Make learning pointer easier (2)

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Preface

In depth study of pointer （ Two ）

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One 、 Function pointer array

An array is a storage space for the same type of data , So we've learned about pointer arrays , You can read my last blog if you don't understand ; Make learning pointer easier （ One ）
such as ：int* arr[10]; // Each element of the array is int *
So we can understand the function pointer array , The first must be an array , Second, the array element type is function pointer （ Address ）; Then save the address of the function in an array , This array is called the function pointer array , How to define the array of function pointers ？
int (* parr1[10])();
int * parr2 [10] ();
int ( * )() parr3[10];
Look at these three , That's a function pointer array ？ The answer is ：parr1 ,parr1 The first and [] combination , explain parr1 It's an array , What is the content of the array ？ yes int (*)() Function pointer of type . Note that the array name should be placed in the band * In parentheses ,int ( *parr1[10])(); So the first one is the function pointer array ; Purpose of function pointer array ： Transfer table ;

1.1 The purpose of the function pointer （ The use of function pointers in the last blog ）

#include <stdio.h>
int add(int a, int b)
{
    
return a + b;
}
int sub(int a, int b)
{
    
return a - b;
}
int mul(int a, int b)
{
    
return a*b;
}
int div(int a, int b)
{
    
return a / b;
}
int main()
{
    
int x, y;
int input = 1;
  int ret = 0;
  do
 {
    
    printf( "*************************\n" );
    printf( " 1:add 2:sub \n" );
    printf( " 3:mul 4:div \n" );
    printf( "*************************\n" );
    printf( " Please select ：" );
    scanf( "%d", &input);
    switch (input)
   {
    
   case 1:
       printf( " Enter the operands ：" );
       scanf( "%d %d", &x, &y);
       ret = add(x, y);
       printf( "ret = %d\n", ret);
       break;
    case 2:
       printf( " Enter the operands ：" );
       scanf( "%d %d", &x, &y);
       ret = sub(x, y);
       printf( "ret = %d\n", ret);
       break;
    case 3:
       printf( " Enter the operands ：" );
       scanf( "%d %d", &x, &y);
       ret = mul(x, y);
       printf( "ret = %d\n", ret);
       break;
    case 4:
       printf( " Enter the operands ：" );
       scanf( "%d %d", &x, &y);
       ret = div(x, y);
       printf( "ret = %d\n", ret);
       break;
    case 0:
        printf(" Exit procedure \n");
breark;
    default:
       printf( " Wrong choice \n" );
       break;
   }
} while (input);
 
  return 0;
}

Select from above 1~4 There are a lot of duplicate or very similar code , If you add more functions to the computer , You have to add a lot of code , There are a lot of the same duplicate code , We can package it into a function to implement the function, so as to avoid writing the same code repeatedly ; So each of these functions is a function , How to integrate into one function to call all function functions ？

void computer(int (*cmp)(int x,int y)){
    
printf( " Enter the operands ：" );
       scanf( "%d %d", &x, &y);
       ret = cmp(x, y);
       printf( "ret = %d\n", ret);
}

I think this function can solve the problem , below switch Statement to call this function directly , Just transfer the function function , One thing to note is that the return value must match , Parameters are of the same type , Here are two int Parameters of type , The return value is int type , Such a function can be passed to perform operations ！！！ This is a function pointer ;

1.2 Purpose of function pointer array

Or the big code above , Use a function pointer array to apply , Use arrays for those function functions , See how the code implements

#include <stdio.h>
int add(int a, int b)
{
    
     return a + b;
}
int sub(int a, int b)
{
    
     return a - b;
}
int mul(int a, int b)
{
    
     return a*b;
}
int div(int a, int b)
{
    
     return a / b;
}
int main()
{
    
int x, y;
  int input = 1;
  int ret = 0;
  int(*p[5])(int x, int y) = {
     0, add, sub, mul, div }; // Transfer table 
  while (input)
  {
    
     printf( "*************************\n" );
     printf( " 1:add 2:sub \n" );
     printf( " 3:mul 4:div \n" );
     printf( "*************************\n" );
     printf( " Please select ：" );
   scanf( "%d", &input);
     if ((input <= 4 && input >= 1))
    {
    
     printf( " Enter the operands ：" );
       scanf( "%d %d", &x, &y);
       ret = (*p[input])(x, y);
    }
     else
       printf( " Incorrect input \n" );
     printf( "ret = %d\n", ret);
  }
   return 0;
}

Do not use the dereference when calling * Symbols are OK , direct p [input]) ( x , y ); This is a use of our function pointer array , Also called transfer table ;

Two , A pointer to an array of function pointers

This is the pointer , Then it saves the address of the function pointer array , This is just to understand , Know how to write , Just understand it ;

void test(const char* str)
{
printf(“%s\n”, str);
}
int main()
{
// A function pointer pfun
void ( * pfun )( const char * ) = test;
// An array of function pointers pfunArr
void ( * pfunArr[5] )( const char * str);
pfunArr[0] = test;
// Pointer to function array pfunArr The pointer to ppfunArr
void ( * ( * ppfunArr)[5] ) (const char * ) = &pfunArr;
return 0;
}

3、 ... and , Callback function

A callback function is a function called through a function pointer . If you put a pointer to a function （ Address ） Pass as a parameter to another
function , When this pointer is used to call the function it points to , Let's just say this is a callback function . The callback function is not controlled by this function
The implementer of directly calls , It's called by another party when a particular event or condition occurs , Used to enter the event or condition
Row response .
That is, we do not call this function directly , Instead, pass the function address , Call this function through the function pointer in another function , Would it be easier to understand ？ We qsort A library function is a typical callback function ;

#include <stdio.h>
//qosrt The user of the function has to implement a comparison function 
int int_cmp(const void * p1, const void * p2)
{
    
 return (*( int *)p1 - *(int *) p2);
}
int main()
{
    
  int arr[] = {
     1, 3, 5, 7, 9, 2, 4, 6, 8, 0 };
  int i = 0;
 
  qsort(arr, sizeof(arr) / sizeof(arr[0]), sizeof (int), int_cmp);
  for (i = 0; i< sizeof(arr) / sizeof(arr[0]); i++)
 {
    
   printf( "%d ", arr[i]);
 }
  printf("\n");
  return 0;
}

Let's simulate the implementation qsort Library function , I have written an article about simulation implementation before qsort Function blog can see ;
qsort Simulation and implementation of library functions

summary

There is also a long topic analysis blog on the interview questions , Then the pointer series blog will come to an end for the time being ;