Is the array name a pointer

Conclusion ： The array name is a pointer constant , It can only point to the address of the first element of the array . When an array is used as a function parameter , Can transform （ Some are also called degradation ） For a pointer , You can operate some pointers , such as a++ To read the next element , But the pointer it is converted to also points to this array .

  When the array name is not used as a function parameter , Although it is a pointer constant , But it cannot be used completely as a pointer . When an array is a function parameter , It is used as a pointer .

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


int main(void)
{
	int a[3] = { 1, 2, 3 };

	printf("%d\n", *a);// The result is 1.
 At this point, the array name looks like a pointer , But it can't be used on this pointer ++ operation .
 Questions may arise at this time , It seems that the array name can also be used as a pointer when it is not used as a function parameter ？ So why does the array name degenerate into a pointer when it is used as a function parameter .
 actually , Array names are used like this ： The array name is a pointer , But a pointer constant , Its essence as a pointer constant has nothing to do with whether it is a function parameter , The pointer constant represented by the array name , Is bound to the corresponding array , Cannot point to other arrays , So when the array name is not used as a function parameter , Can not have a++ Such operation .
 When the formal parameter of a function is an array , In the function body, this formal parameter can be a++ Such operation , Because function parameters are arrays , You can already know a++ It doesn't mean to let a This pointer constant points to other addresses , But through a++ Read the next element , So in the function body with array as function parameter ,a++ This kind of operation is possible , And at this point sizeof（a）, Forever 4.
 When an array is used as a function parameter , There is no need to specify the size of the array , Because no matter how many elements there are in the array , When an argument is passed in , The address of the first element is always passed in .
	printf("%d\n", sizeof(a));//12, At this time, the array name means this array .
	
a++;// This is wrong , Array name a It's a pointer constant , It can't be changed , So the array name is not a pointer .

	
	return 0;
}


void function(int a[])
{

a++;// This is OK again , Because when the array name is used as a parameter , It degenerates into a pointer , Self adding operation can be carried out .

}

 in general , Whether or not it is an argument to a function , The array name is a pointer constant , Only when it is used as a function parameter , Because it has been defined that the parameters of the function are of array type , At this point, the array name will degenerate into a pointer , In the function body, it is treated as a pointer to this array .

Take a look at this article , As a reference .https://www.cnblogs.com/youxin/p/3235862.html

Let's look at the following code ：

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


int main(void)
{

	int a[3] = { 1, 2, 3 };
	
	int* p = a;
	printf("%d\n", p);
	int* p1 = &a;
	printf("%d\n", p1);
	
	return 0;
}
 The two outputs are the same , You can find a That is, it can be regarded as the address of the first element of the array , It can also be regarded as the name of an array variable ,C It's a very flexible language .