Const and the secret of pointers

Preface

The content of the article is read by 《C Experts programming 》 Sort it out . I hope it can help you solve the problem of pointer assignment and const Problems in , I also hope you can correct the mistakes in the article , Common progress .

Assignment of pointer

problem

Set a type to char** The value of is assigned to a const char** Is the object of type legal ？

Let's start with the results , stay vs Under the environment of , The compiler will not report any errors or warnings .
But in linux Environment with gcc The following warning will appear when compiling ：
warning: assignment from incompatible pointer type
Warning : Assignments from incompatible pointer types

For the portability of the code, we obviously can't write such code , Now let's explore the mystery step by step .
First, let's understand ANSI C Relevant standards

ANSI C Criteria for simple assignment

To make the assignment form legal , One of the following conditions must be met ：

1. Both operands are pointers to compatible types of finite or infinite qualifiers
2. The type pointed to by the left pointer must have all qualifiers of the type pointed to by the right pointer

There is also a description of the type ：
const float* Type is not a finite determiner type —— Its type is “ Point to a person with const Qualifier float Pointer to type ”, in other words const Qualifiers are used to decorate the type pointed to by the pointer , Not the pointer itself .

Problem solving

Before solving the problem , Let's first look at a group of simple .

char* and const char*

char* and const char* It is a match. . The reason why it is legal , Because in the following code ：

char* cp;
const char* cpp;
cpp = cp; 

• The left operand is a pointer to const Qualifier char The pointer to ;
• The right operand is a pointer to a non qualified char The pointer to ;
• char The type and char Types are compatible , The type pointed to by the left operand has the qualifier of the type pointed to by the right operand （ nothing ）, Plus its own qualifier （const）.

Be careful , On the contrary, the assignment cannot be made .

char* cp;
const char* cpp;
cp = cpp; // At this time, the left operand does not have the right operand const qualifiers 

char** and const char**

From the above knowledge, we can know ,char** and const char** Are pointer types without qualifiers , But they point to different types （ The former points to char*, The latter points to const char*）, This violates the first rule of the assignment criteria above , So they are incompatible .

It's a little difficult to understand this in this way . It can be understood in the following way ：

char** pp1;
const char** pp2;
pp2 = pp1;

• The type of the left operand is const char**, It's a point const char* Pointer to type , and const char* Is a pointer without qualifier , It points to a with const Limited char type ;
• The type of right operand is char**, It's a point char* The pointer to , and char* Is a pointer without qualifier , It points to a that has no qualifier char type .

const char* and char* It's compatible , And they have no qualifiers of their own , So meet the constraints of the standard , The assignment between the two is legal . but const char** and char** The relationship between them is different , Although neither has a qualifier , But the object types they point to are incompatible , Therefore, assignment cannot be performed .

const modification

const Modifying variables

First , keyword const You can't turn a variable into a constant ！ Add... Before a symbol const The qualifier simply means that the symbol cannot be assigned . in other words const Decorated variables are read-only , Cannot be modified directly , But it cannot prevent being modified indirectly .
for example ：

#include <stdio.h>

int main()
{
    
	const int i = 10;
	int* p = &i;
	printf("before:%d\n", i);
	*p = 20;
	printf("after:%d\n", i);// Here the print value becomes 20, The description can be modified indirectly 

	return 0;
}

const Modify a pointer

const Decorated pointer variables have many positions , Now we will introduce one by one .

const int* p

notes ：const int* p And int const p The writing is different , It's the same thing .
This way of writing means ：const modification p, Cannot dereference （p） Directly modify the pointed variable , But it can be modified by changing the way the pointer points p.
for example ：

#include <stdio.h>

int main()
{
    
	// Modifying the compiler by dereferencing directly below will directly report an error 
	//int i = 10;
	//const int* p = &i;
	//*p = 20;

	int i = 10;
	const int* p = &i;
	printf("before:%d\n", *p);
	int j = 20;
	p = &j;// By changing p The direction of , It can be modified indirectly *p value 
	printf("after:%d\n", *p);
	
	return 0;
}

int* const p

This way of writing means ：const modification p, It cannot be modified by changing the pointer *p Value , But you can dereference （*p） Directly modify the pointed variable .
for example ：

#include <stdio.h>

int main()
{
    
	int i = 10;
	int* const p = &i;
	printf("before:%d\n", *p);
	*p = 20;// Can't change p The direction of , However, you can directly dereference the modified value 
	printf("after:%d\n", *p);

	return 0;
}

const int* const p

This way of writing is to modify p and *p, It can't change p The direction of , Nor can it be directly modified by dereference .