[advanced C language] advanced pointer [Part 1]

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  series ：

 

 C Language programming questions

  Classic question series

List of articles

 

Catalog

List of articles

Preface

Character pointer

Pointer array

Array pointer

Definition of array pointer

& Array name VS Array name

The use of array pointers

Array parameter passing and pointer parameter passing

One dimensional array parameters

Two dimensional array parameters

First level pointer parameter transfer

The secondary pointer transmits parameters

Entrance to exercises

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Preface

This article explains the character pointer in the advanced level of pointer , Array pointer , Pointer array , Array parameter passing and pointer parameter passing , illustrated , So that you can understand and understand

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Tips ： The following is the main body of this article , The following cases can be used for reference

Character pointer

We have learned some knowledge about the primary pointer in the previous learning process , Know the concept of pointer ：

1. A pointer is a variable , It's used to store the address , The address uniquely identifies a piece of memory space .

2. The size of the pointer is fixed 4/8 Bytes （32 Bit platform /64 Bit platform ）.

3. Pointers are typed , The type of pointer determines the type of pointer +- Integer step size , The permission of pointer dereference operation .

4. The operation of the pointer .

Let's move on to the advanced topic of pointers

Among the pointer types, we know that one pointer type is character pointer char*

as follows ：

int main()
{
	char ch = 'w';
	char* pc = &ch;
	*pc = 'b';
	printf("%c\n", ch);
	return 0;
}

  Like the above, a character is put into a pointer , You have seen another kind of

as follows ：

The following is the abcdef The first element of this string a The address of is assigned to p in , Not the whole string p, And the first element address is printed , By the way, I printed out the last string

int main()
{
	char* p = "abcdef";
	// Put the first element of the string a The address of is assigned to p in 
	printf("%s\n", p);
	return 0;
}

  Next, let's look at an interview question to solve our problems , By the way, understand the above ：

int main()
{
	const char* p1 = "abcdef";
	const char* p2 = "abcdef";
	char arr1[] = "abcdef";
	char arr2[] = "abcdef";
	if (p1 == p2)
		printf("p1==p2\n");
	else
		printf("p1!=p2\n");
	if (arr1 == arr2)
		printf("arr1 == arr2\n");
	else
		printf("arr1 != arr2\n");
	return 0;
}

 

  Why is this the case ？

p1 And p2 Are character pointers , and const Modifiers cannot change the pointer , and “abcdef” Is a constant string , Put the first element address a Give it to p1 And p2, Since it's a constant string , Can't modify , Only read , It seems useless to store multiple copies in memory , So there is a space , and p1 And p2 It points to the same space , So the printing is equal

arr1 Is an array , Initialization will open up a space ,arr2 It's also an array , Initialize to open up a space , therefore arr1 And arr2 It's not equal

Pointer array

Pointer arrays are arrays , Is an array used to store pointers

int main()
{
	int arr[10]; // integer array , Store plastic 
	char arr1[10];// A character array , To store characters 
	int* arr2[10];// Integer pointer array , Store the integer pointer 
	char* arr3[10];// Character pointer array , Storing character pointers 
	return 0;
}

  The name of the array is equivalent to the address of the first element , So can we put the array in the array , Accept with a pointer , This is just a pointer array ？ The following code ：

int main()
{
	int arr1[] = { 1,2,3,4,5 };
	int arr2[] = { 2,3,4,5,6 };
	int arr3[] = { 3,4,5,6,7 };
	int* parr[3] = { arr1,arr2,arr3 };
	return 0;
}

  Drawing interpretation ：

  In fact, we found that we simulated a two-dimensional array , We can also take it out , Simulate a two-dimensional array to get the value of a one-dimensional array

int main()
{
	int arr1[] = { 1,2,3,4,5 };
	int arr2[] = { 2,3,4,5,6 };
	int arr3[] = { 3,4,5,6,7 };
	int* parr[3] = { arr1,arr2,arr3 };
	int i = 0;
	for (i = 0; i < 3; i++)
	{
		int j = 0;
		for (j = 0; j < 5; j++)
		{
			// Because we knew before ,*(p+i) -- p[i]
			printf("%d ", *(parr + i) + j);
			//printf("%d ", *(parr[i] + j));
			//printf("%d ", parr[i][j]);
			// So the latter ones are actually equivalent 
		}
	}
	return 0;
}

Array pointer

Definition of array pointer

Is the array pointer a pointer or an array ？

The answer is the pointer

Integer pointer ： Pointer to shaping data

Character pointer ： Pointer to character data

Array pointer ： Pointer to array data

Which of the following is an array pointer ？

int *p1[10];
int (*p2)[10]; 

  explain ：

int *p1[10];

p1 The first and 【】 combination , Priority application , Represents the p1 Is an array , The type is int* Of , It's an array of Pointers

int (*p2)[10];

p2 The first and * The union representation is a pointer , because （） High priority , And then store 10 Elements , The type of each element is int type , Represents an array pointer that holds an array

/ Pay attention here ：[] Priority is higher than * The no. , So we have to add （） To guarantee p The first and * combination .

& Array name VS Array name

Let's talk about array names again ：

For the following array ：

int   arr【10】;

there arr And &arr What's the difference? , What are they ？

We know arr It's an array name , The array name indicates the address of the first element of the array , that &arr What is the array name ？

Let's take a look at a piece of code ：

int main()
{
	int arr[10] = { 0 };
	printf("%p\n", arr);
	printf("%p\n", &arr[0]);
    printf("%p\n", &arr);
	return 0;
}

  The operation results are as follows ：

  We see the array name and & The array name and the address of the first element of the array are exactly the same , Are these three the same ？

The array name usually represents the address of the first element of the array

But there are 2 Exceptions ：

1,sizeof（ Array name ）, The array name here represents the entire array , It calculates the size of the entire array

2,& Array name , The array name here still represents the whole array , therefore & The array name takes out the address of the whole array

Let's take another look at the code ：

int main()
{
	int arr[10] = { 0 };
	printf("%p\n", arr);
	printf("%p\n", arr+1);

	printf("%p\n", &arr[0]);
	printf("%p\n", &arr[0]+1);

	printf("%p\n", &arr);
	printf("%p\n", &arr+1);

	return 0;
}

  The analysis results are as follows ：

  According to the above code, we find that , Actually &arr and arr, Although the values are the same , But the meaning should be different .

actually ：

&arr It means Address of array , Instead of the address of the first element of the array .

In this case &arr The type is ： int(*)[10] , Is an array pointer type

Address of array +1, Skip the size of the entire array , therefore &arr+1 be relative to &arr The difference is 40.

As shown in the following figure ：

int main()
{
	int arr[10] = { 0 };
	int* p = arr;
	int(*p)[10] = &arr;
	//  p With the first * Union is a pointer , Then there is an array of ten elements , The type is int, Deposit is arr Address of array 
	// The integer pointer is used to store the address of the integer 
	// A character pointer is an address used to store characters 
	// Array pointer is the address used to store the array 
	return 0;
}

The use of array pointers

So how exactly are array pointers used ？

Since the array pointer points to an array of numbers , Then the address of the array should be stored in the array pointer

Remember not to confuse it with a second level pointer , The second level pointer is the address where the first level pointer variable is stored

int main()
{
	char* arr[5] = { 0 };
	char* (*pc)[5] = &arr;// Array pointer 
	char ch = 'w';
	char* p1 = &ch;  // First level pointer 
	char** ph = &p1;// The secondary pointer 
	return 0;
}

  The use of an array pointer is as follows ：

#include <stdio.h>
void print_arr1(int arr[3][5], int row, int col) {
	int i = 0;
	for (i = 0; i < row; i++)
	{
		int j = 0;
		for (j = 0; j < col; j++)
		{
			printf("%d ", arr[i][j]);
		}
		printf("\n");
	}
}
void print_arr2(int(*arr)[5], int row, int col) {
	int i = 0;
	for (i = 0; i < row; i++)
	{
		int j = 0;
		for (j = 0; j < col; j++)
		{
			printf("%d ", arr[i][j]);
		}
		printf("\n");
	}
}
int main()
{
	int arr[3][5] = { 1,2,3,4,5,2,3,4,5,6,3,4,5,6,7};
	print_arr1(arr, 3, 5);
	// Array name arr, Represents the address of the first element 
	// But the first element of a two-dimensional array is the first row of a two-dimensional array 
	// So the message here is arr, It's actually equivalent to the address on the first line , Is the address of a one-dimensional array 
	// You can use an array pointer to receive 
	print_arr2(arr, 3, 5);
	return 0;
}

After learning pointer array and array pointer, let's review the meaning together and see the meaning of the following code ：

int main()
{
	int arr[5];
	int* parr1[10];
	int(*parr2)[10];
	int(*parr3[10])[5];
	return 0;
}

 int arr[5];

arr Is an integer array , Yes 5 Elements , The type is int
 int* parr1[10];

parr1 Is an array of integer pointers , Yes 10 Elements , Each type is int*
 int(*parr2)[10];

parr2 Is an integer array pointer , With the first * Combination is a pointer , Then there is 10 Elements , Each type is int
 int(*parr3[10])[5];

parr3 Is an array that holds array pointers ,parr3 With the first 【】 The combination is an array , And then the rest int *  【5】 It's the type , It is an array of pointers , So it's an array that holds array pointers , Array pointer array for short
 

Array parameter passing and pointer parameter passing

When writing code, it's hard to avoid 【 Array 】 perhaps 【 The pointer 】 Pass to function , So how to design the parameters of the function ？

One dimensional array parameters

Code ：

#include <stdio.h>
void test(int arr[])//ok?
{}
void test(int arr[10])//ok?
{}
void test(int* arr)//ok?
{}
void test2(int* arr[20])//ok?
{}
void test2(int** arr)//ok?
{}
int main()
{
	int arr[10] = { 0 };
	int* arr2[20] = { 0 };
	test(arr);
	test2(arr2);
}

Drawing analysis ：

Two dimensional array parameters

Code ：

void test(int arr[3][5])//ok？
{}
void test(int arr[][])//ok？
{}
void test(int arr[][5])//ok？
{}
// summary ： Two dimensional array parameters , Function parameter design can only omit the first [] The number of .
// Because for a two-dimensional array , I don't know how many lines there are , But you have to know how many elements in a row .
// So it's easy to calculate .
void test(int *arr)//ok？
{}
void test(int* arr[5])//ok？
{}
void test(int (*arr)[5])//ok？
{}
void test(int **arr)//ok？
{}
int main()
{
 int arr[3][5] = {0};
 test(arr);
}

Drawing analysis ：

 

First level pointer parameter transfer

Code ：

#include <stdio.h>
void print(int *p, int sz) {
 int i = 0;
 for(i=0; i<sz; i++)
 {
 printf("%d\n", *(p+i));
 }
}
int main()
{
 int arr[10] = {1,2,3,4,5,6,7,8,9};
 int *p = arr;
 int sz = sizeof(arr)/sizeof(arr[0]);
 // First level pointer p, Pass to function 
 print(p, sz);
 return 0; }

Drawing analysis ：

  Think about a problem ： Push backwards , If the parameter part of the function has been fixed, it is a first-order pointer , So what can be used to transmit parameters

Such as void  print（int* p）

{}

If int  a   = 10;    

print（&a）; You can pass it on , Because it's the address , Receive... With a pointer

If int* ptr  = &a;

print（ptr）; You can pass the pointer to receive the pointer

If int  arr【10】

print（arr）; You can pass it on ,arr Represents the address of the first element of the array , Pointer reception

The secondary pointer transmits parameters

Code ：

#include <stdio.h>
void test(int** ptr) 
{
 printf("num = %d\n", **ptr); 
}
int main()
{
 int n = 10;
 int*p = &n;
 int **pp = &p;
 test(pp);
 test(&p);
 return 0; 
}

Drawing analysis ：

  Think about a problem ： Push backwards , If the parameter part of the function has been fixed, it is a second-order pointer , So what can be used to transmit parameters

such as ：test（int** p）

{}

If int *p1;

test（&p1）; The address of the first level pointer and the second level pointer can be used to receive

If int**p2;

test（p2）; The second level pointer can be passed to the second level pointer to receive

If int* arr【10】;

test（arr）;

The pointer array name is the address of the first element , Each element is int* type , The second level pointer reception is OK

Entrance to exercises

After reading these specific operations, it is not allowed , You can click on the top to practice some exercises , You can also have a casual look C Some language exercises , Practice multiple-choice questions and programming questions , Let your knowledge be consolidated , Click directly into the title to go directly to , In addition, if you want to get the qualification of internal promotion of large factories, you can also go and have a look ：

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