The return value of the function is the attention of the pointer, the local variables inside the static limit sub function, and how the pointer to the array represents the array elements

This content mainly studies pointer type functions, which need to pay attention to when returning pointers ： Do not return a pointer to a local variable in a function .

I wanted to simply write a function that modifies the pointer type of the array element value through the pointer , As a result, many problems were exposed , It shows that the basic skills are not solid enough .

So let's see ： When the return value type of a function is a pointer ：

Look at the program first ：

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

int* GetNum()
{
	int array[10];

	for (size_t i = 0; i < 10; i++)
	{
		array[i] = i;
	}


	return array;
    【 Be careful 1】： Here the compiler prompts a warning , Returns the address of a local or temporary variable .
     Here we return the address of the local array variable in this function , But in function return Then this function ends execution ,
 The stack space where the array is located will be emptied , Although the address is still , But the above value is no longer guaranteed , therefore ： Be careful ： Do not return the address of the variable on the internal stack memory of the function ！
}

int main(int agrc, char* argv[])
{
	int* p = GetNum();

	for (size_t i = 0; i < 10; i++)
	{
		printf("p[%d]=%d\n", i, *(p + i));

	}

	for (size_t i = 0; i < 10; i++)
	{
		printf("p[%d]=%d\n", i, *p[i]);
        【 Be careful 2】 The operation of this step is to think of the pointer p The result of the variable is the first address of an array element ,
 So I mistakenly think of the pointer p It can be used completely as an array name , So there was p[i] This way of writing ！ Be careful ,p It's a pointer variable ,
 It's not a pointer array , How can I write p[i] This kind of thing comes . People say that when the pointer points to an array ,
 Yes, it is *（p+i） To address the next element , No *p[i], The second way of writing is completely wrong .
 For array names ,int array[10]; use *（a+i） perhaps array[i] Fine ,
 That's because the array name itself is a pointer ！！！
 remember , For pointer variables that are not array names and point to arrays , Only use *（p+i）!

	}

	return 0;

}

If you want to achieve the above functions , There are two ways to be right ：

1. In the sub function, the forward pointer operates , stay main Define an array in the calling function , hold main The array pointer in the function is passed to the sub function , In this way, array variables are opened up in main In the stack area of the function , The value on the address will not be released until the end of the whole program . as follows ：

#include<stdio.h>


int* GetNum(int* arry)
{


	for (size_t i = 0; i < 10; i++)
	{
		*(arry + i) = i;
	}


	return arry;
}

int main(int agrc, char* argv[])
{
	int a[10];

	int* p = GetNum(a);

	for (size_t i = 0; i < 10; i++)
	{
	
		printf("p[%d]=%d\n", i, *(p + i));

	}

	return 0;

}

2.static, Methods set as static variables . This method can return the address of the local variable of the function , But this local variable should be declared static Static variables , Stored in static storage area , Not released until the end of the program ！ The following code :

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

int* GetNum()
{
	static int array[10];
 although array[] The array is defined inside the function , But because static, It is actually stored in static storage , Even if GetNum Call complete , The value in its memory space will not be released .

	for (size_t i = 0; i < 10; i++)
	{
		array[i] = i;
	}


	return array;
	【 Be careful 3】： Here the compiler will no longer report errors , Because although the address of the internal variable of the function is returned , But this variable is a static variable , Put it in the static storage area ,
	 Even if the function call ends , The memory space will not be released , Until the end of the program , The memory space will be released , So we can still get the correct value .
}

int main(int agrc, char* argv[])
{
	int* p = GetNum();

	for (size_t i = 0; i < 10; i++)
	{
		printf("p[%d]=%d\n", i, *(p + i));

	}

	return 0;

}

In conclusion ： If the return type of a function is a pointer , You cannot return the address of a non static variable of a local variable defined in a function . Either main The address of the local variable in the function is used as the input parameter to the function of pointer type ; Or you can define local variables inside the subfunction , And return the address of the local variable , But the local variable should be declared as static Formal .

// In fact, from the above two methods , The first one is recommended , Because there is no need to store some values in the static storage area , Anyway ,main Local variables in functions always exist on the stack （ It's just personal understanding ）

Or the foundation is not solid , A lot of stupid little questions , Need more practice .