C指针

先看下下面代码：

#include "stdio.h"
#include "conio.h"

void go_south_east(int lat, int lon){
    lat = lat -1;
    lon = lon + 1;
}

int main()
{
    int latitude = 32;
    int longitude = -64;
    go_south_east(latitude, longitude);
    printf("Avast! Now at: [%i, %i]\n", latitude, longitude);

    getch();
    return 0;

}

结果并不是31， -63，而是32， -64

并没有改变，这个时候必须要知道C语言中的指针的概念，而且这个概念是必须应该掌握的

int main()
{
    int x = 8;
    int z=18;
    printf("x is %i and the address is %p\n", x, &x);

    x= z;
    printf("x is %i and the address is %p\n", x, &x);

    return 0;
}

结果为：

[img]http://dl2.iteye.com/upload/attachment/0097/5857/06a39358-cee1-3c66-8a8d-6facdc9654f2.jpg[/img]

[b]%p是输出地址格式，&x是指x的内存地址[/b]

回到第一个程序中，当main方法中调用go_south_east（）方法的时候，两个参数都是传递值，那么实际上lat，lon被赋值了，并且进行了加减操作，而main方法中的latitude，longitude 两个参数地址不会变，值也没有变。

[b]1，获取变量的存储地址，即变量的指针 用&x
2，指针变量是用来存储指针的变量 比如 int * address_x = &x;
3，获取指针变量所指向的值 int y = *address_x;
改变其值就用*address_x=9，即 x=9[/b]


所以第一个程序应该写成

#include "stdio.h"
#include "conio.h"

void go_south_east(int * lat, int * lon){


    *lat = *lat -1;
    *lon = *lon + 1;

}

int main()
{
    int latitude = 32;
    int longitude = -64;
    go_south_east(&latitude, &longitude);
    printf("Avast! Now at: [%i, %i]\n", latitude, longitude);

    getch();
    return 0;

}

[b]
关键点：
Variables are allocated storage in memory.
Local variables live in the stack.
Global variables live in the globals section.
Pointers are just variables that store memory addresses.
The & operator finds the address of a variable.
The * operator can read the contents of a memory address.
The * operator can also set the contents of a memory address.[/b]