[advanced C language] ⑥ detailed explanation of function pointer

One 、 A function pointer

1. Concept

A function pointer ： First of all, it's a pointer , A pointer to a function , The address of the function is stored in the memory space ;

See the example ：

int main(){
    
int a = 10;
int*pa = &a;
char ch = 'c';
char* pc = &ch;
int arr[10] = {
    0};
int (*parr)[10] = &arr;// Take the address of the array 
return 0;
｝

analysis ：parr Is a pointer to an array , It stores the address of the array ;

therefore ：

• Array pointer — Pointer to the address of the array ;
• & Array name — What you get is the address of the array ;

So can we think so ：

• A function pointer — A pointer to the address of the function ;
• & Function name — What you get is the address of a function ;

Is that right ？ So let's test that out , Look at the following examples ：

int Add(int x,int y)
{
    
	return x+y;
 }
int main()
{
    
	printf("%p\n",&Add);// Print the function Add() The address of 
	printf("%p\n",Add);// The array name is equal to the address of the first element of the array , Is the function name equal to the function address ？
	return 0;
}

Please see the result ：

Oh ！ original , The function name is equal to the function address Of ！

1.2 How to use function pointers

Definition of function pointer ： Return value type of function （* Pointer name ）（ The argument list type of the function ）

int Add(int x, int y)
{
    
	return x+y;
}

int main()
{
    
 	int (*pf)(int, int) = &Add;// Function pointer definition , The return value type and parameter type are the same as the function Add（） identical 
}

1.3 Practice consolidating

void test(char* str){
    }
int main (){
    
①(?)pt =&test;
return 0;
}
 Excuse me, ① How should the sentence be perfect ？

answer ：void ( * pt)(char*) = &test;

How to use function pointers to call functions ？
Or the example above :

void Add(int x, int y){
    
return x+y;}
int main(){
    
int (*pf)(int,int)=&Add;
int ret=(*pf)(3,5);

analysis ：
int ret=(*pf)(3,5), At this point, it is equivalent to calling... Through the function name ： int ret=Add(3,5);}, We know ： The function name is equal to & Function name Of , therefore int (*pf)(int,int)=&Add, It can be changed to ：int (*pf)(int,int)=Add; here Add Equivalent to pf, therefore ：int ret=(*pf)(3,5); The statement can be changed to ：int ret=pf(3,5); Equivalent to int ret=Add(3,5), So we know that for ：int ret=(*pf)(3,5); The statement is ,* It doesn't make sense , Having one or more or none does not affect ;

1.4 To sum up

• Array name (arr) != & Array name (&arr)
• Function name (Add) = & Function name (&Add)

Two 、 Read two interesting pieces of code

notes ： originate 《c Pitfalls and pitfalls 》;

1.( *(void( *)( ))0 )( )

analysis ：
The meaning of this code is ：

1. call 0 The function at the address
2. This function has no parameters , The return value is void
3. Split ：
   ●void()() Indicates the function pointer type
   ●( void()() )0 Said to 0 Cast , hold 0 Cast the type to the address of a function ; Such as （int）3.14
   ●* ( void()() )0 Said to 0 The function at the address to understand the reference operation
   ●（ ( void(*)() )0)() It means calling 0 The function at the address
4. Look at the diagram ：
   

2.void (* signal(int,void( * )( int ) ) )(int)

analysis ：

1. signal and （） First combine , explain signal Is a function name
2. signal The type of the first parameter of the function is int, The second parameter is of type function pointer , The function pointer points to a parameter int, The return value is void Function of ;
3. signal The return type of a function is also a function pointer , The function pointer , Point to a parameter that is int, The return value is void function
   4. Look at the diagram ：
   

Sum up ,signal It's a function declaration ;