[use of pointer and pointer and array]

One 、 Pointer to the variable

Pointer variable is the variable that holds the address

​
​
int i;
int *p=&i;

​//* Is a unary operator used to access variables on the address represented by the value of the pointer 

​// It can be left-hand value or right-hand value , That is to say, it can be placed in = On the left of, let's assign values   It can also be placed in = On the right, let's read its value 

So let's take this i The address of is given to the pointer p,p What is kept is i The address of , At this time, we will say p Yes i.

The value of a normal variable is the actual value , The value of the pointer variable is the address of the variable with the actual value .

  Through the pointer, we access i

As a pointer to a parameter

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​
void f(int *p)
// Get the address of a variable when called 
int i;
int *p=&i;
f(p);
// You can access the outside through the pointer in the function i
// Access means reading or writing 
​

​

Let's try this thing

We still think that what happens is the transmission of values ,p（&i） This address value is passed into the function , So this is still the transmission of values , Because the address came in , Through this address, we can access the outside of the function in this way i Variable ,p yes i The address of , therefore *p It stands for i, In this way, we can modify this i, Do the above *p=123 This operation is actually right i do .

Pointer operator & *

The two interact

*&point->*(&point)->*(point The address of )-> Get the variable at that address ->point

&*point->&(*point)->&(point)-> obtain point The address of , That is to say point->point

Two 、 Pointer use

Pointer application scenario 1

Function returns multiple values , Some of them can only be returned by pointer

The parameter passed in is actually the variable that needs to save the returned result

Take a chestnut

​
#include<stdio.h>
void minmax(int a[],int len,int *min,int *max);//len Is the number of array elements 
int main()
{
    int a[]={1,2,3,4,5,6,7,8,9,12,18,55,88};
    int min,max;
    minmax(a,sizeof(a)/sizeof(a[0]),&min,&max);
    printf("min=%d,max=%d",min,max);
	return 0;
}
void minmax(int a[],int len,int *min,int *max)
{
	int i;
	*min=*max=a[0];
	for(i=1;i<len;i++)
	{
		if(*min>a[i])
		*min=a[i];
		if(*max<a[i])
		*max=a[i];
	}
	
}

​

  This is a very common scenario of pointer Application ： The result of my function is more than one , Then I pass in the address of the variable of the result I want to receive through the pointer , Let the function handle these variables for me , Pass the value back , Although these two parameters are from the main function main Passed in , But their function is to bring out the results

Pointer application scenario 2

Function returns the state of the operation , The result returns... Through the pointer

A common routine is to let the function return a special value that does not belong to the valid range to indicate an error ： for instance -1 or 0

But when any number is a valid possible result , We have to go back separately

Take a chestnut

#include<stdio.h>
int divide(int a,int b,int *result);// Define a division function  
int main()
{
	int a,b;
	scanf("%d%d",&a,&b);
	int c;
	if(divide(a,b,&c))
	{
		printf("%d/%d=%d\n",a,b,c);
	 } 
	return 0;

}
int divide(int a,int b,int *result)
{
	int flag=1;
	if(b==0)// Divisor is 0 Then division is meaningless  
	flag=0;
	else
	*result=a/b;
	return flag;
}

The result of division is transmitted through the pointer , Division succeeded flag return 1,

  If the divisor is 0 Then division is meaningless The function returns 0       if（divide（）） No operation

 

The most common error with pointers

That is, a pointer is defined and uninitialized I started using

int *p;
*p=12;// Dabao ！

This is the legendary wild pointer , There is no object ;

  3、 ... and 、 Pointers and arrays

If we pass an array to the function through the parameters of the function , So what does the number of incoming functions consist of ？

If we pass a common variable Then the function receives a value , If you pass a pointer Then the parameter receives a value , Only this value is the address .

How about passing an array ？

We continue to use the just minmax function

#include<stdio.h>
void minmax(int a[],int len,int *min,int *max);
int main()
{
    int a[]={1,2,3,4,5,6,7,8,9,12,18,55,88};
    int min,max;
    printf("in main sizeof(a)=%d\n",sizeof(a));// stay main Function sizeof（a） Value 
    minmax(a,sizeof(a)/sizeof(a[0]),&min,&max);
    printf("min=%d,max=%d",min,max);
	return 0;
}
void minmax(int a[],int len,int *min,int *max)
{
	int i;
	printf("in minmax sizeof(a)=%d\n",sizeof(a));// After passing in the function sizeof（a） Value 
	*min=*max=a[0];
	for(i=1;i<len;i++)
	{
		if(*min>a[i])
		*min=a[i];
		if(*max<a[i])
		*max=a[i];
	}
}

However, a subtle change has taken place

  As can be seen in the main Function and in minmax Function sizeof（a） There is a change

stay minmax Function sizeof（a） The value of is 4, Just the same size as a pointer

Let's try again

​
#include<stdio.h>
void minmax(int a[],int len,int *min,int *max);
int main()
{
    int a[]={1,2,3,4,5,6,7,8,9,12,18,55,88};
    int min,max;
    printf("in main sizeof(a)=%d\n",sizeof(a));


    printf("in main a=%p\n",a);// look here****


    minmax(a,sizeof(a)/sizeof(a[0]),&min,&max);
    printf("min=%d,max=%d",min,max);
	return 0;
}
void minmax(int a[],int len,int *min,int *max)
{
	int i;
	printf("in minmax sizeof(a)=%d\n",sizeof(a));


	printf("in minmax a=%p\n",a);//look here**** 


	*min=*max=a[0];
	for(i=1;i<len;i++)
	{
		if(*min>a[i])
		*min=a[i];
		if(*max<a[i])
		*max=a[i];
	}
}
​

This time we are looking at main Li He minmax in a The address of

  Everyone is exactly the same , The explanation is in minmax This one inside a Arrays are actually main Inside a Array   It's not exactly the same . They are the same

Let's continue the experiment

void minmax(int a[],int len,int *min,int *max)
{
		a[0]=10000;// stay minmax Function lining a[0]=10000;

  So in fact, in the function parameters a[] It's just a pointer , It looks like an array

So even in void minmax(int a[],int len,int *min,int *max) The parameter int a[] Change it to int *a The compilation can still be run correctly

So the array in the function parameter table , It's actually a pointer

// So the following two forms are equivalent ;
int minmax(int *a)
int minmax(int a[])

Array variables are special pointers

Array variables themselves express addresses

// therefore a The address of ==&a[0]
int a[10];
int *p=a;// Don't need to use & Address fetch 
// But the elements in the array represent a single variable   Need to use & Address fetch 
//[]y Operators can do... With arrays   You can also do 
int *p=a;
//p[0]<==>a[0]
//* Operator can also do... On an array 
*a=25;//a[0]==25