Array as function parameter -- pointer constant / constant pointer

It was always thought that array names were pointers , It's not . Only when the array name is used as a function parameter can it degenerate into a pointer of the same type with the same name , Pointer as a function parameter is equivalent to array name .（ Be careful , Equivalence just means the same effect , Not equivalent , You can't think of an array name as a pointer , The pointer is the array name .）

The following procedure ： There is a length of 4 Of float The array holds four numbers , This array is used as an argument to a function that calculates the average .

#include <stdio.h>
#include <string.h>

float Avarage(float array[4])
{
	int i;
	float aver;
	float sum = array[0];

	for ( i = 1; i <= 3; i++)
	{
		sum = sum + array[i];
	}
	
	aver = sum /4;

	return aver;
}

int main(void)
{
	float score[4] = { 10.5, 10.5, 10.5, 10.5};

	
	printf("%.2f\n",Avarage(score));// The parameter here is if 
// It's written in score[4] Will make mistakes , The reason for the mistake is Average The parameter of this function should be a float Pointer to type 
// If the argument is score[4],float The actual participation of type float* Type has incompatible formal parameters .
// It's written in score[] It's not right either , Only score Yes. .

	return 0;
}

In defining Average（） Function , If the function parameter is an array , It can be defined in this way ,Average（） The formal parameters of this function are float An array of types , In fact, this formal parameter array is reduced to pointers in the next function body , therefore Average（float arrar[10] or float array[]） It's actually Average（float* array） ！（ When a function is defined , In fact, there is no need to specify the size of the shape parameter group , Because the allocation of specific memory space is not involved at this time .）

The following definitions are the same , In the following function body ,array This array name is actually a pointer variable with the same name float* array, and float* array The value of this pointer variable is float array[0] The address of .

#include <stdio.h>
#include <string.h>

float Avarage(float* array)// Notice here .
{
	int i;
	float aver;
	float sum = array[0];

	for ( i = 1; i <= 3; i++)
	{
		sum = sum + array[i];
	}
	
	aver = sum /4;


		return aver;
}

int main(void)
{
	float score[10] = { 10.5, 10.5, 10.5, 10.5};

	
	printf("%.2f\n",Avarage(score));
/* You need to enter arguments when calling this function , Actually, it should be float* Variable of type , For example, a float The number of arrays ！ Group ！ name ！, This argument array name is actually equivalent to a pointer */

	return 0;
}

http://blog.sina.com.cn/s/blog_7f69fbf90102whhe.html

Array as a function parameter , Do you want to allocate memory space for the array name when calling the function

 There is no need to .

 The exact understanding of an array name is a pointer , And it's a  Constant pointer , Fixed to the first element of the array it represents .

 When both arguments and formal parameters are arrays , When a call occurs, pass the constant address in the actual parameter array name to the shape parameter group name , That is to say, the shape parameter group name saves the address stored in the actual parameter array name . So inside the called function , The formal parameter array points to the same array as the argument array name .

 Let's put it this way. ： Although the formal parameter is an array , But in fact, the only parameter passed is the pointer constant of argument array name （ Pass this pointer to the shape parameter group name ）. Of course, the formal parameter array name takes up memory space , But there is no need for （ Neither ） You allocate space to the called function （ This is done automatically by the compiler ）. If you want to compile and call correctly , Just make sure that your argument array and formal parameter array are of the same type . That is to say , When you define the function header , There is no need to specify the size of the shape parameter group （ There is no appointment , Because the argument only passes a pointer to the name of the array ）