如何将一个（void *）指针转换为一个字符串数组的指针？

wintersky

1. #include <stdio.h>
   
2. #include <stdlib.h>
   
3. 
   
4. int test(vod *arg)
   
5. {
   
6.         char *local_arg[2];
   
7. 
   
8.         local_arg[0] = ???;    // How to do this here?
   
9.         local_arg[1] = ???:
   
10. 
    
11.         fprintf(stderr, "%s\n", local_arg[0]);
    
12.         fprintf(stderr, "%s\n", local_arg[1]);
    
13. 
    
14.         return 0;
    
15. 
    
16. }
    
17. 
    
18. 
    
19. 
    
20. int main(int argc, char *argv[])
    
21. {
    
22.         char *arg[2] = {"hello", "world"}
    
23.         test(arg);
    
24.         return 0;
    
25. }
    

复制代码

cugb_cat

先强制类型转化为char **型，然后用下标进行赋值

wintersky

Thanks a lot!

1. #include <stdio.h>
   
2. #include <stdlib.h>
   
3. 
   
4. int test(void *arg)
   
5. {
   
6.         char *local_arg[2];
   
7.         local_arg[0] = ((char **)arg)[0];
   
8.         local_arg[1] = ((char **)arg)[1];
   
9. 
   
10.         fprintf(stderr, "%s\n", local_arg[0]);
    
11.         fprintf(stderr, "%s\n", local_arg[1]);
    
12.         return 0;
    
13. 
    
14. }
    
15. 
    
16. int main(int argc, char *argv[])
    
17. {
    
18.         char *arg[2] = {"hello", "world"};
    
19.         test(arg);
    
20.         return 0;
    
21.        
    
22. }
    

复制代码

It is OK now!