The fifth day of learning C language

#include<stdio.h>
void sum(int begin, int end)
{
	int i;
	int sum = 0;
	for (i = begin; i <= end; i++)
	{
		sum + 1;
	}
	printf("%d To %d And is %d\n", begin, end, sum);
}
int main()
{
	sum(1, 10);
	sum(20, 30);
	sum(35, 45);
	return 0;

}

As a pointer to a parameter

void f( int* p)

Get the address of a variable when called ;

int i=0; f(&i);

In the function, you can access the external... Through this pointer i

0 Address

Of course you have 0 Address , however 0 Address is usually an address that can't be touched

So your pointer should not have 0 value

So you can use 0 Address to show special things :

The returned pointer is invalid

The pointer is not really initialized ( Initialize to 0)

NULL Is a predefined symbol , Express 0 Address

Some compilers don't want you to use 0 To express 0 Address

Pointer type conversion

void* A pointer indicating that you don't know what to point to

Calculate with char* identical ( But it's not connected )

Pointers can also convert types  

int*p=&i;void*q=(void*)p;

It doesn't change p Type of variable referred to , But let later generations pass through with different eyes p Look at the variables he refers to

const stay * Before and after

const int*p1=&i;( What he refers to cannot be modified )

int const* p2=&i;( It is also what he refers to that cannot be modified )

int *constp3=&i( The pointer cannot be modified )

count Array

count int a[]={1,2,3,4,5,6,};

The array variable is already const Pointer. , there const Indicates that each cell of the array is const int

So it has to be assigned by initialization