dyli指针初探

1. #include stdio.h>
   
2. 
   
3. int main(int argc,char ** argv)
   
4. {
   
5.         char *p;
   
6.         printf("************************************\n");
   
7.         printf("p=%d,it is an address\n",p);
   
8.         p='\0';
   
9.         printf("p='\'0,print p= %c\n",p);
   
10.         p='0';
    
11.         printf("p='0',print %c! \n\n\n",p);
    
12. 
    
13. 
    
14.         char *p1;
    
15.         printf("************************************\n");
    
16.         p1="adcdefg";
    
17.         printf("p1 can print them '%s'\n\n\n",p1);
    
18. 
    
19. 
    
20.         char *p3;
    
21.         printf("************************************\n");
    
22.         p3=&("abcdefg"[5]);
    
23.         printf("p3=&(''abcdefg''[5]) print: %s\n",p3);
    
24.         char str[] = "adcedfg";
    
25.         p3=&str[5];
    
26.         printf("'p3=&str[5]' will print %s too \n",p3);
    
27. 
    
28. 
    
29. 
    
30.         char *p4;
    
31.         printf("************************************\n");
    
32.         *p4="abcdefg"[5];
    
33.         printf("print p4= %s\n",*p4);
    
34. 
    
35.         char *p5;
    
36.         p5=&("abcdefg"[5]);
    
37.         printf("%s\n",p5);
    
38. 
    
39.         return 0;
    
40. }

运行效果：

************************************
p=1307813,it is an address
p=''0,print p=
p='0',print 0!


************************************
p1 can print them 'adcdefg'


************************************
p3=&(''abcdefg''[5]) print: fg
'p3=&str[5]' will print fg too
************************************
段错误