Learn point C language (37): function

Non-pointer parameters (that is, pass-value parameters) are not modified to the original value, and const does not make sense to it.

Const is used only for pointers.

1. The first usage: const type * variable:

This usage restricts the values that the pointer points to.

#include　<stdio.h>

int　fun(const　int　*p)　{
　　*p　+=　1;　　/*　只有去掉　const　这句才可以执行　*/
　　return　*p;
}

int　main(void)
{
　　int　num　=　3;
　　printf("%d\n",fun(&num));

　　getchar();
　　return　0;
}

2. There are, however, ways to circumvent this limitation:

#include　<stdio.h>

int　fun(const　int　*p)　{
　　int　*p2　=　p;　/*　来个重名指针会绕过　const　的限制　*/
　　*p2　+=　1;
　　return　*p;
}

int　main(void)
{
　　int　num　=　3;
　　printf("%d\n",fun(&num));　/*　4　*/

　　getchar();
　　return　0;
}

2. Second usage: type *const variable:

This usage will limit the pointing of the pointer; The following example attempts to modify the pointer and will not succeed.

#include　<stdio.h>

void　swap(int　*const　p1,int　*const　p2)　{
　　int　*t　=　p1;
　　p2　=　p1;
　　p2　=　t;
}

int　main(void)
{
　　int　x　=　111;
　　int　y　=　222;

　　printf("%d,%d\n",x,y);
　　swap(&x,&y);
　　printf("%d,%d\n",x,y);

　　getchar();
　　return　0;
}