数组其实是一种变相的指针，数组名同时也是指针，eg：

CODE == &CODE[0];

数组的加法：

#include<stdio.h>

#define SIZE 4

int main(void)

{

         shortdates[SIZE];

         short* pti;

         shortindex;

         doublebills[SIZE];

         double* ptf;

         pti= dates;//把数组地址付给指针

         ptf= bills;

         printf("%23s  %10s\n", "short","double");

         for(index= 0; index < SIZE; index++)

                   printf("pointers+ %d: %10p %10p\n", index, pti + index, ptf + index);

         return0;

}

---------------------------------------------------

  short          double

pointerrs + 0:     28ff0c    28fee8

pointerrs + 1:     28ff0e    28fef0

pointerrs + 2:     28ff10    28fef8

pointerrs + 3:     28ff12    28ff00

----------------------------------------------------

1.      指针的数值就是它指向的对象的对象的地址

2.      在指针前运用运算符*就可以得到该指针所指向的对象的数值

3.      对指针+1等价于对指针的值加上他所指的对象的字节的大小

Eg：

Dates + 2 == &date[2]  //相同的地址

*(dates + 2) == dates[2]   //相同的值

总结：可以用指针表示数组的每个元素，并得到每个元素的数值，从本质上说，对同一个对象有两种不同的符号表示方法。

优先级问题：

*的优先级高于+；

所以(*dates) + 2等价于*dates + 2;

函数、数组和指针

int sum(int *ar )

{

         inti;

         inttaotal = 0;

         for(i= 0; i <10; i++)

                   total+= ar[i];

         returntotal;

}

等价于：

int sum(int ar[] )

{

         inti;

         inttaotal = 0;

         for(i= 0; i <10; i++)

                   total+= ar[i];

         returntotal;

}

这段代码说明了：正如可以在指针符号中使用数组名一样，也可以在数组符号中使用指针，

下面代码是对上面的优化，可以灵活的控制变量。

int sum(int * ar, int n)

         {

                   inti;

                   inttotal = 0;

                   for(i= 0; i < n; i++)

                            total+= ar[i];

                   returntotal;

         }

无论在任何形式下，* ar都可以代替ar[];但是ar[]不一定能代替*ar。

数组和指针的不同形式解决同一个问题：

#include<stdio.h>

#define SIZE 10

int sum(int ar[ ], int n)

int main(void)

{

         intmarbles[SIZE] = {20, 10, 5, 39, 4, 16, 19, 26, 31, 20};

         longanswer;

         answer= sum(marbles, SIZE);

         printf("thetotal number of marbles is %ld.\n", answer);

         printf("thesize of marbles is %u bytes.\n", sizeof marbles);

         return0;

}

int sum(int ar[ ], int n)

{

         inti;

         inttotal = 0;

         for(i= 0; i < n; i++)

                   total+= ar[i];

         printf("thesize of ar is %u bytes.\n", sizeof ar);

         returntotal;

}

用指针实现：

#include<stdio.h>

#define SIZE 10

int sum(int ar[ ], int n)

int main(void)

{

         intmarbles[SIZE] = {20,10, 5, 39, 4, 16, 19, 26, 31, 20};

         longanswer;

         answer= sum(marbles, marbles + SIZE);

         printf("thetotal number of marbles is %ld.\n", answer);

         printf("thesize of marbles is %u bytes.\n", sizeof marbles);

         return0;

}

int sum(int * start, int * end)

{

         inttotal = 0;

         while(start< end)

         {

                   total+= *start;

                   start++;

         }

         returntotal;

}

关于对++和*运算符优先级的判断：

#include<stdio.h>

int data[2] = {100, 200};

int moredata[2] = {300, 400};

int main(void)

{

         int*p1, *p2, *p3;

         p1= p2 = data;

         p3= moredata;

         printf("*p1= %d,  *pa = %d,   *p3 = %d\n",*p1, *p2, *p3);

         printf("*p1++= %d, *++p2 = %d, (*p3)++ = %d\n",*p1++, *++p2, (*p3)++);

         printf("*p1 = %d, *p2 = %d, *p3 = %d\n", *p1, *p2, *p3);

         return0;

}

--------------------------------------------------------------

*p1 = 100, *pa = 100,   *p3 = 300

*p1++ = 100, *++p2 = 200, (*p3)++ = 300

 *p1= 200, *p2 = 200, *p3 = 301

---------------------------------------------------------------

*和++具有相同的优先级，只是*是自右向左结合的特性。