<N;j++)
            printf("%d\t",b[i][j]);
    }
}

Method 2 ： Look at the program in the way of one-dimensional array ：

void swap(int *p1,int *p2)
{
    
    int tmp = *p1;
    *p1 = *p2;
    *p2 = tmp;
}

// With  int *a  It's a way to pass the reference , Make the array look at in one dimension in the subroutine ,
// in other words , Think of an array as having  M*N  An element of , Every element is  int  Array of 
void inverse(int *a,int n)
{
    
    int *p,*q;
    for (p=a,q=a+n-1;p<=q;p++,q--)
    {
    
        swap(p,q);
    }
}

int main()
{
    
    int a[M][N];
    int i,j;
    printf(" Please enter  %d * %d  Two dimensional array of ： \n");
    for (i=0;i<M;i++)
        for (j=0;j<M;j++)
            scanf("%d",&a[i][j]);
    inverse(&a[0][0],M*N);
    printf(" The result of storage in reverse order is ： \n");
    for (i=0;i<M;i++)
    {
    
        for (j=0;j<N;j++)
            printf("%d\t",a[i][j]);
    	printf("\n");
    }
    return 0;
}

The program uses M,N Two constants , Such a program has good flexibility .